Does C have a “foreach” loop construct?

Almost all languages have a foreach loop or something similar. Does C have one? Can you post some example code?

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

"foreach" of what?

– alk
Mar 6 '16 at 11:13

How hard would it have been to try writing a foreach loop in a C program?

– MD XF
Oct 17 '16 at 0:04

add a comment |

Almost all languages have a foreach loop or something similar. Does C have one? Can you post some example code?

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

"foreach" of what?

– alk
Mar 6 '16 at 11:13

How hard would it have been to try writing a foreach loop in a C program?

– MD XF
Oct 17 '16 at 0:04

add a comment |

Almost all languages have a foreach loop or something similar. Does C have one? Can you post some example code?

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

Almost all languages have a foreach loop or something similar. Does C have one? Can you post some example code?

c foreach

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

edited May 19 '17 at 1:46

MD XF

4,09652854

edited May 19 '17 at 1:46

MD XF

4,09652854

edited May 19 '17 at 1:46

MD XF

4,09652854

asked Dec 30 '08 at 17:39

Question

532196

asked Dec 30 '08 at 17:39

Question

532196

asked Dec 30 '08 at 17:39

Question

532196

"foreach" of what?

– alk
Mar 6 '16 at 11:13

How hard would it have been to try writing a foreach loop in a C program?

– MD XF
Oct 17 '16 at 0:04

add a comment |

"foreach" of what?

– alk
Mar 6 '16 at 11:13

How hard would it have been to try writing a foreach loop in a C program?

– MD XF
Oct 17 '16 at 0:04

"foreach" of what?

– alk
Mar 6 '16 at 11:13

How hard would it have been to try writing a foreach loop in a C program?

– MD XF
Oct 17 '16 at 0:04

add a comment |

13 Answers
13

active

oldest

votes

177

C doesn't have a foreach, but macros are frequently used to emulate that:

#define for_each_item(item, list) 

    for(T * item = list->head; item != NULL; item = item->next)

And can be used like

for_each_item(i, processes) {

    i->wakeup();

}

Iteration over an array is also possible:

#define foreach(item, array) 

    for(int keep = 1, 

            count = 0,

            size = sizeof (array) / sizeof *(array); 

        keep && count != size; 

        keep = !keep, count++) 

      for(item = (array) + count; keep; keep = !keep)

And can be used like

int values = { 1, 2, 3 };

foreach(int *v, values) {

    printf("value: %dn", *v);

}

Edit: In case you are also interested in C++ solutions, C++ has a native for-each syntax called "range based for"

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

1

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

3

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

1

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

1

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

1

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

|
show 13 more comments

Here is a full program example of a for-each macro in C99:

#include <stdio.h>



typedef struct list_node list_node;

struct list_node {

    list_node *next;

    void *data;

};



#define FOR_EACH(item, list) 

    for (list_node *(item) = (list); (item); (item) = (item)->next)



int

main(int argc, char *argv)

{

    list_node list = {

        { .next = &list[1], .data = "test 1" },

        { .next = &list[2], .data = "test 2" },

        { .next = NULL,     .data = "test 3" }

    };



    FOR_EACH(item, list)

        puts((char *) item->data);



    return 0;

}

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

7

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

1

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

add a comment |

There is no foreach in C.

You can use a for loop to loop through the data but the length needs to be know or the data needs to be terminated by a know value (eg. null).

char* nullTerm;

nullTerm = "Loop through my characters";



for(;nullTerm != NULL;nullTerm++)

{

    //nullTerm will now point to the next character.

}

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

add a comment |

This is a fairly old question, but I though I should post this. It is a foreach loop for GNU C99.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <stdbool.h>



#define FOREACH_COMP(INDEX, ARRAY, ARRAY_TYPE, SIZE) 

  __extension__ 

  ({ 

    bool ret = 0; 

    if (__builtin_types_compatible_p (const char*, ARRAY_TYPE)) 

      ret = INDEX < strlen ((const char*)ARRAY); 

    else 

      ret = INDEX < SIZE; 

    ret; 

  })



#define FOREACH_ELEM(INDEX, ARRAY, TYPE) 

  __extension__ 

  ({ 

    TYPE *tmp_array_ = ARRAY; 

    &tmp_array_[INDEX]; 

  })



#define FOREACH(VAR, ARRAY) 

for (void *array_ = (void*)(ARRAY); array_; array_ = 0) 

for (size_t i_ = 0; i_ && array_ && FOREACH_COMP (i_, array_, 

                                    __typeof__ (ARRAY), 

                                    sizeof (ARRAY) / sizeof ((ARRAY)[0])); 

                                    i_++) 

for (bool b_ = 1; b_; (b_) ? array_ = 0 : 0, b_ = 0) 

for (VAR = FOREACH_ELEM (i_, array_, __typeof__ ((ARRAY)[0])); b_; b_ = 0)



/* example's */

int

main (int argc, char **argv)

{

  int array[10];

  /* initialize the array */

  int i = 0;

  FOREACH (int *x, array)

    {

      *x = i;

      ++i;

    }



  char *str = "hello, world!";

  FOREACH (char *c, str)

    printf ("%cn", *c);



  return EXIT_SUCCESS;

}

This code has been tested to work with gcc, icc and clang on GNU/Linux.

answered Jul 21 '10 at 15:12

Joe D

2,37812425

add a comment |

While C does not have a for each construct, it has always had an idiomatic representation for one past the end of an array (&arr)[1]. This allows you to write a simple idiomatic for each loop as follows:

int arr = {1,2,3,4,5};

for(int *a = arr; a < (&arr)[1]; ++a)

    printf("%dn", *a);

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

3

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

1

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

add a comment |

As u probably already know, there's no "foreach" style loop in C.

Altho there are already tons of great macros provided here to work around this, maybe you'll find this macro useful.

// "length" is the length of the array.   

#define each(item, array, length) 

(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

item is declared and incremented within the for statement (just like
in Python!).

Seems to work on any 1-dimensional array

All variables created in macro (p, item), aren't visible outside the
scope of the loop (since they're declared in the for loop header).

Disadvantages:

Doesn't work for multi-dimensional arrays

Relies on typeof(), which is a gcc extension; NOT part of standard C

Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct _point {

    double x;

    double y;

} Point;



int main(void)

{

    double some_nums = {4.2, 4.32, -9.9, 7.0};

    for each (element, some_nums, 4)

        printf("element = %lfn", element);



    int numbers = {4, 2, 99, -3, 54};

    // Just demonstrating it can be used like a normal for loop

    for each (number, numbers, 5) { 

        printf("number = %dn", number);

        if (number % 2 == 0)

                printf("%d is even.n", number);

    }



    char *dictionary = {"Hello", "World"};

    for each (word, dictionary, 2)

        printf("word = '%s'n", word);



    Point points = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};

    for each (point, points, 3)

        printf("point = (%lf, %lf)n", point.x, point.y);



    // Neither p, element, number or word are visible outside the scope of

    // their respective for loops. Try to see if these printfs work

    // (they shouldn't):

    // printf("*p = %s", *p);

    // printf("word = %s", word);



    return 0;

}

Seems to work on gcc and clang; not sure about other compilers.

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

add a comment |

C has 'for' and 'while' keywords. If a foreach statement in a language like C# looks like this ...

foreach (Element element in collection)

{

}

... then the equivalent of this foreach statement in C might be be like:

for (

    Element* element = GetFirstElement(&collection);

    element != 0;

    element = GetNextElement(&collection, element)

    )

{

    //TODO: do something with this element instance ...

}

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

1

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

add a comment |

Eric's answer doesn't work when you're using "break" or "continue".

This can be fixed by rewriting the first line:

Original line (reformatted):

for (unsigned i = 0, __a = 1; i < B.size(); i++, __a = 1)

Fixed:

for (unsigned i = 0, __a = 1; __a && i < B.size(); i++, __a = 1)

If you compare it to Johannes' loop, you'll see that he's actually doing the same, just a bit more complicated and uglier.

