operator<< and int8_t and uint8_t types....

[Jonathan Wakely]

On Mon, 23 Oct 2023 at 12:53, Jonathan Wakely <jwakely@redhat.com> wrote:
>
> On Mon, 23 Oct 2023 at 12:45, Jonathan Wakely <jwakely@redhat.com> wrote:
> >
> > On Mon, 23 Oct 2023 at 12:36, Théo Papadopoulo <papadopoulo@gmail.com> wrote:
> > >
> > >      Hi,
> > >
> > > I suspect that this will probably receive an answer "this is the
> > > intended/normal/expected  behavior",
> >
> >
> > Correct.
> >
> > > but from what I saw on the internet
> > > this is a kind of gray area in the standard
> >
> > Not really.
> >
> > > and I did not found any
> > > (even closed) report on bugzilla, so I thought of asking here.
> > >
> > > Currently, printing a int8_t of uint8_t number with operator<< basically
> > > prints it as a (u)char. I think this is very unfortunate and is somewhat
> > > deceptive to
> > > users which expect a "int" like behavior.
> >
> > But int8_t is not an int, it's a typedef for an unspecified 8-bit
> > integral type, typically signed char (or potentially, but unlikely,
> > char, if that happens to be signed). The behaviour when printing
> > signed char is 100% specfied, there is no wriggle room for doing
> > anything else.
> >
> > > #include <iostream>
> > > #include <cstdint>
> > >
> > > namespace std {
> > >
> > >      std::ostream& operator<<(std::ostream& os,const int8_t v) {
> > >          return os << static_cast<int>(v);
> > >      }
> > >
> > >      std::ostream& operator<<(std::ostream& os,const uint8_t v) {
> > >          return os << static_cast<int>(v);
> > >      }
> > > }
> > >
> > > int main() {
> > >      std::int8_t a  = 65;
> > >      std::uint8_t b = 66;
> > >      char         c = 'c';
> > >      std::cerr << a << ' ' << b << ' ' << c << std::endl;
> > > }
> > >
> > > Without the operator<< overloads: this code prints:
> > >
> > > A B c
> > >
> > > whereas I think a programmer would expect to see:
> > >
> > > 65 66 c
> > >
> > > which is what is obtained with the overloads.
> > >
> > > There is an old and extended discussion on stack overflow on the
> > > subject, in which from some comment it seems that
> > > it is the promotion to int vs the conversion to char that plays a role
> > > here (and that gcc like many over compiler favors the char
> > > conversion over the int promotion, but I'm not so sure about the
> > > validity of this discussion).
> >
> > The standard specifies exactly how integer promotions work. There is
> > nothing any compiler can do to favour one promotion or another, it's
> > dictated by the standard.
> >
> > But that's irrelevant here, there is no promotion. The standard
> > requires the following overloads which define how signed char and
> > unsigned char are printed:
> >
> > template<class traits>
> > basic_ostream<char, traits>& operator<<(basic_ostream<char, traits>&,
> > signed char);
> > template<class traits>
> > basic_ostream<char, traits>& operator<<(basic_ostream<char, traits>&,
> > unsigned char);
> >
> > The standard requires int8_t and uint8_t to be typedefs, not distinct
> > types. On most platforms, there are no integral types with exactly 8
> > bits except for char, signed char and unsigned char, and since C++20,
> > char8_t. So int8_t and uint8_t MUST be typedefs for one of those. How
> > those are printed with iostreams is clearly specified by the standard.
> >
> >
> > > Certainly from the user's perspective, one would expect to see and
> > > integer value and not a char... and it does not look very difficult to do.
> >
> > The standard already says how to print those types, adding a more
> > specialized overload that only works for std::ostream (and not
> > std::basic_ostream<char, acme::traits<char>>) would be a breaking
> > change, and very surprising to all the users who do actually know how
> > this works.
>
> P.S. the behaviour of printing signed char and unsigned char has been
> the same for 25 years, since the first C++ standard. Therefore the
> behaviour of printing int8_t and uint8_t has been the same for as long
> a those typedefs have existed (they were added to C99 in 1999, and
> then inherited by C++11 in 2011, but were already in common use before
> then). So this behaviour has been the same for a very long time.

P.P.S. The Single UNIX Specification specified those typedefs in
<inttypes.h> in 1997.

> Plenty of users expect exactly what libstdc++ (and other
> implementations) do, because that's what it has always done.
>
> There are many things which are surprising to some people when they
> first encounter them. That doesn't mean we should or can change them
> now. Even if this were to change, it would have to happen in the C++
> standard, not just in libstdc++. We implement the standard.