Path: utzoo!attcan!uunet!world!burley
From: burley@world.std.com (James C Burley)
Newsgroups: comp.lang.c
Subject: Re: Is something wrong with the compiler ?
Message-ID: <BURLEY.90Sep27013312@world.std.com>
Date: 27 Sep 90 05:33:12 GMT
References: <884@gtenmc.UUCP> <1990Sep26.175948.8232@zoo.toronto.edu>
Sender: burley@world.std.com (James C Burley)
Organization: The World
Lines: 23
In-Reply-To: henry@zoo.toronto.edu's message of 26 Sep 90 17:59:48 GMT

In article <1990Sep26.175948.8232@zoo.toronto.edu> henry@zoo.toronto.edu (Henry Spencer) writes:

   In article <884@gtenmc.UUCP> csp@gtenmc.UUCP (Charudutta S Palkar) writes:
   >   ... I wrote the following code , and the results I got were absurd.
   >
   >      printf(" Maxint : %d\na = %d\n", ( int )(( unsigned ) ~0 >> 1 ) , 
   >	      a = ( int )(( unsigned ) ( a = ~0 ) >> 1 ));

   Sure you know what the results should be?  The second expression is
   assigning to `a' twice, which has unpredictable effects.  However, this
   doesn't explain what you're seeing.  It does look to me like a compiler
   bug.

Hmm, I thought it was ok -- doesn't "=" serve as a "sequence point" or
some such thing, and since the lhs of the outer = is so simple it can't
be "evaluated" before the rhs has been, and the rhs contains the other
=, one is guaranteed the latter = is performed before the former, right?

On the other hand, "b = a + (int)((unsigned)(a=~0)>>1));" would be undefined
as to whether the first mention of "a" identifies the value it holds prior
to execution of the statement or the value ~0.

James Craig Burley, Software Craftsperson    burley@world.std.com