Mixing
Differentiate between Array and Pointer in function parameter - C++
0
This is a problem I've been struggling with for a while now.
The codes I have tried below are not compiling either.
The question is: How do I differentiate between a pointer and a fixed array in a function parameter?
// Concepts for the Array I have tried but not succeeded.
template <size_t length, typename type>
const unsigned int len(type arg[static length]) { return length; }
template <size_t length, typename type>
const unsigned int len(type(&)[length]) { return length; }
// This works for Arrays & Pointers
// but should be less prioritized over the function that detects the Array
template <typename type> const unsigned int len(type* arg) { return sizeof(*arg); }
I know how arrays & pointers are basically similar when passed into a function, but it asks the question: is there not a way to tell them apart?
Syntactically yes, but otherwise what other way is there?
Anyway, thanks for reading through and cheers on your response.
c++
asked Nov 21 '18 at 6:24
LapysLapys
3961313
add a comment |
0
This is a problem I've been struggling with for a while now.
The codes I have tried below are not compiling either.
The question is: How do I differentiate between a pointer and a fixed array in a function parameter?
// Concepts for the Array I have tried but not succeeded.
template <size_t length, typename type>
const unsigned int len(type arg[static length]) { return length; }
template <size_t length, typename type>
const unsigned int len(type(&)[length]) { return length; }
// This works for Arrays & Pointers
// but should be less prioritized over the function that detects the Array
template <typename type> const unsigned int len(type* arg) { return sizeof(*arg); }
I know how arrays & pointers are basically similar when passed into a function, but it asks the question: is there not a way to tell them apart?
Syntactically yes, but otherwise what other way is there?
Anyway, thanks for reading through and cheers on your response.
c++
asked Nov 21 '18 at 6:24
LapysLapys
3961313
2
static lengthis not valid C++. That particular C99 feature is not available.
– StoryTeller
Nov 21 '18 at 6:30
add a comment |
0
0
0
This is a problem I've been struggling with for a while now.
The codes I have tried below are not compiling either.
The question is: How do I differentiate between a pointer and a fixed array in a function parameter?
// Concepts for the Array I have tried but not succeeded.
template <size_t length, typename type>
const unsigned int len(type arg[static length]) { return length; }
template <size_t length, typename type>
const unsigned int len(type(&)[length]) { return length; }
// This works for Arrays & Pointers
// but should be less prioritized over the function that detects the Array
template <typename type> const unsigned int len(type* arg) { return sizeof(*arg); }
I know how arrays & pointers are basically similar when passed into a function, but it asks the question: is there not a way to tell them apart?
Syntactically yes, but otherwise what other way is there?
Anyway, thanks for reading through and cheers on your response.
c++
asked Nov 21 '18 at 6:24
LapysLapys
3961313
This is a problem I've been struggling with for a while now.
The codes I have tried below are not compiling either.
The question is: How do I differentiate between a pointer and a fixed array in a function parameter?
// Concepts for the Array I have tried but not succeeded.
template <size_t length, typename type>
const unsigned int len(type arg[static length]) { return length; }
template <size_t length, typename type>
const unsigned int len(type(&)[length]) { return length; }
// This works for Arrays & Pointers
// but should be less prioritized over the function that detects the Array
template <typename type> const unsigned int len(type* arg) { return sizeof(*arg); }
I know how arrays & pointers are basically similar when passed into a function, but it asks the question: is there not a way to tell them apart?
Syntactically yes, but otherwise what other way is there?
Anyway, thanks for reading through and cheers on your response.
c++
c++
asked Nov 21 '18 at 6:24
LapysLapys
3961313
asked Nov 21 '18 at 6:24
LapysLapys
3961313
asked Nov 21 '18 at 6:24
LapysLapys
3961313
asked Nov 21 '18 at 6:24
LapysLapys
3961313
asked Nov 21 '18 at 6:24
LapysLapys
3961313
3961313
2
static lengthis not valid C++. That particular C99 feature is not available.
– StoryTeller
Nov 21 '18 at 6:30
add a comment |
2
static lengthis not valid C++. That particular C99 feature is not available.
