Question

#include <iostream>
using namespace std;

void printarray (int arg[], int length) {
    for (int n = 0; n < length; n++) {
        cout << arg[n] << " ";
        cout << "\n";
    }
}

int main ()
{
     int firstarray[] = {5, 10, 15};
     int secondarray[] = {2, 4, 6, 8, 10};
     printarray(firstarray, 3);
     printarray(secondarray, 5);

     return 0;
}

This code works, but I want to understand how is the array being passed.

When a call is made to the printarray function from the main function, the name of the array is being passed. The name of the array refers to the address of the first element of the array. How does this equate to int arg[]?

Answer 1

The syntaxes

int[]

and

int[X] // Where X is a compile-time positive integer

are exactly the same as

int*

when in a function parameter list (I left out the optional names).

Additionally, an array name decays to a pointer to the first element when passed to a function (and not passed by reference) so both int firstarray[3] and int secondarray[5] decay to int*s.

It also happens that both an array dereference and a pointer dereference with subscript syntax (subscript syntax is x[y]) yield an lvalue to the same element when you use the same index.

These three rules combine to make the code legal and work how you expect; it just passes pointers to the function, along with the length of the arrays which you cannot know after the arrays decay to pointers.