print dynamic array in c language

Question

i try to get numbers from my user and this is my function,

my function get arr as pointer and set it to a new array and return the counter of the number that i get for printing.

but when i try to print the array i get an ERROR that ther is noting

int GetNumber(int *arr)
{
    int n,i=0;
    int *temp;
    temp = (int*)calloc(1,sizeof(int));
    assert(temp);
    scanf("%d",&n);
    while(n != -1)
    {
        i++;
        temp = (int*) realloc(temp,i*sizeof(int));
        assert(temp);
        temp[i-1] = n;
        scanf("%d",&n);
    }
    arr = temp;
    return i;
}

Answer 1

The problem is that you modify a local variable.

In C, all arguments are passed "by value", that means the value is copied into the scope of the function. This also happens with your pointer arr . If you modify arr in the function, this will never affect the caller.

The solution is to pass a pointer to what you want to modify, so your signature should look like:

int GetNumber(int **arr)

still, this pointer is passed by value, but it points to the other pointer you want to modify.

On a side note, don't cast void * in C. It's implicitly convertible to any pointer type.