[Bug c++/94061] New: defaulted member operator <=> defined as deleted if a base has protected member operator <=>

[okannen at gmail dot com]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94061

            Bug ID: 94061
           Summary: defaulted member operator <=> defined as deleted if a
                    base has protected member operator <=>
           Product: gcc
           Version: 10.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: okannen at gmail dot com
  Target Milestone: ---

Version: GCC 10.0.1 20200229

When a base class declares the three way comparison member operator as
protected, a defaulted three way comparison member operator in the derived is
defined as deleted, while it shall obviously not:

Code:

#include <compare>

struct A{
    protected:
    auto operator <=> (const A&) const = default;
    };

struct B
    : A
    {
    auto operator <=> (const B&) const = default;
    };

void f(B b){
    b <=> b; //error see bellow
}

Error message:
<source>: In function 'void f(B)':
<source>:15:11: error: use of deleted function 'constexpr auto
B::operator<=>(const B&) const'
   15 |     b <=> b;
      |           ^
<source>:11:10: note: 'constexpr auto B::operator<=>(const B&) const' is
implicitly deleted because the default definition would be ill-formed:
   11 |     auto operator <=> (const B&) const = default;
      |          ^~~~~~~~
<source>:11:10: error: 'auto A::operator<=>(const A&) const' is protected
within this context
<source>:5:10: note: declared protected here
    5 |     auto operator <=> (const A&) const = default;
      |          ^~~~~~~~
Compiler returned: 1