[Bug c++/103563] New: ostream operator<< resolved to variant containing type resulting in stack overflow

[Bug c++/103563] New: ostream operator<< resolved to variant containing type resulting in stack overflow

[samuel.hangouet at gmail dot com]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103563

            Bug ID: 103563
           Summary: ostream operator<< resolved to variant containing type
                    resulting in stack overflow
           Product: gcc
           Version: 11.1.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: samuel.hangouet at gmail dot com
  Target Milestone: ---

Created attachment 51928
  --> https://gcc.gnu.org/bugzilla/attachment.cgi?id=51928&action=edit
Minimal test case.

$ g++-11 --version
g++-11 (Ubuntu 11.1.0-1ubuntu1~20.04) 11.1.0
(...)
$ g++-11 -std=c++2a test.cpp -o test
$ ./test
Segmentation fault (core dumped)

In my opinion, this file should not compile as no operator<< is provided for
custom type V. 

Instead, the operator<< taking the variant which contains V is called,
resulting in stack overflow at runtime.

[Bug c++/103563] ostream operator<< resolved to variant containing type resulting in stack overflow

[pinskia at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103563

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|---                         |INVALID

--- Comment #1 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
No because std::variant has the following constructor:
template< class T >
constexpr variant( T&& t ) noexcept(/* see below */);

So it is calling that.

This constructor is not marked as explicit either.

[Bug c++/103563] ostream operator<< resolved to variant containing type resulting in stack overflow

[pinskia at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103563

--- Comment #2 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
The following code is same example but removed the headers and such:
namespace t
{
   class g{}cout;
   template<class ...T>
   struct v
   {
     template <class T1> v(T1&&){}
   };
  class monostate {};
  g &operator <<(g &out, const char *a);
}
struct V
{
    int a;
};
inline t::g& operator << (t::g& os, const t::v<t::monostate, V>& a)
{
    V v;
    os << v;
    return os;
}

int main()
{
    V v={4};
    t::cout << v;
    return 0;
}

[Bug c++/103563] ostream operator<< resolved to variant containing type resulting in stack overflow

[m.cencora at gmail dot com]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103563

m.cencora at gmail dot com changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |m.cencora at gmail dot com

--- Comment #3 from m.cencora at gmail dot com ---
This is a known issue with operators in C++ in general.
That's why coding guidelines suggest that operators should be declared as
hidden friends (to prevent unexpected implicit conversions).

In your scenario you cannot make it a hidden friend, so constraint it with a
concept or
SFINAE:
https://godbolt.org/z/de869vW6h

[Bug c++/103563] ostream operator<< resolved to variant containing type resulting in stack overflow

[samuel.hangouet at gmail dot com]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=103563

--- Comment #4 from Samuel Hangouët <samuel.hangouet at gmail dot com> ---
Ok. Thanks for the tip and sorry for the wrong report.