– StoryTeller
Nov 21 '18 at 6:30
2
2
static length is not valid C++. That particular C99 feature is not available.– StoryTeller
Nov 21 '18 at 6:30
static length is not valid C++. That particular C99 feature is not available.– StoryTeller
Nov 21 '18 at 6:30
add a comment |
2 Answers
2
active
oldest
votes
1
This approach works for me:
#include <stdio.h>
template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is: %in", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zun", sizeof(*p)); return sizeof(*p);}
int main(int, char **)
{
int myArray[10];
int * myPointer = myArray;
(void) len(myArray);
(void) len(myPointer);
return 0;
}
... when I run it, it prints out:
The number of items in your array is: 10
The size of the item your pointer points to is: 4
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
add a comment |
1
The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
// ^^^
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
// ^^^
Full demo:
#include <cstdlib>
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
#include <iostream>
int main(int, char **)
{
const int array[7] = {};
const int *pointer = array;
std::cout << "array has " << len(array) << " itemsn"
<< "and pointer is to " << len(pointer) << " charsn";
}
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
This approach works for me:
#include <stdio.h>
template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is: %in", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zun", sizeof(*p)); return sizeof(*p);}
int main(int, char **)
{
int myArray[10];
int * myPointer = myArray;
(void) len(myArray);
(void) len(myPointer);
return 0;
}
... when I run it, it prints out:
The number of items in your array is: 10
The size of the item your pointer points to is: 4
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
add a comment |
1
This approach works for me:
#include <stdio.h>
template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is: %in", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zun", sizeof(*p)); return sizeof(*p);}
int main(int, char **)
{
int myArray[10];
int * myPointer = myArray;
(void) len(myArray);
(void) len(myPointer);
return 0;
}
... when I run it, it prints out:
The number of items in your array is: 10
The size of the item your pointer points to is: 4
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
add a comment |
1
1
1
This approach works for me:
#include <stdio.h>
template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is: %in", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zun", sizeof(*p)); return sizeof(*p);}
int main(int, char **)
{
int myArray[10];
int * myPointer = myArray;
(void) len(myArray);
(void) len(myPointer);
return 0;
}
... when I run it, it prints out:
The number of items in your array is: 10
The size of the item your pointer points to is: 4
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
This approach works for me:
#include <stdio.h>
template<typename T, int size> unsigned int len(const T(&)[size]) {printf("The number of items in your array is: %in", size); return size;}
template<typename T> unsigned int len(const T * p) {printf("The size of the item your pointer points to is: %zun", sizeof(*p)); return sizeof(*p);}
int main(int, char **)
{
int myArray[10];
int * myPointer = myArray;
(void) len(myArray);
(void) len(myPointer);
return 0;
}
... when I run it, it prints out:
The number of items in your array is: 10
The size of the item your pointer points to is: 4
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
answered Nov 21 '18 at 6:30
Jeremy FriesnerJeremy Friesner
39.3k1080161
39.3k1080161
add a comment |
add a comment |
1
The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
// ^^^
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
// ^^^
Full demo:
#include <cstdlib>
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
#include <iostream>
int main(int, char **)
{
const int array[7] = {};
const int *pointer = array;
std::cout << "array has " << len(array) << " itemsn"
<< "and pointer is to " << len(pointer) << " charsn";
}
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
add a comment |
1
The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
// ^^^
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
// ^^^
Full demo:
#include <cstdlib>
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
#include <iostream>
int main(int, char **)
{
const int array[7] = {};
const int *pointer = array;
std::cout << "array has " << len(array) << " itemsn"
<< "and pointer is to " << len(pointer) << " charsn";
}
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
add a comment |
1
1
1
The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
// ^^^
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
// ^^^
Full demo:
#include <cstdlib>
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
#include <iostream>
int main(int, char **)
{
const int array[7] = {};
const int *pointer = array;
std::cout << "array has " << len(array) << " itemsn"
<< "and pointer is to " << len(pointer) << " charsn";
}
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
The length of an array can only be deduced if the array is passed by reference, otherwise it decays to pointer:
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
// ^^^
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
// ^^^
Full demo:
#include <cstdlib>
template<typename T, std::size_t length>
constexpr std::size_t len(const T(&)[length]) { return length; }
template<typename T>
constexpr std::size_t len(const T *&p) { return sizeof *p; }
#include <iostream>
int main(int, char **)
{
const int array[7] = {};
const int *pointer = array;
std::cout << "array has " << len(array) << " itemsn"
<< "and pointer is to " << len(pointer) << " charsn";
}
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
answered Nov 21 '18 at 9:43
Toby SpeightToby Speight
16.7k134165
16.7k134165
add a comment |
add a comment |
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2
static lengthis not valid C++. That particular C99 feature is not available.– StoryTeller
Nov 21 '18 at 6:30