c++ function call within cout statement

Question

I'm learning c++, and recently run into a confusing problem, here's the code:

#include <iostream>

using namespace std;

class A {
    public:
        A() { a[0] = 1; a[1] = 0; }
        int a[2];
        int b(void) { int x=a[0];a[0]=a[1];a[1]=x; return x; }
};

int main(void) {
    A a;
    cout<<a.a[0]<<a.a[1]<<endl; //outputs 10
    a.b();
    cout<<a.a[0]<<a.a[1]<<endl; //outputs 01
    a.b();
    cout<<a.a[0]<<a.a[1]<<endl; //outputs 10

    cout << a.b() << //outputs 1
    endl<< a.a[0]<<a.a[1] << endl; //outputs 10???

    cout<<a.a[0]<<a.a[1]<<endl; //outputs 01???
    return 0;
}

The first two calls of b() behaves as expected, but when i call b() within the cout statement, it doesn't switch the two elements of the array right away, but later i check it, it's already switched.

Can you help me understand this behavior? Thank you.

Answer 1

std::cout << f() << g();

The order of evaluation of the two function calls is unspecified; the compiler can call g() then f() , or it can call f() then g() .

Same thing in your code; the compiler can squirrel away the value of aa[0] the call ab() , or it can call ab() then grab the value of aa[0] .

Answer 2

Function evaluation in an expression depends on compiler stream operations such as << and >> . The same problem that happens when you evaluate x = f1() + f2() . You don't know which will be evaluated first because it's compiler dependent.

It would be more safe to call these on separate lines to rule out the ambiguity.