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

add a comment |

Here's a simple one, single for loop:

#define FOREACH(type, array, size) do { 

        type it = array[0]; 

        for(int i = 0; i < size; i++, it = array[i])

#define ENDFOR  } while(0);



int array = { 1, 2, 3, 4, 5 };



FOREACH(int, array, 5)

{

    printf("element: %d. index: %dn", it, i);

}

ENDFOR

Gives you access to the index should you want it (i) and the current item we're iterating over (it). Note you might have naming issues when nesting loops, you can make the item and index names be parameters to the macro.

Edit: Here's a modified version of the accepted answer foreach. Lets you specify the start index, the size so that it works on decayed arrays (pointers), no need for int* and changed count != size to i < size just in case the user accidentally modifies 'i' to be bigger than size and get stuck in an infinite loop.

#define FOREACH(item, array, start, size)

    for(int i = start, keep = 1;

        keep && i < size;

        keep = !keep, i++)

    for (item = array[i]; keep; keep = !keep)



int array = { 1, 2, 3, 4, 5 };

FOREACH(int x, array, 2, 5)

    printf("index: %d. element: %dn", i, x);

Output:

index: 2. element: 3

index: 3. element: 4

index: 4. element: 5

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

add a comment |

C does not have an implementation of for-each. When parsing an array as a point the receiver does not know how long the array is, thus there is no way to tell when you reach the end of the array.
Remember, in C int* is a point to a memory address containing an int. There is no header object containing information about how many integers that are placed in sequence. Thus, the programmer needs to keep track of this.

However, for lists, it is easy to implement something that resembles a for-each loop.

for(Node* node = head; node; node = node.next) {

   /* do your magic here */

}

To achieve something similar for arrays you can do one of two things.

use the first element to store the length of the array.

wrap the array in a struct which holds the length and a pointer to the array.

The following is an example of such struct:

typedef struct job_t {

   int count;

   int* arr;

} arr_t;

answered Mar 14 '18 at 23:12

Jonas

355416

add a comment |

Here is what I use when I'm stuck with C. You can't use the same item name twice in the same scope, but that's not really an issue since not all of us get to use nice new compilers :(

#define FOREACH(type, item, array, size) 

    size_t X(keep), X(i); 

    type item; 

    for (X(keep) = 1, X(i) = 0 ; X(i) < (size); X(keep) = !X(keep), X(i)++) 

        for (item = (array)[X(i)]; X(keep); X(keep) = 0)



#define _foreach(item, array) FOREACH(__typeof__(array[0]), item, array, length(array))

#define foreach(item_in_array) _foreach(item_in_array)



#define in ,

#define length(array) (sizeof(array) / sizeof((array)[0]))

#define CAT(a, b) CAT_HELPER(a, b) /* Concatenate two symbols for macros! */

#define CAT_HELPER(a, b) a ## b

#define X(name) CAT(__##name, __LINE__) /* unique variable */

Usage:

int ints = {1, 2, 0, 3, 4};

foreach (i in ints) printf("%i", i);

/* can't use the same name in this scope anymore! */

foreach (x in ints) printf("%i", x);

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

add a comment |

If you're planning to work with function pointers

#define lambda(return_type, function_body)

    ({ return_type __fn__ function_body __fn__; })



#define array_len(arr) (sizeof(arr)/sizeof(arr[0]))



#define foreachnf(type, item, arr, arr_length, func) {

    void (*action)(type item) = func;

    for (int i = 0; i<arr_length; i++) action(arr[i]);

}



#define foreachf(type, item, arr, func)

    foreachnf(type, item, arr, array_len(arr), func)



#define foreachn(type, item, arr, arr_length, body)

    foreachnf(type, item, arr, arr_length, lambda(void, (type item) body))



#define foreach(type, item, arr, body)

    foreachn(type, item, arr, array_len(arr), body)

Usage:

int ints = { 1, 2, 3, 4, 5 };

foreach(int, i, ints, {

    printf("%dn", i);

});



char* strs = { "hi!", "hello!!", "hello world", "just", "testing" };

foreach(char*, s, strs, {

    printf("%sn", s);

});



char** strsp = malloc(sizeof(char*)*2);

strsp[0] = "abcd";

strsp[1] = "efgh";

foreachn(char*, s, strsp, 2, {

    printf("%sn", s);

});



void (*myfun)(int i) = somefunc;

foreachf(int, i, ints, myfun);

But I think this will work only on gcc (not sure).

answered May 15 '17 at 20:01

Naheel

362210

add a comment |

Thanks to accepted answer I was able to fix my C++ macro. It's not clean, but it supports nesting, break and continue, and for a list<t> it allows to iterate over the t& instead of list<t>::elem*, so you don't have to dereference.

template <typename t> struct list { // =list=

  struct elem {

    t Data;

    elem* Next;

  };

  elem* Head;

  elem* Tail;

};

#define for_list3(TYPE, NAME, NAME2, LIST) 

  bool _##NAME##NAME2 = true;

  for (list<TYPE>::elem* NAME##_E = LIST.Head;

      NAME##_E && _##NAME##NAME2;

      _##NAME##NAME2 = !_##NAME##NAME2, NAME##_E = NAME##_E->Next) 

    for (auto& NAME = NAME##_E->Data; _##NAME##NAME2; _##NAME##NAME2=false)

#define for_list2(TYPE, NAME, NAME2, LIST) for_list3(TYPE, NAME, NAME2, LIST)

#define for_list(TYPE, NAME, LIST) for_list2(TYPE, NAME, __COUNTER__, LIST)



void example() {

  list<string> Words;

  for_list(string, Word, Words) {

    print(Word);

  }

}

With the for_each_item macro from the accepted answer, you'd have to do this:

void example2() {

  list<string> Words;

  for_each_item(string, Elem, Words) {

    string& Word = Elem->Data;

    print(Word);

  }

}

If you want to rewrite it into a clean version, feel free to edit.

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

add a comment |

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13 Answers
13

active

oldest

votes

13 Answers
13

active

oldest

votes

177

C doesn't have a foreach, but macros are frequently used to emulate that:

#define for_each_item(item, list) 

    for(T * item = list->head; item != NULL; item = item->next)

And can be used like

for_each_item(i, processes) {

    i->wakeup();

}

Iteration over an array is also possible:

#define foreach(item, array) 

    for(int keep = 1, 

            count = 0,

            size = sizeof (array) / sizeof *(array); 

        keep && count != size; 

        keep = !keep, count++) 

      for(item = (array) + count; keep; keep = !keep)

And can be used like

int values = { 1, 2, 3 };

foreach(int *v, values) {

    printf("value: %dn", *v);

}

Edit: In case you are also interested in C++ solutions, C++ has a native for-each syntax called "range based for"

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

1

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

3

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

1

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

1

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

1

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

|
show 13 more comments

177

C doesn't have a foreach, but macros are frequently used to emulate that:

#define for_each_item(item, list) 

    for(T * item = list->head; item != NULL; item = item->next)

And can be used like

for_each_item(i, processes) {

    i->wakeup();

}

Iteration over an array is also possible:

#define foreach(item, array) 

    for(int keep = 1, 

            count = 0,

            size = sizeof (array) / sizeof *(array); 

        keep && count != size; 

        keep = !keep, count++) 

      for(item = (array) + count; keep; keep = !keep)

And can be used like

int values = { 1, 2, 3 };

foreach(int *v, values) {

    printf("value: %dn", *v);

}

Edit: In case you are also interested in C++ solutions, C++ has a native for-each syntax called "range based for"

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

1

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

3

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

1

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

1

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

1

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

|
show 13 more comments

177

C doesn't have a foreach, but macros are frequently used to emulate that:

#define for_each_item(item, list) 

    for(T * item = list->head; item != NULL; item = item->next)

And can be used like

for_each_item(i, processes) {

    i->wakeup();

}

Iteration over an array is also possible:

#define foreach(item, array) 

    for(int keep = 1, 

            count = 0,

            size = sizeof (array) / sizeof *(array); 

        keep && count != size; 

        keep = !keep, count++) 

      for(item = (array) + count; keep; keep = !keep)

And can be used like

int values = { 1, 2, 3 };

foreach(int *v, values) {

    printf("value: %dn", *v);

}

Edit: In case you are also interested in C++ solutions, C++ has a native for-each syntax called "range based for"

