Re: moused(8): char signed-ness problem with gcc 3.1

> 
> I observed gcc 2.95.4 and gcc 3.1 interpret (or maybe optimize) the
> following code differently (CFLAGS=-O):
> 
> int main(void)
> {
>   unsigned char i = 127;
>   printf("%d\n", ((char)(i << 1)) / 2);
>   return 0;
> }
> 
> gcc 2.95.4 says it's -1, whereas gcc 3.1 says it's 127.  On FreeBSD
> char should be signed, so I suspect it's a (optimization) bug of gcc
> 3.1 which should be fixed.  Or we'll have to do a mass audit of the
> whole src tree to check and fix the similar expressions.

 Let's examine the what the "right" answer should be:

 First - in the expression (i << 1) - the unsigned char `i' will
 be promoted to a signed int through the correct integral promotion
 rules, then left-shifted 1 bit.  The result of that is an int.

 So - this becomes:

	((char)(254)) / 2 ;

 The expression:
	(char)(254) 
 is then also promoted to int when it participates
 in the division operation.  So, the value 254 is
 converted into a (signed) char, and then converted
 to an int.   Converting 254 to a signed character
 should result in an integer value of -2.

 Then, -2 / 2 becomes -1.

 If characters were unsigned by default, you do
 get the value 127...

 So - yes - it seems gcc 3.1 does have a problem...

	- Dave Rivers -

--
[email protected]                        Work: (919) 676-0847
Get your mainframe programming tools at http://www.dignus.com

To Unsubscribe: send mail to [email protected]
with "unsubscribe freebsd-audit" in the body of the message