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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?
0
farzanj
Asked:
farzanj
5 Solutions
 
Kyle AbrahamsSenior .Net DeveloperCommented:
It boils down to the sequence order is not defined.

There's some great reading about this here:
stackoverflow.com/questions/1504752/c-output-evaluation-order-with-embedded-function-calls
0
 
farzanjAuthor Commented:
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;
}

Open in new window

0
 
Kyle AbrahamsSenior .Net DeveloperCommented:
0
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farzanjAuthor Commented:
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.
0
 
sarabandeCommented:
cout << f(a) << f(b) << endl;

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the statement evaluates to

operator<<(operator<<(operator<<(cout, f(a)), f(b)), endl);

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that means the arguments of the middle function call are operator<<(cout, f(a)) and f(b). VC compiler would pass the arguments from right to left regardless of the calling convention (see http://msdn.microsoft.com/en-us/library/vstudio/46t77ak2(v=vs.120).aspx) what also means they were evaluated in that order.

so, your observation that f(b) was evaluated before f(a) is true for vc compiler and probably true for most other compilers as well.

Sara
0
 
ambienceCommented:
The answer is what ged325 said:  "sequence order is not defined."

cout << f(a) << f(b) << endl;

is a single expression, and whereas the associativity of the operator is defined, the order of evaluation of functions, temporaries - in a single expression - is up to the compiler (implementation). Even if all compiler writer chose to evaluate in a specific order, it still remains undefined and some day a compiler may choose to do it differently and still not break any rules.

It is therefore bad practice to write code that relies on sequencing of a particular compiler.

Similar reasoning applies to this for example

func( f(a), f(b), f(c) )

There is no defined order in which f(X) would be evaluated.
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peprCommented:
In other words, the a.b() is used as a member function for returning a value but with side effects. This is a general problem in any language that allows that.

If you want to be sure that the things are called in the right order, call them as commands outside the cout << ... << end; and assign the results to temporary variables. Then do output the content of the variables. Or design the dedicated member function that has no side effects.
0
 
farzanjAuthor Commented:
Thank you guys and sorry for the delay.
0

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