On Tue, 27 Aug 2019 at 14:48, Jonny Grant <jg@xxxxxxxx> wrote: > > > > On 27/08/2019 13:42, Jonathan Wakely wrote: > > On Fri, 9 Aug 2019 at 21:32, Jonny Grant wrote: > >> > >> Hi > >> Looks like the bool can't be converted to size_t > > > > Of course it can. You're using options which enable some noisy > > warnings and then turn those warnings into errors. But your example > > doesn't try to conbert bool to size_t anyway, it performs arithmetic > > on bools, which produces an int, then tries to convert that to size_t. > > Obviously converting int to size_t changes from signed to unsigned, > > and that's why you get a -Wsign-conversion diagnostic. > > > > Questionable code (like performing arithmetic on bool values) will > > often trigger noisy warnings that have many false positives. There's a > > reason -Wsign-conversion isn't included in -Wall or -Wextra. > > > > It's easy to avoid the warning if you really need to perform arithmetic on bool: > > > > size_t i = size_t(a + b); > > > > or: > > > > size_t i = size_t(a) + b; > > I see. I had been expected the bool to be treated as an unsigned number, It *is* treated as an unsigned number. But C and C++ have a rule called integer promotion which affects this. Operands smaller than 'int' will be promoted to 'int', so adding two bools is equivalent to int(a) + int(b). Obviously that is a signed type now. See https://en.cppreference.com/w/cpp/language/implicit_conversion#Integral_promotion
