From: "News Reader" <nospam AT aol DOT com>
Newsgroups: comp.os.msdos.djgpp
Subject: Re: integer overflow
Date: Mon, 28 Jul 2003 16:44:36 +0200
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To: djgpp AT delorie DOT com
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According to my DJGPP documentation (C-library, HTML)
the format string "%U" instead of "%lu" should work,
but it doesn't.

Any clue why?



"Paul Cousoulis" <paulcsouls AT worldnet DOT att DOT net> wrote in message
news:3F24AA4B DOT 589D3482 AT worldnet DOT att DOT net...
> 1LU ? I guess the LU means Long Unsigned. Is that now part of standard
> C? I would of guessed the preprocessor would have handled the math first
> and the storage later.
>
> Thanks
> Paul
>
> Martin Ambuhl wrote:
> >
> > Paul Cousoulis wrote:
> > > Why does this code result in an integer overflow?
> > >
> > > unsigned long res;
> > >
> > > res = (1<<31)-1;
> >
> > Because (1 << 31) is a signed expression. Never use signed integers for
> > bit twiddling. Here's an example without such overflow.
> >
> > #include <stdio.h>
> >
> > int main(void)
> > {
> >      unsigned long res;
> >      res = 1LU << 31;
> >      printf("1LU << 31     : %lu\n", res);
> >      res = (1LU << 31)-1;
> >      printf("(1LU << 31)-1 : %lu\n", res);
> >      return 0;
> > }
> >
> > 1LU << 31     : 2147483648
> > (1LU << 31)-1 : 2147483647