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

C doesn't have a foreach, but macros are frequently used to emulate that:

#define for_each_item(item, list) 

    for(T * item = list->head; item != NULL; item = item->next)

And can be used like

for_each_item(i, processes) {

    i->wakeup();

}

Iteration over an array is also possible:

#define foreach(item, array) 

    for(int keep = 1, 

            count = 0,

            size = sizeof (array) / sizeof *(array); 

        keep && count != size; 

        keep = !keep, count++) 

      for(item = (array) + count; keep; keep = !keep)

And can be used like

int values = { 1, 2, 3 };

foreach(int *v, values) {

    printf("value: %dn", *v);

}

Edit: In case you are also interested in C++ solutions, C++ has a native for-each syntax called "range based for"

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

edited Sep 22 '15 at 16:39

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

answered Dec 30 '08 at 17:51

Johannes Schaub - litb

405k1007731107

1

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

3

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

1

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

1

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

1

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

|
show 13 more comments

1

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

3

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

1

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

1

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

1

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

If you've got the "typeof" operator (gcc extension; pretty common on many other compilers) you can get rid of that "int *". The inner for loop becomes something like "for(typeof((array)+0) item = ..." Then you can call as "foreach( v, values ) ..."

– leander
Aug 6 '09 at 4:46

@eSKay yes consider if(...) foreach(int *v, values) ... . If they are outside the loop it expands to if(...) int count = 0 ...; for(...) ...; and will break.

– Johannes Schaub - litb
May 9 '10 at 11:37

@fred to make "break" work

– Johannes Schaub - litb
Feb 15 '12 at 10:34

@rem it does not break the outer loop if you use "break"

– Johannes Schaub - litb
Sep 26 '16 at 11:05

@rem you can however simplify my code if you change the inner "keep = !keep" into "keep = 0". I liked the "symmetry" so i just used negation and not straight assignment.

– Johannes Schaub - litb
Sep 26 '16 at 11:12

|
show 13 more comments

Here is a full program example of a for-each macro in C99:

#include <stdio.h>



typedef struct list_node list_node;

struct list_node {

    list_node *next;

    void *data;

};



#define FOR_EACH(item, list) 

    for (list_node *(item) = (list); (item); (item) = (item)->next)



int

main(int argc, char *argv)

{

    list_node list = {

        { .next = &list[1], .data = "test 1" },

        { .next = &list[2], .data = "test 2" },

        { .next = NULL,     .data = "test 3" }

    };



    FOR_EACH(item, list)

        puts((char *) item->data);



    return 0;

}

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

7

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

1

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

add a comment |

Here is a full program example of a for-each macro in C99:

#include <stdio.h>



typedef struct list_node list_node;

struct list_node {

    list_node *next;

    void *data;

};



#define FOR_EACH(item, list) 

    for (list_node *(item) = (list); (item); (item) = (item)->next)



int

main(int argc, char *argv)

{

    list_node list = {

        { .next = &list[1], .data = "test 1" },

        { .next = &list[2], .data = "test 2" },

        { .next = NULL,     .data = "test 3" }

    };



    FOR_EACH(item, list)

        puts((char *) item->data);



    return 0;

}

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

7

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

1

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

add a comment |

Here is a full program example of a for-each macro in C99:

#include <stdio.h>



typedef struct list_node list_node;

struct list_node {

    list_node *next;

    void *data;

};



#define FOR_EACH(item, list) 

    for (list_node *(item) = (list); (item); (item) = (item)->next)



int

main(int argc, char *argv)

{

    list_node list = {

        { .next = &list[1], .data = "test 1" },

        { .next = &list[2], .data = "test 2" },

        { .next = NULL,     .data = "test 3" }

    };



    FOR_EACH(item, list)

        puts((char *) item->data);



    return 0;

}

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

Here is a full program example of a for-each macro in C99:

#include <stdio.h>



typedef struct list_node list_node;

struct list_node {

    list_node *next;

    void *data;

};



#define FOR_EACH(item, list) 

    for (list_node *(item) = (list); (item); (item) = (item)->next)



int

main(int argc, char *argv)

{

    list_node list = {

        { .next = &list[1], .data = "test 1" },

        { .next = &list[2], .data = "test 2" },

        { .next = NULL,     .data = "test 3" }

    };



    FOR_EACH(item, list)

        puts((char *) item->data);



    return 0;

}

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

answered Dec 30 '08 at 18:13

Judge Maygarden

21.4k76591

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

7

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

1

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

add a comment |

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

7

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

1

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

What does the dot do in the list definition? Couldn't you simply write next instead of .next?

– Rizo
Jun 28 '10 at 10:25

@Rizo No, the dot is a part of the syntax for C99 designated initializers. See en.wikipedia.org/wiki/C_syntax#Initialization

– Judge Maygarden
Jun 28 '10 at 15:27

@Rizo: Note also that that's a really hacky way of building a linked list. It'll do for this demo but don't do it that way in practice!

– Donal Fellows
Jul 21 '10 at 15:18

@Donal What makes it "hacky"?

– Judge Maygarden
Jul 22 '10 at 13:17

@Judge: Well, for one thing it has “surprising” lifetime (if you're working with code which removes elements, chances are you'll crash in free()) and for another it has a reference to the value inside its definition. It's really an example of something that's just too damn clever; code's complex enough without purposefully adding cleverness to it. Kernighan's aphorism (stackoverflow.com/questions/1103299/…) applies!

– Donal Fellows
Jul 22 '10 at 23:25

add a comment |

There is no foreach in C.

You can use a for loop to loop through the data but the length needs to be know or the data needs to be terminated by a know value (eg. null).

char* nullTerm;

nullTerm = "Loop through my characters";



for(;nullTerm != NULL;nullTerm++)

{

    //nullTerm will now point to the next character.

}

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

add a comment |

There is no foreach in C.

You can use a for loop to loop through the data but the length needs to be know or the data needs to be terminated by a know value (eg. null).

char* nullTerm;

nullTerm = "Loop through my characters";



for(;nullTerm != NULL;nullTerm++)

{

    //nullTerm will now point to the next character.

}

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

add a comment |

There is no foreach in C.

You can use a for loop to loop through the data but the length needs to be know or the data needs to be terminated by a know value (eg. null).

char* nullTerm;

nullTerm = "Loop through my characters";



for(;nullTerm != NULL;nullTerm++)

{

    //nullTerm will now point to the next character.

}

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

There is no foreach in C.

You can use a for loop to loop through the data but the length needs to be know or the data needs to be terminated by a know value (eg. null).

char* nullTerm;

nullTerm = "Loop through my characters";



for(;nullTerm != NULL;nullTerm++)

{

    //nullTerm will now point to the next character.

}

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

edited Dec 30 '08 at 18:06

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

answered Dec 30 '08 at 17:43

Adam Peck

4,28631827

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

add a comment |

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

You should add the initialization of the nullTerm pointer to the beginning of the data set. The OP might be confused about the incomplete for loop.

– cschol
Dec 30 '08 at 17:54

Fleshed out the example a little.

