Re: Unexpected behaviour using unsigned char in comparison

["Falk S." <falk-o-mat at gmx dot de>]

Axel Freyn wrote:
> Hi Falk,
> On Sun, Jan 09, 2011 at 07:00:04PM +0100, Falk S. wrote:
>   
> > Hi,
> >
> > in the following program the return value of the test2uc() function is  
> > always false, however I would expect it to be true.
> > #include <iostream>
> > using namespace std;
> >
> > bool test2uc() {
> >        typedef unsigned char uchar;
> >        uchar mask = ~(uchar(0)) << 1;
> >        return (mask | uchar(true)) == (~ uchar(0));
> > }
> >
> > int main(int argc, char** args) {
> >        cout << test2uc() << endl;
> >        return 0;
> > }
> >
> > Compiled with -ansi -Wall -g -gdwarf-2 -pedantic
> > g++ (GCC) 3.4.4 (cygming special, gdc 0.12, using dmd 0.125)
> > g++.exe (GCC) 4.5.0 (MingW)
> > on Windows 7
> >
> > I get a "warning: comparison is always false due to limited range of  
> > data type" when compiling with g++ a c++ file, which I do not understand.
> > If I change the type of uchar to char, i.e. signed char, I get the  
> > expected result.
> >     
> The problem is that the operator "~" when applied upon the unsigned
> char returns always a signed char. uchar(0) are 8 bits of "0",
> "~uchar(0)" are 8 bits of "1" -- when interpreted as signed char, this
> gives the value "-1". So effectively you check for "255 == -1", which is
> false.
>   
Thanks for the explanation. Is this a gcc-specific behaviour?
> You can correct it by:
>  - comparing the bits and not the "full" values, e.g. by using
>    return  ( (mask | uchar(true)) xor uchar(~ uchar(0)) ) == 0; 
>  - convert explicitely the right side to an unsigned value
>    return (mask | uchar(true)) == uchar(~ uchar(0));
>
> However, I'm not 100% sure what the C++-standard says to the question
> (whether ~uchar(0) should be an signed or an unsigned integer...)
>   
Ok, that would be the "unexpected". I thought the unary operator~ needs 
no conversion (promotion) and therefore the return type is the same as 
the argument type.
bye,
Falk