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?
---
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