– Adam Peck
Dec 30 '08 at 18:13

you are changing your original pointer, I would do something like: char* s;s="...";for(char *it=s;it!=NULL;it++){/*it point to the character*/}

– hiena
Aug 6 '09 at 4:32

add a comment |

This is a fairly old question, but I though I should post this. It is a foreach loop for GNU C99.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <stdbool.h>



#define FOREACH_COMP(INDEX, ARRAY, ARRAY_TYPE, SIZE) 

  __extension__ 

  ({ 

    bool ret = 0; 

    if (__builtin_types_compatible_p (const char*, ARRAY_TYPE)) 

      ret = INDEX < strlen ((const char*)ARRAY); 

    else 

      ret = INDEX < SIZE; 

    ret; 

  })



#define FOREACH_ELEM(INDEX, ARRAY, TYPE) 

  __extension__ 

  ({ 

    TYPE *tmp_array_ = ARRAY; 

    &tmp_array_[INDEX]; 

  })



#define FOREACH(VAR, ARRAY) 

for (void *array_ = (void*)(ARRAY); array_; array_ = 0) 

for (size_t i_ = 0; i_ && array_ && FOREACH_COMP (i_, array_, 

                                    __typeof__ (ARRAY), 

                                    sizeof (ARRAY) / sizeof ((ARRAY)[0])); 

                                    i_++) 

for (bool b_ = 1; b_; (b_) ? array_ = 0 : 0, b_ = 0) 

for (VAR = FOREACH_ELEM (i_, array_, __typeof__ ((ARRAY)[0])); b_; b_ = 0)



/* example's */

int

main (int argc, char **argv)

{

  int array[10];

  /* initialize the array */

  int i = 0;

  FOREACH (int *x, array)

    {

      *x = i;

      ++i;

    }



  char *str = "hello, world!";

  FOREACH (char *c, str)

    printf ("%cn", *c);



  return EXIT_SUCCESS;

}

This code has been tested to work with gcc, icc and clang on GNU/Linux.

answered Jul 21 '10 at 15:12

Joe D

2,37812425

add a comment |

This is a fairly old question, but I though I should post this. It is a foreach loop for GNU C99.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <stdbool.h>



#define FOREACH_COMP(INDEX, ARRAY, ARRAY_TYPE, SIZE) 

  __extension__ 

  ({ 

    bool ret = 0; 

    if (__builtin_types_compatible_p (const char*, ARRAY_TYPE)) 

      ret = INDEX < strlen ((const char*)ARRAY); 

    else 

      ret = INDEX < SIZE; 

    ret; 

  })



#define FOREACH_ELEM(INDEX, ARRAY, TYPE) 

  __extension__ 

  ({ 

    TYPE *tmp_array_ = ARRAY; 

    &tmp_array_[INDEX]; 

  })



#define FOREACH(VAR, ARRAY) 

for (void *array_ = (void*)(ARRAY); array_; array_ = 0) 

for (size_t i_ = 0; i_ && array_ && FOREACH_COMP (i_, array_, 

                                    __typeof__ (ARRAY), 

                                    sizeof (ARRAY) / sizeof ((ARRAY)[0])); 

                                    i_++) 

for (bool b_ = 1; b_; (b_) ? array_ = 0 : 0, b_ = 0) 

for (VAR = FOREACH_ELEM (i_, array_, __typeof__ ((ARRAY)[0])); b_; b_ = 0)



/* example's */

int

main (int argc, char **argv)

{

  int array[10];

  /* initialize the array */

  int i = 0;

  FOREACH (int *x, array)

    {

      *x = i;

      ++i;

    }



  char *str = "hello, world!";

  FOREACH (char *c, str)

    printf ("%cn", *c);



  return EXIT_SUCCESS;

}

This code has been tested to work with gcc, icc and clang on GNU/Linux.

answered Jul 21 '10 at 15:12

Joe D

2,37812425

add a comment |

This is a fairly old question, but I though I should post this. It is a foreach loop for GNU C99.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <stdbool.h>



#define FOREACH_COMP(INDEX, ARRAY, ARRAY_TYPE, SIZE) 

  __extension__ 

  ({ 

    bool ret = 0; 

    if (__builtin_types_compatible_p (const char*, ARRAY_TYPE)) 

      ret = INDEX < strlen ((const char*)ARRAY); 

    else 

      ret = INDEX < SIZE; 

    ret; 

  })



#define FOREACH_ELEM(INDEX, ARRAY, TYPE) 

  __extension__ 

  ({ 

    TYPE *tmp_array_ = ARRAY; 

    &tmp_array_[INDEX]; 

  })



#define FOREACH(VAR, ARRAY) 

for (void *array_ = (void*)(ARRAY); array_; array_ = 0) 

for (size_t i_ = 0; i_ && array_ && FOREACH_COMP (i_, array_, 

                                    __typeof__ (ARRAY), 

                                    sizeof (ARRAY) / sizeof ((ARRAY)[0])); 

                                    i_++) 

for (bool b_ = 1; b_; (b_) ? array_ = 0 : 0, b_ = 0) 

for (VAR = FOREACH_ELEM (i_, array_, __typeof__ ((ARRAY)[0])); b_; b_ = 0)



/* example's */

int

main (int argc, char **argv)

{

  int array[10];

  /* initialize the array */

  int i = 0;

  FOREACH (int *x, array)

    {

      *x = i;

      ++i;

    }



  char *str = "hello, world!";

  FOREACH (char *c, str)

    printf ("%cn", *c);



  return EXIT_SUCCESS;

}

This code has been tested to work with gcc, icc and clang on GNU/Linux.

answered Jul 21 '10 at 15:12

Joe D

2,37812425

This is a fairly old question, but I though I should post this. It is a foreach loop for GNU C99.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <stdbool.h>



#define FOREACH_COMP(INDEX, ARRAY, ARRAY_TYPE, SIZE) 

  __extension__ 

  ({ 

    bool ret = 0; 

    if (__builtin_types_compatible_p (const char*, ARRAY_TYPE)) 

      ret = INDEX < strlen ((const char*)ARRAY); 

    else 

      ret = INDEX < SIZE; 

    ret; 

  })



#define FOREACH_ELEM(INDEX, ARRAY, TYPE) 

  __extension__ 

  ({ 

    TYPE *tmp_array_ = ARRAY; 

    &tmp_array_[INDEX]; 

  })



#define FOREACH(VAR, ARRAY) 

for (void *array_ = (void*)(ARRAY); array_; array_ = 0) 

for (size_t i_ = 0; i_ && array_ && FOREACH_COMP (i_, array_, 

                                    __typeof__ (ARRAY), 

                                    sizeof (ARRAY) / sizeof ((ARRAY)[0])); 

                                    i_++) 

for (bool b_ = 1; b_; (b_) ? array_ = 0 : 0, b_ = 0) 

for (VAR = FOREACH_ELEM (i_, array_, __typeof__ ((ARRAY)[0])); b_; b_ = 0)



/* example's */

int

main (int argc, char **argv)

{

  int array[10];

  /* initialize the array */

  int i = 0;

  FOREACH (int *x, array)

    {

      *x = i;

      ++i;

    }



  char *str = "hello, world!";

  FOREACH (char *c, str)

    printf ("%cn", *c);



  return EXIT_SUCCESS;

}

This code has been tested to work with gcc, icc and clang on GNU/Linux.

answered Jul 21 '10 at 15:12

Joe D

2,37812425

answered Jul 21 '10 at 15:12

Joe D

2,37812425

answered Jul 21 '10 at 15:12

Joe D

2,37812425

answered Jul 21 '10 at 15:12

Joe D

2,37812425

add a comment |

int arr = {1,2,3,4,5};

for(int *a = arr; a < (&arr)[1]; ++a)

    printf("%dn", *a);

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

3

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

1

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

add a comment |

int arr = {1,2,3,4,5};

for(int *a = arr; a < (&arr)[1]; ++a)

    printf("%dn", *a);

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

3

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

1

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

add a comment |

int arr = {1,2,3,4,5};

for(int *a = arr; a < (&arr)[1]; ++a)

    printf("%dn", *a);

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

int arr = {1,2,3,4,5};

for(int *a = arr; a < (&arr)[1]; ++a)

    printf("%dn", *a);

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

answered Jun 2 '16 at 14:32

Steve Cox

1,329811

3

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

1

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

add a comment |

3

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

1

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

If not so sure this is well-defined. (&arr)[1] doesn't mean one array item past the end of the array, it means one array past the end of the array. (&arr)[1] is not the last item of array [0], it is the array [1], which decays into a pointer to the first element (of array [1]). I believe it would be much better, safer and idiomatic to do const int* begin = arr; const int* end = arr + sizeof(arr)/sizeof(*arr); and then for(const int* a = begin; a != end; a++).

– Lundin
Oct 5 '16 at 9:22

@Lundin This is well defined. You're right, it's one array past the end of the array, but that array type converts to a pointer in this context (an expression), and that pointer is one past the end of the array.

– Steve Cox
Oct 5 '16 at 14:09

add a comment |

As u probably already know, there's no "foreach" style loop in C.

Altho there are already tons of great macros provided here to work around this, maybe you'll find this macro useful.

// "length" is the length of the array.   

