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

Question

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.

try using typedef when you create your structure then use the type name in your sizeof. — SPlatten
– SPlatten, Commented Nov 21, 2018 at 7:37

StoryTeller - Unslander Monica · Accepted Answer · 2018-11-21 07:44:57Z

0

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, 2018 at 7:44

StoryTeller - Unslander Monica

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Comments

Sourav Ghosh · Accepted Answer · 2018-11-21 07:47:26Z

0

You need to change

  sizeof(struct test_struct)

to

   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, 2018 at 7:47

answered Nov 21, 2018 at 7:42

Sourav Ghosh

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3 Comments

Shiva Kumar Over a year ago

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

Sourav Ghosh Over a year ago

@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?

Shiva Kumar Over a year ago

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.

kiran Biradar · Accepted Answer · 2018-11-21 07:56:35Z

0

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, 2018 at 7:56

answered Nov 21, 2018 at 7:40

kiran Biradar

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3 Comments

Sourav Ghosh Over a year ago

And why flexible array is not suitable?

kiran Biradar Over a year ago

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

Gerhardh Over a year ago

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

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