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From: Igor Tandetnik on 19 Jan 2010 13:11
Ulrich Eckhardt <eckhardt(a)satorlaser.com> wrote:
> #include <stdio.h>
> int g = 0;
> void foo(int x)
> {
> printf("x=%d, g=%d\n", x, g);
> }
> int main()
> {
> foo(g++);
> }
> // output: x=0, g=1
>
> Note: I'm not 100% sure if this is a proof, it might actually invoke
> implementation-specific behaviour or even undefined behaviour.

Looks OK to me. There is a sequence point after all function parameters are evaluated, and before the function's body is entered. All side effects from parameter evaluation must compete by then.
--
With best wishes,
Igor Tandetnik

With sufficient thrust, pigs fly just fine. However, this is not necessarily a good idea. It is hard to be sure where they are going to land, and it could be dangerous sitting under them as they fly overhead. -- RFC 1925

From: Barry Schwarz on 30 Jan 2010 19:53
On Tue, 19 Jan 2010 00:32:58 -0500, "Chris Saunders"
<evas(a)mountaincable.net> wrote:

>
>Thanks for the reply Scott. Unfortunatly your answer was not clear to me.
>Perhaps my question wasn't clear enough. I'll try to be clearer. Is the ++
>performed before or after x is set? I don't think x = mt[mti++] is the same
>as x = mt[++mti] is it?

In your original example, mti++ is evaluated. The result is the
current unmodified value of mti. Then two additional things happen. A
value is stored in x and the incremented value is stored in mti. These
last two can happen in any order the compiler chooses.

In your second example, ++mti is evaluated. The result is the current
value of mti+1. Then two additional things happen. A (different)
value is stored in x and the incremented value is stored in mti. These
last two also can happen in any order the compiler chooses.

The pre- and post-fix increment operators do not tell you when the
operand is incremented. They only tell you whether the result of the
evaluation is the incremented value or not.

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