From: Jonathan Wakely <jwakely.gcc@gmail.com>
To: Jason Mancini <jayrusman@hotmail.com>
Cc: gcc-help@gcc.gnu.org
Subject: Re: infinite for-loop and related question
Date: Wed, 16 Feb 2011 20:41:00 -0000 [thread overview]
Message-ID: <AANLkTim=BeKq2wpiVs3DeouVqAc-QrgtnvSUSWSFRCZN@mail.gmail.com> (raw)
In-Reply-To: <SNT106-W3B382A5342E9FE1D26967ABD20@phx.gbl>
On 16 February 2011 20:09, Jason Mancini wrote:
>
> Hello,
> So as I recall, the following can be an infinite loop now with optimizations, right?
>
> for (int i(1); i!=0; ++i) { ... }
Right.
> What about:
>
> unsigned int x = 0xFFFFFFFFU;
> x = x+1;
> if (x) { ... can we get here because "positive x + 1 must still positive"? ... }
>
> If not, given the first, why not?
No. The C and C++ standards define that unsigned integers do not
overflow, they wrap, with well-defined behaviour.
They do not define what happens if a signed integer overflows, so your
first loop results in undefined behaviour, and so you cannot
reasonably expect any particular behaviour. The compiler can do
whatever it likes with your code.
Put another way:
There is no way for a correct C or C++ program to increment a signed
integer greater than zero such that the result is zero. Because a
correct C or C++ program does not contain integer overflows.
next prev parent reply other threads:[~2011-02-16 20:17 UTC|newest]
Thread overview: 12+ messages / expand[flat|nested] mbox.gz Atom feed top
2011-02-16 20:17 Jason Mancini
2011-02-16 20:41 ` Jonathan Wakely [this message]
2011-02-16 21:37 ` Jason Mancini
2011-02-16 21:49 ` Bob Plantz
2011-02-16 23:00 ` Jonathan Wakely
2011-02-16 23:22 ` Thomas Martitz
2011-02-17 6:31 ` Jonathan Wakely
2011-02-17 10:36 ` Axel Freyn
2011-02-17 12:23 ` Jonathan Wakely
2011-02-17 13:16 ` Axel Freyn
2011-02-17 14:08 ` Jonathan Wakely
2011-02-16 23:00 Bill McEnaney
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