Re: c/10226: unsigned short promotion with bitwise inversion

[Glen Nakamura <glen@imodulo.com>]

The following reply was made to PR c/10226; it has been noted by GNATS.

From: Glen Nakamura <glen@imodulo.com>
To: Michael Marks <Michael.Marks@internetmachines.com>
Cc: Falk Hueffner <falk.hueffner@student.uni-tuebingen.de>,
	gcc-gnats@gcc.gnu.org
Subject: Re: c/10226: unsigned short promotion with bitwise inversion
Date: Wed, 26 Mar 2003 12:13:33 -1000

 On Wed, Mar 26, 2003 at 01:50:29PM -0800, Michael Marks wrote:
 > Either is far clearer than the original.  
 > I like the second one better.
 
 On second thought, my suggestions won't work for the other comparisons:
 e.g. <= >= < > !=
 I'm starting to see why it's worded the way it is...
 
 > Just one other comment, are both promoted, and are both unsigned after the
 > promotion.
 > I thought they are promoted to int. As per you earlier...
 > 
 > The ~ operator is subject to integer promotion,
 > so with the implicit conversions the expression becomes:
 >   if ((int) B == ~((int) A))
 > which is indeed false in the example above.
 
 I think it is promoted to int, but that really doesn't matter since:
   if ((unsigned int) B == ~((unsigned int) A))
 would still be false.
 The problem is widening an unsigned type and then doing a ~ operation,
 forces the high bits to 1.  In contrast, the other unsigned operand is just
 widened so the high bits are guaranteed 0.  Comparison of these two operands
 can give unexpected results, which is what the warning is trying to illustrate.
 
 - glen