Changing array inside function in C

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In c you can’t pass a variable by reference, the array variable that you assign inside the function contains initially the same address as the passed pointer, but it’s a copy of it so modifying it will not alter the passed pointer.

You need to pass the address of the pointer in order to be able to alter it, like this

// Change the pointer of the array
void change(int **array, int length)
{
    *array = malloc(length * sizeof(int));
    if (*array == NULL)
        return;
    for (int i = 0 ; i < length ; i++)
        (*array)[i] = 1;
}

Then in main() you cannot assign to an array, doing so through this kind of function is surely undefined behavior. The array defined in main() is allocated on the stack and you cannot assign anything to an array since they are non-writeable lvalues so you cannot make it point to a heap memory location obtained with malloc(), you need to pass a pointer like this

int *array;
change(&array, length);
free(array);

If you want the function to replace the previous array, it will have to free() the malloc()ed data (note that passing NULL to free() is well defined), so

// Change the pointer of the array
void change(int **array, int length)
{
    free(*array);

    *array = malloc(length * sizeof(int));
    if (*array == NULL)
        return;
    for (int i = 0 ; i < length ; i++)
        (*array)[i] = 1;
}

then in main()

int *array;
array = NULL;
change(&array, length);
change(&array, length);
change(&array, length);
change(&array, length);
free(array);

will do what you apparently want.

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