Re: Evaluation order

=?ISO-8859-1?Q?Marcel_M=FCller?= <>
Thu, 12 Sep 2013 11:17:09 +0200
On 12.09.2013 10:46, Stefan Ram wrote:

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

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




   , but could it also be possible that it prints



Your expression evaluates to


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

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.


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