Why does realloc() and free() fail in my code?

Question

I have some problem with realloc():

int main(int argc, char* argv[])
{
    int* amis;
    int saisie, cpt = 1;

    while(saisie != -1) {
        printf("Entrer les notes -1 pour quitter :");
        scanf("%d", &saisie);
        if (cpt == 1) {
            amis = malloc(sizeof(int));
            if(amis == NULL) {
                printf("amis == NULL");
                exit(0);
            }
        }
        if(saisie != -1) {
           amis = realloc(amis, sizeof (int) + sizeof (amis));
           if(amis == NULL) {
                printf("amis == NULL       cpt= %d", cpt);
                exit(0);
            }
           amis[cpt] = saisie;
           printf("size = %d, saisie = %d, tab = %d \n", cpt * sizeof(int), saisie, amis[cpt]);
           cpt++;
        }
    }
    printf("%d",1==0);

    afficherTab(amis,cpt);
    printf("END\n");

    free(amis);
    return 0;
}

Why does realloc() cause an error when I use sizeof(int) * cpt instead of sizeof(amis) + sizeof(int)?

free(amis) also doesn't work in that case.

`sizeof(amis)` is `sizeof(int*)`. also `amis[cpt] = saisie;` out of range when use `sizeof(int)*cpt`. — BLUEPIXY, Sep 16 '14 at 11:52
A pointer to a block of memory doesn't know how big that block is. A pointer only knows how big a mem address is, so `sizeof` a pointer will always be the same (4, or 8 most of the time). You'll have to use a `size_t` variable to keep track of the size of `amis` — Elias Van Ootegem, Sep 16 '14 at 12:07

score 2 · Accepted Answer · edited May 23 '17 at 12:29

The biggest problem you have is that you seem to confuse a pointer with an array. If you use an array, then:

int foo[10];
printf("%zu\n", sizeof foo/ sizeof *foo);//sizeof foo/sizeof(int)

Will give you the length of the array, but a pointer is not an array. As I've explained here:

A pointer is not an array, so it doesn't need to know what the size of the array is. A pointer can point to a single value, so a pointer can exist without there even being an array. It doesn't even care where the memory it points to is situated (Read only, heap or stack... doesn't matter). A pointer doesn't have a length other than itself. A pointer just is...

So sizeof amis will always be the same value: the size of a memory address (4 on 32 bit, 8 on 64 bit). To address this, you're going to have to track the size of the allocated block yourself:

size_t amis_size = 0;//use this

scanf(" %d",&saisie);//note the space before %d
amis_size += saisie;
amis = realloc(amis, sizeof *amis * amis_size);

And so on.
Other things you should do is: initialize your variables:

int *amis = NULL,
    saisie = 0;

Fix the format of scanf, and check the value of saisie for negative values other than -1...

Last but not least: exit(0); means you're terminating the execution, with an exit status of 0. 0 means that the process terminated without error, whereas malloc or realloc failing is an error, use the stdlib macro exit( EXIT_FAILURE );, or exit with a non-zero value.

Calling free on a pointer, just before the main function returns is a rather pointless thing to do, but calling free is a good habit to get into, so you can leave it there either way.
However, try to get used to assigning NULL to any pointer you free:

free(amis);
amis = NULL;

Why does realloc() and free() fail in my code?

1 Answers1