On Thu, Jan 26, 2012 at 12:53 PM, Konstantin Vladimirov <konstantin.vladimi...@gmail.com> wrote: > Hi, > > If I know what I am doing, and my code itself guarantees, that there > will be no overflows and UB here, can I switch off this signed char to > unsigned char expansion in favor of signed char to signed int > expansion?
No, you can't. Richard. > --- > With best regards, Konstantin > > On Thu, Jan 26, 2012 at 3:04 PM, Jakub Jelinek <ja...@redhat.com> wrote: >> On Thu, Jan 26, 2012 at 02:27:45PM +0400, Konstantin Vladimirov wrote: >>> Consider code: >>> >>> char A; >>> char B; >>> >>> char sum_A_B ( void ) >>> { >>> char sum = A + B; >>> >>> return sum; >>> } >>> [repro.c : 6:8] A.0 = A; >>> [repro.c : 6:8] A.1 = (unsigned char) A.0; >>> [repro.c : 6:8] B.2 = B; >>> [repro.c : 6:8] B.3 = (unsigned char) B.2; >>> [repro.c : 6:8] D.1990 = A.1 + B.3; >>> [repro.c : 6:8] sum = (char) D.1990; >>> [repro.c : 8:3] D.1991 = sum; >>> [repro.c : 8:3] return D.1991; >>> } >>> >>> It looks really weird. Why gcc promotes char to unsigned char internally? >> >> To avoid triggering undefined behavior. >> A + B in C for char A and B is (int) A + (int) B, so either we'd have to >> promote it to int and then demote, or we just cast it to unsigned and do the >> addition in 8-bit. If we don't do that, e.g. for >> A = 127 and B = 127 we'd trigger undefined behavior of signed addition. >> In unsigned char 127 + 127 is valid. >> >> Jakub