bool to size_t warning

[Jonathan Wakely]

On Tue, 27 Aug 2019 at 14:48, Jonny Grant <jg@jguk.org> 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