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cout << operator associativity

All the sources that I have seen say that << operator in cout uses left to right associativity. But this it doesn't explain the following example.

#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;
	a.b();
	cout << a.b() << endl; 
		cout << a.a[1] << endl;
	return 0;
}

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Output is only explained if it is evaluated right to left.

So I made another example and used debugger to see the execution.

#include <iostream>

using namespace std;

int f(int& x)
{
	x++;
    x *= x;
    return x;
}

int main(void)
{
    int a = 2, b=3;
    cout << f(a) << f(b) << endl;
    return 0;
}

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I put a break point on function f and sure enough it called f(b) and then f(a).
Why?

ASKER CERTIFIED SOLUTION

Kyle Abrahams, PMP

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farzanj

ASKER

Sorry my example above incorrect. Try this:

#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;
	a.b();
	cout << a.b() << a.a[1] << endl;
	return 0;
}

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SOLUTION

Kyle Abrahams, PMP

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farzanj

ASKER

I have tried Visual Studio and g++ (GCC) on Linux as well and both have the same output. If the order is not defined, it is pretty consistent with the compilers.

SOLUTION