On Wed, Aug 21, 2002 at 11:55:27AM +0300, Alexei Khlebnikov wrote:
> > I think this program should not terminate at all because i will
> > always be one greater than oldi.
> > I think gcc3.0 has a problem with no optimization then but since
> > there is later version that works gcc 3.1.1, upgrade.
>
> With no optimization the program runs correctly by the rules of integers
> representation in memory. See the explanation below.
>
Changed my mind. After a posting from Linus on dri-devel and a discussion
about integer overflow (undefined) in C the following came out:
---
unsigned int i = 0;
unsigned int oldi = 0;
while ((int) ++i > (int) oldi) oldi = i;
return oldi;
---
Looks good... incrementing unsigned (defined overflow) and doing an
unsigned check (both operands).
When compiling the above code with -ON (N>0), everyone gets it wrong.
gcc-2.9 optimizes it away:
080483c0 <main>:
80483c0: 55 push %ebp
80483c1: 89 e5 mov %esp,%ebp
80483c3: eb fe jmp 80483c3 <main+0x3>
80483c5: 90 nop
...
gcc-3.0 and 3.1 optimize it away:
08048304 <main>:
8048304: 55 push %ebp
8048305: 89 e5 mov %esp,%ebp
8048307: 83 ec 08 sub $0x8,%esp
804830a: 83 e4 f0 and $0xfffffff0,%esp
804830d: 8d 76 00 lea 0x0(%esi),%esi
8048310: eb fe jmp 8048310 <main+0xc>
8048312: 90 nop
8048313: 90 nop
The following works, but the evaluated expression is no longer part
of the while.
unsigned int i = 0;
unsigned int oldi = 0;
while(1) if ( (int) ++i < (int) oldi ) { break; } else oldi=i;
return (int)oldi;
This possibly breaks the loop optimization?
Anyone care to jump in? :)
c.
