Strange cout output (code blocks codeblocks-16.01mingw-setup.exe)

Question

When I run this program I get strange output 1132. Can somebody explan me why is 1132?

//main.cpp
#include <iostream>
using namespace std;

int f1(int &a){return a++;}
int f2(int &a){return ++a;}

int main(){
    int x, y;
    int *px, *py;
    x = 1;
    y = 2;
    px = &x;
    py = &y;
    cout << f1(*px) << *px << f2(*py) << *py << "\n";

    return 0;
}

There is no guarantee that the expressions are evaluated left to right. — Shloim, Sep 08 '16 at 12:18
It is totally depand on compiler. In your case expression is evaluated from right to left . :) — Undefined Behaviour, Sep 08 '16 at 12:20
Here's a similar question: http://stackoverflow.com/questions/25669493/c-why-does-llvm-evaluate-printf-left-to-right-when-gcc-evaluates-right-to-left — Shloim, Sep 08 '16 at 12:23
@Shloim: Of course, thanks! I forgot that while this guarantees the order of the calls to `< — rainer, Sep 08 '16 at 12:25
Well, anyone who writes code like this gets what they deserve. This does not seem like a "practical programming problem" to me. — Cody Gray, Sep 08 '16 at 12:37

score 2 · Answer 1 · answered Sep 08 '16 at 12:20

2

The order of evaluation in your case was right-to-left.

Note that there is no guarantee for a left-to-right evaluation.

So order was:

*py
f2(*py)
*px
f1(*px)

Only then the << operators ran (and were evaluated left-to-right as expected)

answered Sep 08 '16 at 12:20

Shloim

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score 1 · Answer 2 · answered Sep 08 '16 at 12:25

In this line:

cout << f1(*px) << *px << f2(*py) << *py << "\n";

The compiler is free to evaluate each expression in any order (even though the order of execution of operator<< is left to right).

The expressions are:

f1(*px)
f2(*py)
*px
*py

The evaluation order may be any one of the npr(4,4) permutations.

npr(4,4) = 4! / (4-4)! = 24

Do you feel lucky?

Strange cout output (code blocks codeblocks-16.01mingw-setup.exe)

2 Answers2