#define each(item, array, length) 

(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

item is declared and incremented within the for statement (just like
in Python!).

Seems to work on any 1-dimensional array

All variables created in macro (p, item), aren't visible outside the
scope of the loop (since they're declared in the for loop header).

Disadvantages:

Doesn't work for multi-dimensional arrays

Relies on typeof(), which is a gcc extension; NOT part of standard C

Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct _point {

    double x;

    double y;

} Point;



int main(void)

{

    double some_nums = {4.2, 4.32, -9.9, 7.0};

    for each (element, some_nums, 4)

        printf("element = %lfn", element);



    int numbers = {4, 2, 99, -3, 54};

    // Just demonstrating it can be used like a normal for loop

    for each (number, numbers, 5) { 

        printf("number = %dn", number);

        if (number % 2 == 0)

                printf("%d is even.n", number);

    }



    char *dictionary = {"Hello", "World"};

    for each (word, dictionary, 2)

        printf("word = '%s'n", word);



    Point points = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};

    for each (point, points, 3)

        printf("point = (%lf, %lf)n", point.x, point.y);



    // Neither p, element, number or word are visible outside the scope of

    // their respective for loops. Try to see if these printfs work

    // (they shouldn't):

    // printf("*p = %s", *p);

    // printf("word = %s", word);



    return 0;

}

Seems to work on gcc and clang; not sure about other compilers.

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

add a comment |

As u probably already know, there's no "foreach" style loop in C.

Altho there are already tons of great macros provided here to work around this, maybe you'll find this macro useful.

// "length" is the length of the array.   

#define each(item, array, length) 

(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

item is declared and incremented within the for statement (just like
in Python!).

Seems to work on any 1-dimensional array

All variables created in macro (p, item), aren't visible outside the
scope of the loop (since they're declared in the for loop header).

Disadvantages:

Doesn't work for multi-dimensional arrays

Relies on typeof(), which is a gcc extension; NOT part of standard C

Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct _point {

    double x;

    double y;

} Point;



int main(void)

{

    double some_nums = {4.2, 4.32, -9.9, 7.0};

    for each (element, some_nums, 4)

        printf("element = %lfn", element);



    int numbers = {4, 2, 99, -3, 54};

    // Just demonstrating it can be used like a normal for loop

    for each (number, numbers, 5) { 

        printf("number = %dn", number);

        if (number % 2 == 0)

                printf("%d is even.n", number);

    }



    char *dictionary = {"Hello", "World"};

    for each (word, dictionary, 2)

        printf("word = '%s'n", word);



    Point points = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};

    for each (point, points, 3)

        printf("point = (%lf, %lf)n", point.x, point.y);



    // Neither p, element, number or word are visible outside the scope of

    // their respective for loops. Try to see if these printfs work

    // (they shouldn't):

    // printf("*p = %s", *p);

    // printf("word = %s", word);



    return 0;

}

Seems to work on gcc and clang; not sure about other compilers.

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

add a comment |

As u probably already know, there's no "foreach" style loop in C.

Altho there are already tons of great macros provided here to work around this, maybe you'll find this macro useful.

// "length" is the length of the array.   

#define each(item, array, length) 

(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

item is declared and incremented within the for statement (just like
in Python!).

Seems to work on any 1-dimensional array

All variables created in macro (p, item), aren't visible outside the
scope of the loop (since they're declared in the for loop header).

Disadvantages:

Doesn't work for multi-dimensional arrays

Relies on typeof(), which is a gcc extension; NOT part of standard C

Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct _point {

    double x;

    double y;

} Point;



int main(void)

{

    double some_nums = {4.2, 4.32, -9.9, 7.0};

    for each (element, some_nums, 4)

        printf("element = %lfn", element);



    int numbers = {4, 2, 99, -3, 54};

    // Just demonstrating it can be used like a normal for loop

    for each (number, numbers, 5) { 

        printf("number = %dn", number);

        if (number % 2 == 0)

                printf("%d is even.n", number);

    }



    char *dictionary = {"Hello", "World"};

    for each (word, dictionary, 2)

        printf("word = '%s'n", word);



    Point points = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};

    for each (point, points, 3)

        printf("point = (%lf, %lf)n", point.x, point.y);



    // Neither p, element, number or word are visible outside the scope of

    // their respective for loops. Try to see if these printfs work

    // (they shouldn't):

    // printf("*p = %s", *p);

    // printf("word = %s", word);



    return 0;

}

Seems to work on gcc and clang; not sure about other compilers.

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

As u probably already know, there's no "foreach" style loop in C.

Altho there are already tons of great macros provided here to work around this, maybe you'll find this macro useful.

// "length" is the length of the array.   

#define each(item, array, length) 

(typeof(*(array)) *p = (array), (item) = *p; p < &((array)[length]); p++, (item) = *p)

...which can be used with for (as in for each (...)).

Advantages of this approach:

item is declared and incremented within the for statement (just like
in Python!).

Seems to work on any 1-dimensional array

All variables created in macro (p, item), aren't visible outside the
scope of the loop (since they're declared in the for loop header).

Disadvantages:

Doesn't work for multi-dimensional arrays

Relies on typeof(), which is a gcc extension; NOT part of standard C

Since it declares variables in the for loop header, it only works in C11 or later.

Just to save you some time, here's how you could test it:

typedef struct _point {

    double x;

    double y;

} Point;



int main(void)

{

    double some_nums = {4.2, 4.32, -9.9, 7.0};

    for each (element, some_nums, 4)

        printf("element = %lfn", element);



    int numbers = {4, 2, 99, -3, 54};

    // Just demonstrating it can be used like a normal for loop

    for each (number, numbers, 5) { 

        printf("number = %dn", number);

        if (number % 2 == 0)

                printf("%d is even.n", number);

    }



    char *dictionary = {"Hello", "World"};

    for each (word, dictionary, 2)

        printf("word = '%s'n", word);



    Point points = {{3.4, 4.2}, {9.9, 6.7}, {-9.8, 7.0}};

    for each (point, points, 3)

        printf("point = (%lf, %lf)n", point.x, point.y);



    // Neither p, element, number or word are visible outside the scope of

    // their respective for loops. Try to see if these printfs work

    // (they shouldn't):

    // printf("*p = %s", *p);

    // printf("word = %s", word);



    return 0;

}

Seems to work on gcc and clang; not sure about other compilers.

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

edited Jul 7 '18 at 23:43

answered Nov 6 '17 at 9:47

Saeed Baig

35926

answered Nov 6 '17 at 9:47

Saeed Baig

35926

answered Nov 6 '17 at 9:47

Saeed Baig

35926

add a comment |

C has 'for' and 'while' keywords. If a foreach statement in a language like C# looks like this ...

foreach (Element element in collection)

{

}

... then the equivalent of this foreach statement in C might be be like:

for (

    Element* element = GetFirstElement(&collection);

    element != 0;

    element = GetNextElement(&collection, element)

    )

{

    //TODO: do something with this element instance ...

}

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

1

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

add a comment |

C has 'for' and 'while' keywords. If a foreach statement in a language like C# looks like this ...

foreach (Element element in collection)

{

}

... then the equivalent of this foreach statement in C might be be like:

for (

    Element* element = GetFirstElement(&collection);

    element != 0;

    element = GetNextElement(&collection, element)

    )

{

    //TODO: do something with this element instance ...

}

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

1

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

add a comment |

C has 'for' and 'while' keywords. If a foreach statement in a language like C# looks like this ...

foreach (Element element in collection)

{

}

... then the equivalent of this foreach statement in C might be be like:

for (

    Element* element = GetFirstElement(&collection);

    element != 0;

    element = GetNextElement(&collection, element)

    )

{

    //TODO: do something with this element instance ...

}

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

C has 'for' and 'while' keywords. If a foreach statement in a language like C# looks like this ...

foreach (Element element in collection)

{

}

... then the equivalent of this foreach statement in C might be be like:

for (

    Element* element = GetFirstElement(&collection);

    element != 0;

    element = GetNextElement(&collection, element)

    )

{

    //TODO: do something with this element instance ...

