stack quirk?

[brian_helterline]

>Hi everyone.  Here's one.  I hope it is not as simple as my last one,
>you remember( gets() ).  I'm running the following program, and I'm not 
>getting out what I think I should be( so what else is new!).  Any ideas?

>#include <stdio.h>
>main()
>{
>int i= 0;
>printf("\n%d %d\n", ++i, ++i);	/* output is: 2 1, I expected 1 2 */
>i= 0;
>printf("%d %d\n", ++i, i++);	/* output is: 2 0, I expected 1 1 */
>i= 0;
>printf("%d %d\n", i++, ++i);	/* output is: 1 1, I expected 0 2 */
>i= 0;
>printf("%d %d\n", i++, i++);	/* output is: 1 0, I expected 0 1 */
>}
>
>Please send e-mail:  ndimas at pikes.denver.colorado.edu
>
>
>				Thank You 
>				Nicholas Dimas
>----------

	The output you got should is correct.  The arguments to
	printf() [or any function] are pushed on to the stack
	right to left so they are in the correct order for
	printf() to pop them off when needed.  If you evaluate
	each expression above right-to-left, you will "expect"
	exactly what the output was.  Also note that each argument
	to printf() is a single expression, so once it is evaluated,
	the ++ operator can do the increment and then move on to the
	next expression.  As an example, printf( "%d %d\n", i++, ++i );

	arg1 = "%d %d\n"
	arg2 = i++;
	arg3 = ++i;

	if i=0 to start, and we evaluate right-to-left,
	arg3 = 1 and i = 1
	arg2 = 1 and _then i = 2
	arg1 = "%d %d\n";

	I am not certain as to what ANSI specifies about whether this
	behavior is correct or not.  I am just trying to describe
	what is happening.  It is usually a bad idea to use ++ operators
	on the same variable more than once in _ANY_ expression.

	Hope this helps,