How do I dynamically allocate memory to a pointer array inside a structure

here is what i have in done so far

struct test_case {

  int n;

  int *test;

};





struct test_case *test_case_struct = (struct test_case *)malloc(

      sizeof(struct test_struct) + 100 * sizeof(int));

I need to allocate n pointers in the "test" pointer array. As far as i know i need to allocate space to the structure and then some more for the pointer array, but when i try to compile this, i get the error
invalid use of sizeof operator for to incomplete type struct test_struct

if someone could please inform me how i can take the value of n as a user input and have int *test [n] made possible.

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

try using typedef when you create your structure then use the type name in your sizeof.

– SPlatten
Nov 21 '18 at 7:37

Side note, but consider not casting the result of malloc.

– StoryTeller
Nov 21 '18 at 7:40

add a comment |

here is what i have in done so far

struct test_case {

  int n;

  int *test;

};





struct test_case *test_case_struct = (struct test_case *)malloc(

      sizeof(struct test_struct) + 100 * sizeof(int));

if someone could please inform me how i can take the value of n as a user input and have int *test [n] made possible.

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

try using typedef when you create your structure then use the type name in your sizeof.

– SPlatten
Nov 21 '18 at 7:37

Side note, but consider not casting the result of malloc.

– StoryTeller
Nov 21 '18 at 7:40

add a comment |

here is what i have in done so far

struct test_case {

  int n;

  int *test;

};





struct test_case *test_case_struct = (struct test_case *)malloc(

      sizeof(struct test_struct) + 100 * sizeof(int));

if someone could please inform me how i can take the value of n as a user input and have int *test [n] made possible.

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

here is what i have in done so far

struct test_case {

  int n;

  int *test;

};





struct test_case *test_case_struct = (struct test_case *)malloc(

      sizeof(struct test_struct) + 100 * sizeof(int));

if someone could please inform me how i can take the value of n as a user input and have int *test [n] made possible.

c data-structures dynamic-memory-allocation

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

edited Nov 21 '18 at 7:36

Sourav Ghosh

110k14130188

asked Nov 21 '18 at 7:33

Shiva Kumar

asked Nov 21 '18 at 7:33

Shiva Kumar

asked Nov 21 '18 at 7:33

Shiva Kumar

try using typedef when you create your structure then use the type name in your sizeof.

– SPlatten
Nov 21 '18 at 7:37

Side note, but consider not casting the result of malloc.

– StoryTeller
Nov 21 '18 at 7:40

add a comment |

try using typedef when you create your structure then use the type name in your sizeof.

– SPlatten
Nov 21 '18 at 7:37

Side note, but consider not casting the result of malloc.

– StoryTeller
Nov 21 '18 at 7:40

try using typedef when you create your structure then use the type name in your sizeof.

– SPlatten
Nov 21 '18 at 7:37

Side note, but consider not casting the result of malloc.

– StoryTeller
Nov 21 '18 at 7:40

add a comment |

3 Answers
3

active

oldest

votes

Don't repeat type names. You already stumbled over your own code twice because you did that. You made the mistake of typing the wrong struct tag and confusing int* for int.

A more hardy allocation would look like this

struct test_case *test_case_struct =

  malloc(sizeof (*test_case_struct) + sizeof (test_case_struct->test[0]) * 100);

This here will allocate the size of whatever test_case_struct points at, plus 100 more of whatever test_case_struct->test[0] should be. Now you can play with the structure definition without breaking this call to malloc. And if you do perform a breaking change (like renaming test), you'll be notified by your compiler promptly.

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

add a comment |

You need to change

  sizeof(struct test_struct)

   sizeof(struct test_case)

as test_struct is not the correct structure type.

In a better way, you can also use the already-declared variable name, like

struct test_case *test_case_struct = malloc(

         sizeof (*test_case_struct) + n * sizeof(int*));

That said, you need to allocate memory worth of int *s, not ints, for the flexible member.

Also, below is a snippet which shows the count is taken as user input

int main(void)

{

    int n = 0;

    puts("Enter the count of pointers");

    if (scanf("%d", &n) != 1) {

        puts("Got a problem in the input");

        exit (-1);

    }

    struct test_case *test_case_struct = malloc( sizeof(struct test_case) + n * sizeof(int*));

    printf("Hello, world!n");

    return 0;

}

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

add a comment |

Currently you are using flexible array(aka zero length array).
Which can be allocated as below.

struct test_case *test_case_struct =

   malloc(sizeof (*test_case_struct) + 100 * sizeof (int *));

Note missing * for int and typo sizeof(struct test_struct) in your code.

