Multiple statements in a single cout produce different results

Question

int x;
int fun1()
{
    x=x+10;
    return x;
}
int main()
{
    x=5;
    cout<<x;
    cout<<fun1();
}

this produces 5 and 15 while

cout<<x<<fun1();

this produces 15 and 15. Please explain. Thankyou

See this answer for the sequence of evaluation: http://stackoverflow.com/questions/10782863/what-is-the-correct-answer-for-cout-c-c — Gerriet, May 04 '17 at 06:34
Insert newlines in your output for clarity. Right now the output will be e.g. 515 without \n. — Brandin, May 04 '17 at 06:37
Maybe it will be clearer if you write out the actual [`operator <<`](http://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt2) invokes as they're defined. Try writing them as nested function calls. The selected answer for the question linked by Gerriet goes into detail in doing so. — WhozCraig, May 04 '17 at 06:45

v78 · Accepted Answer · 2017-05-04T06:43:21.917

0

In c++ reference, order of evaluation of arguments of std::cout is unspecified. It is not left-to-right, right-to-left, or anything else.

Please avoid this. Instead use separate calls.

edited May 04 '17 at 06:43

answered May 04 '17 at 06:34

v78

1

cout << a << b << c has a well-defined order. The problem is the function call, not the ordering per se. – Brandin May 04 '17 at 06:36
@Brandin, what about ``std::cout << c++ << c ;`` [link](http://stackoverflow.com/questions/10782863/what-is-the-correct-answer-for-cout-c-c) . Not function just call, but evaluations in general. – v78 May 04 '17 at 06:38
@BenjaminLindley, thanks. corrected the ambiguous sentence. – v78 May 04 '17 at 06:39

1 Answers1