C problem, order of evaluation

[Karl Heuer]

In article <19539 at grebyn.com> ckp at grebyn.UUCP (Checkpoint Technologies) writes:
>In article <1990Apr8.185047.7385 at diku.dk> null at diku.dk (Niels Ull Jacobsen) writes:
>>"( stack[++sp]= exp , stack[sp -= N])", where exp contains N "stack[++sp]"'s.
>
>As long as that comma is really the comma *operator* and not the
>function argument separator, then the order of evaluation of your
>subexpressions *is* guaranteed....

I think the problem was with the multiple references to `sp' on the left side
of the comma.  Niels: am I correct in assuming that
	(stack[++sp] = stack[++sp] + stack[++sp], stack[sp -= 2])
is an example of the generated code?

My recommendation is to force your translator to generate code like
	(++sp, stack[sp] = stack[sp+1] + stack[sp+2])
instead.  That's the only way to be sure that the compiler knows what you're
talking about.

Karl W. Z. Heuer (karl at ima.ima.isc.com or harvard!ima!karl), The Walking Lint