what is happening exactly when i use pointers the following way?

Question

int * func()
{
    static int array[10];
    int i = 0;
    while(i<10){
        array[i] = i + 1;
        i++;
    }
    return array;
}

int main()
{
    int *pointerFunc;
    pointerFunc = func();

    for (int i = 0; i < 10; i++ ) {
      printf("*(pointerFunc[%d]) : %d\n", i, *(pointerFunc + i));
    }

    return 0;
}

I have a function which creates an array. In main, I want to print every element from that array. I used a pointer to get "access" to my array in the function. It does work the way it is, but I don't really understand why. To be precise, I wonder what is happening here (sure it prints the studd, but the pointer thing...): printf("*(pointerFunc[%d]) : %d\n", i, *(pointerFunc + i)); I don't quite understand this: *(pointerFunc + i) It works, but how and why? It does print every element of my array, but I dont't get it. What is the pointer "seeing"?

Answer 1

printf("*(pointerFunc[%d]) : %d\n", i, *(pointerFunc + i));

Here, you are using pointer arithmetic to access each element of the array.

The expression *(pointerFunc + i) adds i to the memory address stored in pointerFunc, and then dereferences the resulting memory address to access the value stored in that memory location. Since pointerFunc points to the first element of the array, *(pointerFunc + i) gives you the value stored in the i-th element of the array.

Alternatively, you could also use the array subscript operator [] to access the elements of the array using the pointer variable, like this:

printf("pointerFunc[%d] : %d\n", i, pointerFunc[i]);

Both expressions are equivalent and will give you the value stored in the i-th element of the array.