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

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.

Hello,

Take a look here -> http://www.c-faq.com/expr/comma.html

Regards,

Mariusz

Function calls (for all functions, not just variadic functions such as printf) are typically implemented by evaluating arguments right-to-left, pushing each value onto the stack as it is calculated, then calling the function's start address with a CALL/JSR/BL/etc instruction (which saves the current PC/IP somewhere then jumps to the specified address).

The result is that the function's arguments are stored at increasing addresses when read left-to-right.

This allows for variadic functions, where the function knows where to find the first argument, but has to determine for itself how many arguments are present.

Note that this evaluation order isn't dictated by any standard; it's an implementation detail. If you use side-effects in the arguments to a function call such that the values depend upon the evaluation order, the result is implementation-dependent.

Glynn Clements <[email protected]>