GNU bug report logs - #8545
issues with recent doprnt-related changes

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Message #99 received at 8545 <at> debbugs.gnu.org (full text, mbox):

From: Richard Stallman <rms <at> gnu.org>
To: Paul Eggert <eggert <at> cs.ucla.edu>
Cc: lekktu <at> gmail.com, 8545 <at> debbugs.gnu.org
Subject: Re: bug#8545: issues with recent doprnt-related changes
Date: Sat, 30 Apr 2011 17:03:47 -0400

    >> If you assign i = INT_MAX + 1, the resulting behavior is undefined.
    >
    > The result is INT_MIN.  We don't try to support any theoretical machine
    > where this would not be so.

    Those machines used to be theoretical, but they're in common
    use now.

I assumed we were talking about type `int', but you did not explicitly
say so.  Touché -- but that just means we are talking at cross
purposes.  What I said about addition on type int is still valid.

     printf ("%d", INT_MAX+1);

will output INT_MIN.

      Practical C code can no longer assume that integers
    always wrap around when doing integer arithmetic.

I think that is the wrong interpretation of the facts.

       long
       foo (char *p, int i)
       {
	 return &p[i + 1] - &p[i];
       }
    On typical hosts where int is 32 bits, and long and char * are
    both 64 bits, most compilers optimize that "return" statement
    to "return 1;", even when I is INT_MAX and I + 1 therefore
    overflows.  These compilers are therefore rejecting the notion
    that INT_MAX + 1 must always equal INT_MIN.

i+1 is computed as an integer, but then it gets converted to a long.
What happens here seems to be an issue about type conversion combined
with addition -- not addition itself.

These compilers are taking a strange liberty.
Why isn't that a bug?

-- 
Dr Richard Stallman
President, Free Software Foundation
51 Franklin St
Boston MA 02110
USA
www.fsf.org, www.gnu.org
Skype: No way! That's nonfree (freedom-denying) software.
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