Alternatively you can use pointer to pointer as below.

struct test_case {

  int n;

  int **test;

};





struct test_case *test_case_struct = malloc(

      sizeof(*test_case_struct));

test_case_struct->test = malloc(100 * sizeof(int *)); // Allocates 100 pointers

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53407212%2fhow-do-i-dynamically-allocate-memory-to-a-pointer-array-inside-a-structure%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Don't repeat type names. You already stumbled over your own code twice because you did that. You made the mistake of typing the wrong struct tag and confusing int* for int.

A more hardy allocation would look like this

struct test_case *test_case_struct =

  malloc(sizeof (*test_case_struct) + sizeof (test_case_struct->test[0]) * 100);

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

add a comment |

Don't repeat type names. You already stumbled over your own code twice because you did that. You made the mistake of typing the wrong struct tag and confusing int* for int.

A more hardy allocation would look like this

struct test_case *test_case_struct =

  malloc(sizeof (*test_case_struct) + sizeof (test_case_struct->test[0]) * 100);

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

add a comment |

Don't repeat type names. You already stumbled over your own code twice because you did that. You made the mistake of typing the wrong struct tag and confusing int* for int.

A more hardy allocation would look like this

struct test_case *test_case_struct =

  malloc(sizeof (*test_case_struct) + sizeof (test_case_struct->test[0]) * 100);

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

Don't repeat type names. You already stumbled over your own code twice because you did that. You made the mistake of typing the wrong struct tag and confusing int* for int.

A more hardy allocation would look like this

struct test_case *test_case_struct =

  malloc(sizeof (*test_case_struct) + sizeof (test_case_struct->test[0]) * 100);

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

answered Nov 21 '18 at 7:44

StoryTeller

98k12200267

add a comment |

You need to change

  sizeof(struct test_struct)

   sizeof(struct test_case)

as test_struct is not the correct structure type.

In a better way, you can also use the already-declared variable name, like

struct test_case *test_case_struct = malloc(

         sizeof (*test_case_struct) + n * sizeof(int*));

That said, you need to allocate memory worth of int *s, not ints, for the flexible member.

Also, below is a snippet which shows the count is taken as user input

int main(void)

{

    int n = 0;

    puts("Enter the count of pointers");

    if (scanf("%d", &n) != 1) {

        puts("Got a problem in the input");

        exit (-1);

    }

    struct test_case *test_case_struct = malloc( sizeof(struct test_case) + n * sizeof(int*));

    printf("Hello, world!n");

    return 0;

}

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

add a comment |

You need to change

  sizeof(struct test_struct)

   sizeof(struct test_case)

as test_struct is not the correct structure type.

In a better way, you can also use the already-declared variable name, like

struct test_case *test_case_struct = malloc(

         sizeof (*test_case_struct) + n * sizeof(int*));

That said, you need to allocate memory worth of int *s, not ints, for the flexible member.

Also, below is a snippet which shows the count is taken as user input

int main(void)

{

    int n = 0;

    puts("Enter the count of pointers");

    if (scanf("%d", &n) != 1) {

        puts("Got a problem in the input");

        exit (-1);

    }

    struct test_case *test_case_struct = malloc( sizeof(struct test_case) + n * sizeof(int*));

    printf("Hello, world!n");

    return 0;

}

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

add a comment |

You need to change

  sizeof(struct test_struct)

   sizeof(struct test_case)

as test_struct is not the correct structure type.

In a better way, you can also use the already-declared variable name, like

struct test_case *test_case_struct = malloc(

         sizeof (*test_case_struct) + n * sizeof(int*));

That said, you need to allocate memory worth of int *s, not ints, for the flexible member.

Also, below is a snippet which shows the count is taken as user input

int main(void)

{

    int n = 0;

    puts("Enter the count of pointers");

    if (scanf("%d", &n) != 1) {

        puts("Got a problem in the input");

        exit (-1);

    }

    struct test_case *test_case_struct = malloc( sizeof(struct test_case) + n * sizeof(int*));

    printf("Hello, world!n");

    return 0;

}

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

You need to change

  sizeof(struct test_struct)

   sizeof(struct test_case)

as test_struct is not the correct structure type.

