Question

EDIT: Thank you very much for your responses. I understand this properly now!

I am trying to learn more on C pointers. Tinkering around, I am questioning the difference between two actions I am using.

This code seems to work at first glance, but I am not sure of what’s the difference, and if any of these two approaches is wrong in some way.

I’d like to know what’s the difference between the two pieces of code, when I should I pass the adress, and when a pointer to an array?

Is any of the pieces wrong? If so, what would be the proper way?

having a simple struct grid pretty much like struct grid { int val; } (for demonstration purposes)

First piece of code. Passing address of the pointer to the array.

    void set (mygrid *grid, int foo){
        grid->bar = foo; //should this be '*grid->bar?' But this seems to work properly.
    }

    void main(){
        int i;
        int* array;
        int max = 24;

        array = malloc(sizeof(grid) * max);

        for(i = 0; i < max; i++){
            set(&array[i], 0);
        }
    }

Second piece of code. I am not entirely sure why this works, but the compiler doesn’t output any warning.
I am supposed to be passing the pointer to the start of the array like this?

    void set(mygrid *grid, int foo){
        int i; int max = 24; //so this example code compiles :P

        for(i = 0; i < max; i++){
            grid[i].bar = foo;
        }
    }

    void main(){
        int* array;
        int max = 24;

        array = malloc(sizeof(grid) * max);
        set(array, 0); //Why not &array?
    }

Answer 1

Passing an array decays into a pointer that points to the first member of the array, just like &array[0].