strange behavior of std::cout in c++

Question

#include <iostream>

int a(int &x) {
    x = -1;
    return x;
}

int main () {
    int x = 5;
    std::cout << a(x) << " " << x << std::endl;
}

Why output is "-1 5"?

PS: compiler is:

i686-apple-darwin11-llvm-g++-4.2 (GCC) 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2336.11.00)

PS: compiled without any optimization.

The standard term for these is function side effects – fkl Oct 07 '13 at 18:35 — fkl, Oct 07 '13 at 18:35

Daniel Frey · Accepted Answer · 2013-10-07T18:25:54.270

15

In this line:

std::cout << a(x) << " " << x << std::endl;

the order of evaluation of a(x) and x is not specified. It's unspecified behaviour what happens, in your case the compiler decided to evaluate x first, a(x) afterwards.

edited Oct 07 '13 at 18:25

answered Oct 07 '13 at 18:17

Daniel Frey

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It is *undefined behavior*? I think it is *unspecified behavior*. – Nawaz Oct 07 '13 at 18:21
@Nawaz Hm, could be unspecified behaviour (§1.3.15) I think. Edited. – Daniel Frey Oct 07 '13 at 18:25
This is strange in my opinion because function has worked already. And i hoped that value in the cell in a memory was changed. Maybe somebody can explain me how it can be translate in an asm code? – Sukhanov Niсkolay Oct 07 '13 at 18:26
1

@SukhanovNiсkolay It is changed, if you look at it *after* this line. The compiler turned this line into `auto tmp1 = x; auto tmp2 = a(x); std::cout << tmp2 << " " << tmp1 << std::endl;` (without the additional sequence points, of course) - which is legal although, as already mentioned, not guaranteed. – Daniel Frey Oct 07 '13 at 18:29

LihO · Answer 2 · 2013-10-07T18:32:07.917

3

The order, in which a(x) and x are being evaluated is unspecified^[1]. To make sure that x will not be modified within the same expression (and avoiding unspecified behavior by doing so), you could do:

int x = 5;
int y = a(x);
std::cout << y << " " << x << std::endl;

[1] "Except where noted, the order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is unspecified. " _{5 Expressions, §4}

edited Oct 07 '13 at 18:32

answered Oct 07 '13 at 18:21

LihO

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I think it is *unspecified*, not *undefined*. – Nawaz Oct 07 '13 at 18:24

Shafik Yaghmour · Answer 3 · 2013-10-07T21:47:55.687

The order of evaluation is unsequenced and hence it is unspecified behavior either a(x) or x could be evaluated first. With respect to the C++ draft standard section 1.9 Program execution paragraph 15 says(emphasis mine):

[...]Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [ Note: In an expression that is evaluated more than once during the execution of a program, unsequenced and indeterminately sequenced evaluations of its subexpressions need not be performed consistently in different evaluations. —end note ] [...]

and it may even differ between different evaluations and furthmore if we go back to paragraph 13 it says:

[...]If A is not sequenced before B and B is not sequenced before A, then A and B are unsequenced. [ Note: The execution of unsequenced evaluations can overlap. —end note ] Evaluations A and B are indeterminately sequenced when either A is sequenced before B or B is sequenced before A, but it is unspecified which.[...]

which explains that this is unspecified behavior.

strange behavior of std::cout in c++

3 Answers3