C realloc segmentation fault

Question

I've got a very simple piece of C code that uses malloc and realloc, but it induces a seg fault if I change a value that was part of the first array.

#include <stdlib.h>

void increase(int** array) 
{
    int * new_array;
    new_array = realloc(*array, 10 * sizeof(int));
    if (new_array != NULL) {
        *array = new_array;
    }
    else {
        // Error in reallocation
    }
    int i = 3;
    *array[i] = 2; // Seg fault if i = 0, 1, 2, 3
}

main()
{
    int *array = malloc(4 * sizeof(int));
    increase(&array);
    free(array);
}

Is my understanding of pointers at fault? Can anyone explain what's happening and how I can use realloc properly?

Many thanks!

dohashi · Accepted Answer · 2014-07-30 13:35:38Z

6

You probably need:

(*array)[i] = 2;

The [] operator binds before the *, so your version was doing *(array[i]) which is, well, wrong.

answered Jul 30, 2014 at 13:35

dohashi

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4 Comments

user2765038 Over a year ago

Yup, that'll do it, thanks. Can you explain why this is needed?

Drew McGowen Over a year ago

Operator precedence - the array indexing is happening before the pointer dereference, but you want the other way around.

tohava Over a year ago

To give a bit more detail. The previous version without () is interpreted as *(array[i]).

MaxC Over a year ago

You saved my day... after I've already lost yesterday night. Thanks

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