Evaluation order

[Stefan Ram]

#include <iostream> /* ::std::cout */
#include <ostream> /* << */
#include <cstdio> /* ::std:utchar */
int main()
{ ::std::cout << ::std:utchar( 65 )<< ::std:utchar( 66 )<< '\n'; }

When using an ASCII execution character set, the above
can print

AB6566

or

BA6566

, but could it also be possible that it prints

A65B66

?

 

[Marcel Müller]

#include <iostream> /* ::std::cout */
#include <ostream> /* << */
#include <cstdio> /* ::std:utchar */
int main()
{ ::std::cout << ::std:utchar( 65 )<< ::std:utchar( 66 )<< '\n'; }

When using an ASCII execution character set, the above
can print

AB6566

or

BA6566

, but could it also be possible that it prints

A65B66

?

Your expression evaluates to

cout.operator<<(putchar(65)).operator<<(putchar(66)).operator<<('\n')

including this pointers it looks like

operator<<(operator<<(operator<<(cout, putchar(65)), putchar(66)), '\n')

defining an evaluation tree

a = putchar(65)
|
b = operator<<(cout, a) c = putchar(66)
| |
d = operator<<(b, c) --------------+
|
operator<<(d, '\n')

The calls to putchar(65) and putchar(66) have no defined sequence.
But all calls to operator<< are in order.

Unfortunately putchar does not output to cout. It uses stdout. Both,
cout and stdout are allowed to use their own buffers. So although the
function calls have a defined order the output of operator<< and putchar
is not synchronized in any way. You might even see
6566
AB

Since most iostream implementations simply wrap to the stdio API
(everything else could be too fast, I guess) it is not very likely that
the output from cout precedes the direct output over the stdio API.


Marcel

 

[Stefan Ram]

....
the
function calls have a defined order

I used to believe that the order of evaluation of the two
function calls ::std:utchar( 65 ) and ::std:utchar( 66 )
was implementation specified.

the output of operator<< and putchar
is not synchronized in any way

I understand your words to allow for the possibility that

A65B66

might be printed.