}

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

edited Dec 30 '08 at 17:55

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

answered Dec 30 '08 at 17:45

ChrisW

45.5k983185

1

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

add a comment |

1

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

You should mention that your example code is not written in C syntax.

– cschol
Dec 30 '08 at 17:48

> You should mention that your example code is not written in C syntax You're right, thank you: I'll edit the post.

– ChrisW
Dec 30 '08 at 17:53

@monjardin-> sure you can just define pointer to function in the struct and there is no problem to make the call like this.

– Ilya
Dec 31 '08 at 7:53

add a comment |

Eric's answer doesn't work when you're using "break" or "continue".

This can be fixed by rewriting the first line:

Original line (reformatted):

for (unsigned i = 0, __a = 1; i < B.size(); i++, __a = 1)

Fixed:

for (unsigned i = 0, __a = 1; __a && i < B.size(); i++, __a = 1)

If you compare it to Johannes' loop, you'll see that he's actually doing the same, just a bit more complicated and uglier.

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

add a comment |

Eric's answer doesn't work when you're using "break" or "continue".

This can be fixed by rewriting the first line:

Original line (reformatted):

for (unsigned i = 0, __a = 1; i < B.size(); i++, __a = 1)

Fixed:

for (unsigned i = 0, __a = 1; __a && i < B.size(); i++, __a = 1)

If you compare it to Johannes' loop, you'll see that he's actually doing the same, just a bit more complicated and uglier.

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

add a comment |

Eric's answer doesn't work when you're using "break" or "continue".

This can be fixed by rewriting the first line:

Original line (reformatted):

for (unsigned i = 0, __a = 1; i < B.size(); i++, __a = 1)

Fixed:

for (unsigned i = 0, __a = 1; __a && i < B.size(); i++, __a = 1)

If you compare it to Johannes' loop, you'll see that he's actually doing the same, just a bit more complicated and uglier.

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

Eric's answer doesn't work when you're using "break" or "continue".

This can be fixed by rewriting the first line:

Original line (reformatted):

for (unsigned i = 0, __a = 1; i < B.size(); i++, __a = 1)

Fixed:

for (unsigned i = 0, __a = 1; __a && i < B.size(); i++, __a = 1)

If you compare it to Johannes' loop, you'll see that he's actually doing the same, just a bit more complicated and uglier.

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

edited May 23 '17 at 12:03

Community♦

edited May 23 '17 at 12:03

Community♦

edited May 23 '17 at 12:03

Community♦

answered Jul 14 '11 at 22:04

Michael Blurb

283

answered Jul 14 '11 at 22:04

Michael Blurb

283

answered Jul 14 '11 at 22:04

Michael Blurb

283

add a comment |

Here's a simple one, single for loop:

#define FOREACH(type, array, size) do { 

        type it = array[0]; 

        for(int i = 0; i < size; i++, it = array[i])

#define ENDFOR  } while(0);



int array = { 1, 2, 3, 4, 5 };



FOREACH(int, array, 5)

{

    printf("element: %d. index: %dn", it, i);

}

ENDFOR

#define FOREACH(item, array, start, size)

    for(int i = start, keep = 1;

        keep && i < size;

        keep = !keep, i++)

    for (item = array[i]; keep; keep = !keep)



int array = { 1, 2, 3, 4, 5 };

FOREACH(int x, array, 2, 5)

    printf("index: %d. element: %dn", i, x);

Output:

index: 2. element: 3

index: 3. element: 4

index: 4. element: 5

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

add a comment |

Here's a simple one, single for loop:

#define FOREACH(type, array, size) do { 

        type it = array[0]; 

        for(int i = 0; i < size; i++, it = array[i])

#define ENDFOR  } while(0);



int array = { 1, 2, 3, 4, 5 };



FOREACH(int, array, 5)

{

    printf("element: %d. index: %dn", it, i);

}

ENDFOR

#define FOREACH(item, array, start, size)

    for(int i = start, keep = 1;

        keep && i < size;

        keep = !keep, i++)

    for (item = array[i]; keep; keep = !keep)



int array = { 1, 2, 3, 4, 5 };

FOREACH(int x, array, 2, 5)

    printf("index: %d. element: %dn", i, x);

Output:

index: 2. element: 3

index: 3. element: 4

index: 4. element: 5

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

add a comment |

Here's a simple one, single for loop:

#define FOREACH(type, array, size) do { 

        type it = array[0]; 

        for(int i = 0; i < size; i++, it = array[i])

#define ENDFOR  } while(0);



int array = { 1, 2, 3, 4, 5 };



FOREACH(int, array, 5)

{

    printf("element: %d. index: %dn", it, i);

}

ENDFOR

#define FOREACH(item, array, start, size)

    for(int i = start, keep = 1;

        keep && i < size;

        keep = !keep, i++)

    for (item = array[i]; keep; keep = !keep)



int array = { 1, 2, 3, 4, 5 };

FOREACH(int x, array, 2, 5)

    printf("index: %d. element: %dn", i, x);

Output:

index: 2. element: 3

index: 3. element: 4

index: 4. element: 5

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

Here's a simple one, single for loop:

#define FOREACH(type, array, size) do { 

        type it = array[0]; 

        for(int i = 0; i < size; i++, it = array[i])

#define ENDFOR  } while(0);



int array = { 1, 2, 3, 4, 5 };



FOREACH(int, array, 5)

{

    printf("element: %d. index: %dn", it, i);

}

ENDFOR

#define FOREACH(item, array, start, size)

    for(int i = start, keep = 1;

        keep && i < size;

        keep = !keep, i++)

    for (item = array[i]; keep; keep = !keep)



int array = { 1, 2, 3, 4, 5 };

FOREACH(int x, array, 2, 5)

    printf("index: %d. element: %dn", i, x);

Output:

index: 2. element: 3

index: 3. element: 4

index: 4. element: 5

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

edited Aug 29 '15 at 13:12

answered Aug 27 '15 at 17:45

vexe

2,17273360

answered Aug 27 '15 at 17:45

vexe

2,17273360

answered Aug 27 '15 at 17:45

vexe

2,17273360

add a comment |

However, for lists, it is easy to implement something that resembles a for-each loop.

for(Node* node = head; node; node = node.next) {

   /* do your magic here */

}

To achieve something similar for arrays you can do one of two things.

use the first element to store the length of the array.

wrap the array in a struct which holds the length and a pointer to the array.

The following is an example of such struct:

typedef struct job_t {

   int count;

   int* arr;

} arr_t;

answered Mar 14 '18 at 23:12

Jonas

355416

add a comment |

However, for lists, it is easy to implement something that resembles a for-each loop.

for(Node* node = head; node; node = node.next) {

   /* do your magic here */

}

To achieve something similar for arrays you can do one of two things.

use the first element to store the length of the array.

wrap the array in a struct which holds the length and a pointer to the array.

The following is an example of such struct:

typedef struct job_t {

   int count;

   int* arr;

} arr_t;

answered Mar 14 '18 at 23:12

Jonas

355416

add a comment |

However, for lists, it is easy to implement something that resembles a for-each loop.

for(Node* node = head; node; node = node.next) {

   /* do your magic here */

}

To achieve something similar for arrays you can do one of two things.

use the first element to store the length of the array.

wrap the array in a struct which holds the length and a pointer to the array.

The following is an example of such struct:

typedef struct job_t {

   int count;

   int* arr;

} arr_t;

answered Mar 14 '18 at 23:12

Jonas

355416

However, for lists, it is easy to implement something that resembles a for-each loop.

for(Node* node = head; node; node = node.next) {

   /* do your magic here */

}

To achieve something similar for arrays you can do one of two things.

use the first element to store the length of the array.

wrap the array in a struct which holds the length and a pointer to the array.

