Hello,
> For you people this is probably old hat but none of my book sources
> explain the C / C++ idiosyncrasy of evaluating arguments of printf (and
> maybe other functions too, I don't know) from right-to-left. For
> instance, the following code:
>
> # include <stdio.h>
> void main ( void )
> {
> int i = 0 ;
> printf ( "%d %d\n", i ++, i ++ ) ;
> }
>
> prints out:
>
> 1 0
>
> instead of:
>
> 0 1
>
> as expected. I remember having heard of this years ago and got around to
> testing it today, but I don't get the logic behind this behaviour. Is
> there one? If yes, what is it?
Take a look here -> http://www.c-faq.com/expr/comma.html
Regards,
Mariusz
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