Changing array inside function in C

Question

I am learning C and confused why a array created in the main wont change inside the function, i am assuming the array passed is a pointer, and changing the pointer should\'v

Answer 1

Your array in main is an array. It will decay to a pointer, to produce the behavior you expect, but it is not a pointer.

int a[10];
int* p = a; // equivalent to &a[0]; create a pointer to the first element
a = p;      // illegal, a is NOT a pointer.

What your code is doing is copying the address of a into a function-local variable. Modifying it will have no more difference outside than changing length.

void change(int* local_ptr, size_t length)
{
    local_ptr = 0;
    length = 0;
}

int main()
{
    int a[10];
    int length = 10;
    printf("before a=%p, length=%d\n", a, length);
    change(a, length);  // we copied 'a' into 'local_ptr'. 
    printf("after a=%p, length=%d\n", a, length);
}

If you wish to modify a pointer from the caller, you will need to use pointer-to-pointer syntax:

void change(int** ptr, size_t length)
{
    // change the first element:
    **ptr = 0;
    // change the pointer:
    *ptr = 0;
    // but ptr itself is a function-local variable
    ptr = 0;  // local effect
}

However: There is a problem with what you are trying to do that goes deeper than this.

In your code, "int a" is an array on the stack, not an allocated pointer; you cannot free it and you should avoid mixing heap/stack pointers this way because eventually you'll free the wrong thing.