The following is an example of such struct:

typedef struct job_t {

   int count;

   int* arr;

} arr_t;

answered Mar 14 '18 at 23:12

Jonas

355416

answered Mar 14 '18 at 23:12

Jonas

355416

answered Mar 14 '18 at 23:12

Jonas

355416

answered Mar 14 '18 at 23:12

Jonas

355416

add a comment |

Here is what I use when I'm stuck with C. You can't use the same item name twice in the same scope, but that's not really an issue since not all of us get to use nice new compilers :(

#define FOREACH(type, item, array, size) 

    size_t X(keep), X(i); 

    type item; 

    for (X(keep) = 1, X(i) = 0 ; X(i) < (size); X(keep) = !X(keep), X(i)++) 

        for (item = (array)[X(i)]; X(keep); X(keep) = 0)



#define _foreach(item, array) FOREACH(__typeof__(array[0]), item, array, length(array))

#define foreach(item_in_array) _foreach(item_in_array)



#define in ,

#define length(array) (sizeof(array) / sizeof((array)[0]))

#define CAT(a, b) CAT_HELPER(a, b) /* Concatenate two symbols for macros! */

#define CAT_HELPER(a, b) a ## b

#define X(name) CAT(__##name, __LINE__) /* unique variable */

Usage:

int ints = {1, 2, 0, 3, 4};

foreach (i in ints) printf("%i", i);

/* can't use the same name in this scope anymore! */

foreach (x in ints) printf("%i", x);

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

add a comment |

Here is what I use when I'm stuck with C. You can't use the same item name twice in the same scope, but that's not really an issue since not all of us get to use nice new compilers :(

#define FOREACH(type, item, array, size) 

    size_t X(keep), X(i); 

    type item; 

    for (X(keep) = 1, X(i) = 0 ; X(i) < (size); X(keep) = !X(keep), X(i)++) 

        for (item = (array)[X(i)]; X(keep); X(keep) = 0)



#define _foreach(item, array) FOREACH(__typeof__(array[0]), item, array, length(array))

#define foreach(item_in_array) _foreach(item_in_array)



#define in ,

#define length(array) (sizeof(array) / sizeof((array)[0]))

#define CAT(a, b) CAT_HELPER(a, b) /* Concatenate two symbols for macros! */

#define CAT_HELPER(a, b) a ## b

#define X(name) CAT(__##name, __LINE__) /* unique variable */

Usage:

int ints = {1, 2, 0, 3, 4};

foreach (i in ints) printf("%i", i);

/* can't use the same name in this scope anymore! */

foreach (x in ints) printf("%i", x);

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

add a comment |

Here is what I use when I'm stuck with C. You can't use the same item name twice in the same scope, but that's not really an issue since not all of us get to use nice new compilers :(

#define FOREACH(type, item, array, size) 

    size_t X(keep), X(i); 

    type item; 

    for (X(keep) = 1, X(i) = 0 ; X(i) < (size); X(keep) = !X(keep), X(i)++) 

        for (item = (array)[X(i)]; X(keep); X(keep) = 0)



#define _foreach(item, array) FOREACH(__typeof__(array[0]), item, array, length(array))

#define foreach(item_in_array) _foreach(item_in_array)



#define in ,

#define length(array) (sizeof(array) / sizeof((array)[0]))

#define CAT(a, b) CAT_HELPER(a, b) /* Concatenate two symbols for macros! */

#define CAT_HELPER(a, b) a ## b

#define X(name) CAT(__##name, __LINE__) /* unique variable */

Usage:

int ints = {1, 2, 0, 3, 4};

foreach (i in ints) printf("%i", i);

/* can't use the same name in this scope anymore! */

foreach (x in ints) printf("%i", x);

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

Here is what I use when I'm stuck with C. You can't use the same item name twice in the same scope, but that's not really an issue since not all of us get to use nice new compilers :(

#define FOREACH(type, item, array, size) 

    size_t X(keep), X(i); 

    type item; 

    for (X(keep) = 1, X(i) = 0 ; X(i) < (size); X(keep) = !X(keep), X(i)++) 

        for (item = (array)[X(i)]; X(keep); X(keep) = 0)



#define _foreach(item, array) FOREACH(__typeof__(array[0]), item, array, length(array))

#define foreach(item_in_array) _foreach(item_in_array)



#define in ,

#define length(array) (sizeof(array) / sizeof((array)[0]))

#define CAT(a, b) CAT_HELPER(a, b) /* Concatenate two symbols for macros! */

#define CAT_HELPER(a, b) a ## b

#define X(name) CAT(__##name, __LINE__) /* unique variable */

Usage:

int ints = {1, 2, 0, 3, 4};

foreach (i in ints) printf("%i", i);

/* can't use the same name in this scope anymore! */

foreach (x in ints) printf("%i", x);

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

edited Nov 5 '15 at 23:08

answered Sep 29 '15 at 7:34

Watercycle

43137

answered Sep 29 '15 at 7:34

Watercycle

43137

answered Sep 29 '15 at 7:34

Watercycle

43137

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

add a comment |

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

Note: VAR(i) < (size) && (item = array[VAR(i)]) would stop once the array element had a value of 0. So using this with double Array may not iterate through all elements. Seems like the loop test should be one or the other: i<n or A[i]. Maybe add sample use cases for clarity.

– chux
Oct 16 '15 at 17:32

Even with pointers in my previous approach the result seems to be 'undefined behavior'. Oh well. Trust the double for loop approach!

– Watercycle
Oct 22 '15 at 1:54

This version pollutes the scope and will fail if used twice in the same scope. Also doesn't work as an un-braced block (e.g. if ( bla ) FOREACH(....) { } else....

– M.M
Nov 5 '15 at 22:15

1, C is the language of scope pollution, some of us are limited to older compilers. 2, Don't Repeat Yourself / be descriptive. 3, yeah, unfortunately you MUST have braces if it is going to be a conditional for loop (people usually do anyway). If you have access to a compiler that supports variable declarations in a for loop, by all means do it.

– Watercycle
Nov 5 '15 at 23:13

add a comment |

If you're planning to work with function pointers

#define lambda(return_type, function_body)

    ({ return_type __fn__ function_body __fn__; })



#define array_len(arr) (sizeof(arr)/sizeof(arr[0]))



#define foreachnf(type, item, arr, arr_length, func) {

    void (*action)(type item) = func;

    for (int i = 0; i<arr_length; i++) action(arr[i]);

}



#define foreachf(type, item, arr, func)

    foreachnf(type, item, arr, array_len(arr), func)



#define foreachn(type, item, arr, arr_length, body)

    foreachnf(type, item, arr, arr_length, lambda(void, (type item) body))



#define foreach(type, item, arr, body)

    foreachn(type, item, arr, array_len(arr), body)

Usage:

int ints = { 1, 2, 3, 4, 5 };

foreach(int, i, ints, {

    printf("%dn", i);

});



char* strs = { "hi!", "hello!!", "hello world", "just", "testing" };

foreach(char*, s, strs, {

    printf("%sn", s);

});



char** strsp = malloc(sizeof(char*)*2);

strsp[0] = "abcd";

strsp[1] = "efgh";

foreachn(char*, s, strsp, 2, {

    printf("%sn", s);

});



void (*myfun)(int i) = somefunc;

foreachf(int, i, ints, myfun);

But I think this will work only on gcc (not sure).

answered May 15 '17 at 20:01

Naheel

362210

add a comment |

If you're planning to work with function pointers

#define lambda(return_type, function_body)

    ({ return_type __fn__ function_body __fn__; })



#define array_len(arr) (sizeof(arr)/sizeof(arr[0]))



#define foreachnf(type, item, arr, arr_length, func) {

    void (*action)(type item) = func;

    for (int i = 0; i<arr_length; i++) action(arr[i]);

}



#define foreachf(type, item, arr, func)

    foreachnf(type, item, arr, array_len(arr), func)



#define foreachn(type, item, arr, arr_length, body)

    foreachnf(type, item, arr, arr_length, lambda(void, (type item) body))



#define foreach(type, item, arr, body)

    foreachn(type, item, arr, array_len(arr), body)

Usage:

int ints = { 1, 2, 3, 4, 5 };

foreach(int, i, ints, {

    printf("%dn", i);

});



char* strs = { "hi!", "hello!!", "hello world", "just", "testing" };

foreach(char*, s, strs, {

    printf("%sn", s);

});



char** strsp = malloc(sizeof(char*)*2);

strsp[0] = "abcd";

strsp[1] = "efgh";

foreachn(char*, s, strsp, 2, {

    printf("%sn", s);

