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
