why right-to-left evaluation in printf argument list?

[Shriramana Sharma <jamadagni@xxxxxxxxx>]

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?

Thanks.

Shriramana Sharma.

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