});



void (*myfun)(int i) = somefunc;

foreachf(int, i, ints, myfun);

But I think this will work only on gcc (not sure).

answered May 15 '17 at 20:01

Naheel

362210

add a comment |

If you're planning to work with function pointers

#define lambda(return_type, function_body)

    ({ return_type __fn__ function_body __fn__; })



#define array_len(arr) (sizeof(arr)/sizeof(arr[0]))



#define foreachnf(type, item, arr, arr_length, func) {

    void (*action)(type item) = func;

    for (int i = 0; i<arr_length; i++) action(arr[i]);

}



#define foreachf(type, item, arr, func)

    foreachnf(type, item, arr, array_len(arr), func)



#define foreachn(type, item, arr, arr_length, body)

    foreachnf(type, item, arr, arr_length, lambda(void, (type item) body))



#define foreach(type, item, arr, body)

    foreachn(type, item, arr, array_len(arr), body)

Usage:

int ints = { 1, 2, 3, 4, 5 };

foreach(int, i, ints, {

    printf("%dn", i);

});



char* strs = { "hi!", "hello!!", "hello world", "just", "testing" };

foreach(char*, s, strs, {

    printf("%sn", s);

});



char** strsp = malloc(sizeof(char*)*2);

strsp[0] = "abcd";

strsp[1] = "efgh";

foreachn(char*, s, strsp, 2, {

    printf("%sn", s);

});



void (*myfun)(int i) = somefunc;

foreachf(int, i, ints, myfun);

But I think this will work only on gcc (not sure).

answered May 15 '17 at 20:01

Naheel

362210

If you're planning to work with function pointers

#define lambda(return_type, function_body)

    ({ return_type __fn__ function_body __fn__; })



#define array_len(arr) (sizeof(arr)/sizeof(arr[0]))



#define foreachnf(type, item, arr, arr_length, func) {

    void (*action)(type item) = func;

    for (int i = 0; i<arr_length; i++) action(arr[i]);

}



#define foreachf(type, item, arr, func)

    foreachnf(type, item, arr, array_len(arr), func)



#define foreachn(type, item, arr, arr_length, body)

    foreachnf(type, item, arr, arr_length, lambda(void, (type item) body))



#define foreach(type, item, arr, body)

    foreachn(type, item, arr, array_len(arr), body)

Usage:

int ints = { 1, 2, 3, 4, 5 };

foreach(int, i, ints, {

    printf("%dn", i);

});



char* strs = { "hi!", "hello!!", "hello world", "just", "testing" };

foreach(char*, s, strs, {

    printf("%sn", s);

});



char** strsp = malloc(sizeof(char*)*2);

strsp[0] = "abcd";

strsp[1] = "efgh";

foreachn(char*, s, strsp, 2, {

    printf("%sn", s);

});



void (*myfun)(int i) = somefunc;

foreachf(int, i, ints, myfun);

But I think this will work only on gcc (not sure).

answered May 15 '17 at 20:01

Naheel

362210

answered May 15 '17 at 20:01

Naheel

362210

answered May 15 '17 at 20:01

Naheel

362210

answered May 15 '17 at 20:01

Naheel

362210

add a comment |

template <typename t> struct list { // =list=

  struct elem {

    t Data;

    elem* Next;

  };

  elem* Head;

  elem* Tail;

};

#define for_list3(TYPE, NAME, NAME2, LIST) 

  bool _##NAME##NAME2 = true;

  for (list<TYPE>::elem* NAME##_E = LIST.Head;

      NAME##_E && _##NAME##NAME2;

      _##NAME##NAME2 = !_##NAME##NAME2, NAME##_E = NAME##_E->Next) 

    for (auto& NAME = NAME##_E->Data; _##NAME##NAME2; _##NAME##NAME2=false)

#define for_list2(TYPE, NAME, NAME2, LIST) for_list3(TYPE, NAME, NAME2, LIST)

#define for_list(TYPE, NAME, LIST) for_list2(TYPE, NAME, __COUNTER__, LIST)



void example() {

  list<string> Words;

  for_list(string, Word, Words) {

    print(Word);

  }

}

With the for_each_item macro from the accepted answer, you'd have to do this:

void example2() {

  list<string> Words;

  for_each_item(string, Elem, Words) {

    string& Word = Elem->Data;

    print(Word);

  }

}

If you want to rewrite it into a clean version, feel free to edit.

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

add a comment |

template <typename t> struct list { // =list=

  struct elem {

    t Data;

    elem* Next;

  };

  elem* Head;

  elem* Tail;

};

#define for_list3(TYPE, NAME, NAME2, LIST) 

  bool _##NAME##NAME2 = true;

  for (list<TYPE>::elem* NAME##_E = LIST.Head;

      NAME##_E && _##NAME##NAME2;

      _##NAME##NAME2 = !_##NAME##NAME2, NAME##_E = NAME##_E->Next) 

    for (auto& NAME = NAME##_E->Data; _##NAME##NAME2; _##NAME##NAME2=false)

#define for_list2(TYPE, NAME, NAME2, LIST) for_list3(TYPE, NAME, NAME2, LIST)

#define for_list(TYPE, NAME, LIST) for_list2(TYPE, NAME, __COUNTER__, LIST)



void example() {

  list<string> Words;

  for_list(string, Word, Words) {

    print(Word);

  }

}

With the for_each_item macro from the accepted answer, you'd have to do this:

void example2() {

  list<string> Words;

  for_each_item(string, Elem, Words) {

    string& Word = Elem->Data;

    print(Word);

  }

}

If you want to rewrite it into a clean version, feel free to edit.

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

add a comment |

template <typename t> struct list { // =list=

  struct elem {

    t Data;

    elem* Next;

  };

  elem* Head;

  elem* Tail;

};

#define for_list3(TYPE, NAME, NAME2, LIST) 

  bool _##NAME##NAME2 = true;

  for (list<TYPE>::elem* NAME##_E = LIST.Head;

      NAME##_E && _##NAME##NAME2;

      _##NAME##NAME2 = !_##NAME##NAME2, NAME##_E = NAME##_E->Next) 

    for (auto& NAME = NAME##_E->Data; _##NAME##NAME2; _##NAME##NAME2=false)

#define for_list2(TYPE, NAME, NAME2, LIST) for_list3(TYPE, NAME, NAME2, LIST)

#define for_list(TYPE, NAME, LIST) for_list2(TYPE, NAME, __COUNTER__, LIST)



void example() {

  list<string> Words;

  for_list(string, Word, Words) {

    print(Word);

  }

}

With the for_each_item macro from the accepted answer, you'd have to do this:

void example2() {

  list<string> Words;

  for_each_item(string, Elem, Words) {

    string& Word = Elem->Data;

    print(Word);

  }

}

If you want to rewrite it into a clean version, feel free to edit.

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

template <typename t> struct list { // =list=

  struct elem {

    t Data;

    elem* Next;

  };

  elem* Head;

  elem* Tail;

};

#define for_list3(TYPE, NAME, NAME2, LIST) 

  bool _##NAME##NAME2 = true;

  for (list<TYPE>::elem* NAME##_E = LIST.Head;

      NAME##_E && _##NAME##NAME2;

      _##NAME##NAME2 = !_##NAME##NAME2, NAME##_E = NAME##_E->Next) 

    for (auto& NAME = NAME##_E->Data; _##NAME##NAME2; _##NAME##NAME2=false)

#define for_list2(TYPE, NAME, NAME2, LIST) for_list3(TYPE, NAME, NAME2, LIST)

#define for_list(TYPE, NAME, LIST) for_list2(TYPE, NAME, __COUNTER__, LIST)



void example() {

  list<string> Words;

  for_list(string, Word, Words) {

    print(Word);

  }

}

With the for_each_item macro from the accepted answer, you'd have to do this:

void example2() {

  list<string> Words;

  for_each_item(string, Elem, Words) {

    string& Word = Elem->Data;

    print(Word);

  }

}

If you want to rewrite it into a clean version, feel free to edit.

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

edited Nov 19 '18 at 22:40

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

answered Nov 19 '18 at 22:31

Andreas Haferburg

3,33411947

add a comment |

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