Re: c++/7536: C++ front end does not warn of promotion of negative ints to unsigneds

[Gabriel Dos Reis <gdr at integrable-solutions dot net>]

The following reply was made to PR c++/7536; it has been noted by GNATS.

From: Gabriel Dos Reis <gdr@integrable-solutions.net>
To: Wolfgang Bangerth <bangerth@ticam.utexas.edu>
Cc: gcc-bugs@gcc.gnu.org, <jss@ast.cam.ac.uk>, <gcc-gnats@gcc.gnu.org>
Subject: Re: c++/7536: C++ front end does not warn of promotion of negative ints to unsigneds
Date: 22 Nov 2002 04:24:31 +0100

 Wolfgang Bangerth <bangerth@ticam.utexas.edu> writes:
 
 | > |     double dostuff(int i)
 | > |     {
 | > |       const unsigned cx = 10;
 | > |       return i-cx;
 | > |     }
 | > 
 | > Here is how things happen (as per the standard):
 | > 
 | >   1) since cx is unsigned, "i" is converted to unsigned int
 | >      before the operation, which yeilds an unsigned value.  Note that,
 | >      the above may be written equivalently explicilty as 
 | >        (unsigned)i + (unsigned)-10
 | > 
 | >   2) the result is converted to double in the return.
 | > 
 | > That is, the destination type does not "hijack" into the
 | > sub-expressions.
 | 
 | That was clear. The question I was unsure about was whether the expression 
 | is converted to the type of the left, or of the right operand, or in the 
 | case of differing signedness, to a signed integer.
 
 No matter who is on the right or on the left, if one is int an and the
 other is unsigned then the int is converted to unsigned.  The relevant
 rule is dictated at 5/9
 
  [...]
   --- Otherwise, if either operand is unsigned, the other shall be
       converted to unsigned.
 
 | Given your explanation, should we expect a warning? (On this depends 
 | whether I re-classify the report as "change-request" or whether I close 
 | it).
 
 Well, I don't know what the C front-end does.  I would suggest we do
 whatever the does on this.  If it does nothing, I'm not sure the
 issue is frequent enought to warrant a warning.
 
 -- Gaby