Allocate memory for an array inside a structure

Question

I want to allocate memory to an array in a structure.

struct str
{
    int *num;

};

creat()
{
    str s = malloc(sizeof(str));
    s->num = (int*)malloc(5*sizeof(int));
}

But after execution of line s->num = (int*)malloc(5*sizeof(int));, if I check, sizeof(s->num) is still the same.

Am I missing something?

You probably meant `str *s` instead of `str s`. Please consider editing your question. — Maxime Chéramy, Nov 14 '13 at 16:55

score 1 · Answer 1 · answered Nov 14 '13 at 16:47

1

s->num is a pointer to integer and so its size will never change. You will have to keep track of the allocated block of memory that begins at s->num in some other way.

answered Nov 14 '13 at 16:47

Ivaylo Strandjev

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score 1 · Answer 2 · edited May 23 '17 at 12:05

1

sizeof(s->num) gives you the size of the pointer of type int* which won't change after your malloc. More precisely, sizeof is not a function but an operator and the value is given by your compiler. It's not a function that returns the size of the array allocated with malloc.

edited May 23 '17 at 12:05

Community

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answered Nov 14 '13 at 16:53

Maxime Chéramy

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score 0 · Answer 3 · edited May 23 '17 at 12:28

Its because s->num is not an array created dynamically but a pointer. Therefore it gives you the sizeof pointer.

To find the size of dynamically allocated array, a previous SO answer gives a clue as below,

As you're trying to allocate array of 5 integers, add 1 more integer to it and make it 6.
Hide the size in the first integer and return ptr + 1 as the pointer to the array.
Now you can get the size at *(ptr - 1), which is hidden first int.

Allocate memory for an array inside a structure

3 Answers3