C++ Array passed by reference, but how to understand this?

Question

The arrays are passed by reference. Any changes made to the array within the function changeArray will be observed in the calling scope ( main function here).

However the codes below print 0 1 in the 1st cout , and print 2 in the 2nd "cout". What I don't understand is that why the first cout prints the original value of array[0]=1 instead of the changed value of array[0]=2 ?

Thanks a lot.

#include <iostream>

using namespace std;

int changeArray(int array[]) {
    array[0]=2*array[0];
    return 0;
}

int main() {
    int array[]={1,2,3,4};
    cout << changeArray(array) << " " << array[0] << endl;
    cout << array[0] << endl;
    return 0;
}

Answer 1

To make sure that the compiler doesn't reorder the execution:

cout << array[0] << endl;
changeArray(array);
cout << array[0] << endl;

This prints 1 and then 2.

The C++ compiler is allowed to optimize the code by reordering the execution of code within a single expression (eg cout << changeArray(array) << " " << array[0] << endl ). To avoid that, and to make sure changeArray gets called first, you need to split your expression to separate statements, eg by using the semicolon ( ; ). Everything before the semicolon gets executed before anything after the semicolon can start.