In a better way, you can also use the already-declared variable name, like

struct test_case *test_case_struct = malloc(

         sizeof (*test_case_struct) + n * sizeof(int*));

That said, you need to allocate memory worth of int *s, not ints, for the flexible member.

Also, below is a snippet which shows the count is taken as user input

int main(void)

{

    int n = 0;

    puts("Enter the count of pointers");

    if (scanf("%d", &n) != 1) {

        puts("Got a problem in the input");

        exit (-1);

    }

    struct test_case *test_case_struct = malloc( sizeof(struct test_case) + n * sizeof(int*));

    printf("Hello, world!n");

    return 0;

}

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

edited Nov 21 '18 at 7:47

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

answered Nov 21 '18 at 7:42

Sourav Ghosh

110k14130188

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

add a comment |

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

my intention is to have a pointer array that can point to n elements. where n is a user input. both n and pointer array being inside the structure. sorry for the typo in the snippet there

– Shiva Kumar
Nov 21 '18 at 9:07

@ShivaKumar So, you are saying, you don't need an array of pointers to ints, rather an array of ints, the count of the ints would be stored in n, right?

– Sourav Ghosh
Nov 21 '18 at 9:19

nope. if i were to have a pointer array int *test[n], n would be a user input (this alone is not challenging for me, i gave done it before), the complexity is where i need to put the pointer array int *test[ ] inside a structure, and also its size int n inside a structure. the use case i am needing this for is i have the user inputting different square matrices of different dimensions (int n). each pointer in the test[0], test[1] ... test[n] is pointing to an int array which inturn have n elements in it. i hope i have made the picture clear enough.

– Shiva Kumar
Nov 22 '18 at 5:23

add a comment |

Currently you are using flexible array(aka zero length array).
Which can be allocated as below.

struct test_case *test_case_struct =

   malloc(sizeof (*test_case_struct) + 100 * sizeof (int *));

Note missing * for int and typo sizeof(struct test_struct) in your code.

Alternatively you can use pointer to pointer as below.

struct test_case {

  int n;

  int **test;

};





struct test_case *test_case_struct = malloc(

      sizeof(*test_case_struct));

test_case_struct->test = malloc(100 * sizeof(int *)); // Allocates 100 pointers

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

add a comment |

Currently you are using flexible array(aka zero length array).
Which can be allocated as below.

struct test_case *test_case_struct =

   malloc(sizeof (*test_case_struct) + 100 * sizeof (int *));

Note missing * for int and typo sizeof(struct test_struct) in your code.

Alternatively you can use pointer to pointer as below.

struct test_case {

  int n;

  int **test;

};





struct test_case *test_case_struct = malloc(

      sizeof(*test_case_struct));

test_case_struct->test = malloc(100 * sizeof(int *)); // Allocates 100 pointers

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

add a comment |

Currently you are using flexible array(aka zero length array).
Which can be allocated as below.

struct test_case *test_case_struct =

   malloc(sizeof (*test_case_struct) + 100 * sizeof (int *));

Note missing * for int and typo sizeof(struct test_struct) in your code.

Alternatively you can use pointer to pointer as below.

struct test_case {

  int n;

  int **test;

};





struct test_case *test_case_struct = malloc(

      sizeof(*test_case_struct));

test_case_struct->test = malloc(100 * sizeof(int *)); // Allocates 100 pointers

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

Currently you are using flexible array(aka zero length array).
Which can be allocated as below.

struct test_case *test_case_struct =

   malloc(sizeof (*test_case_struct) + 100 * sizeof (int *));

Note missing * for int and typo sizeof(struct test_struct) in your code.

Alternatively you can use pointer to pointer as below.

struct test_case {

  int n;

  int **test;

};





struct test_case *test_case_struct = malloc(

      sizeof(*test_case_struct));

test_case_struct->test = malloc(100 * sizeof(int *)); // Allocates 100 pointers

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

edited Nov 21 '18 at 7:56

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

answered Nov 21 '18 at 7:40

kiran Biradar

5,2582926

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

add a comment |

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

And why flexible array is not suitable?

– Sourav Ghosh
Nov 21 '18 at 7:45

@SouravGhosh I just ass*u*me*d OP might not intended to use flexible array and accidentally declared structure like that.

– kiran Biradar
Nov 21 '18 at 7:52

@kiranBiradar The title suggests that an array of pointers in a struct is desired.

– Gerhardh
Nov 21 '18 at 8:54

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu