Can I use the result of a C++17 captureless lambda constexpr conversion operator as a function pointer template non-type argument?

Question

While answering How do I write a lambda expression that looks like a method?, I tried to turn a captureless lambda into a member function pointer by exploiting the fact that, since C++17, captureless lambdas have a constexpr conversion operator to their function pointer type.

So I came up with an issue boiling down to:

template<void(*)()> struct A{};

int main()
{
  A<static_cast<void(*)()>([]{})>{}; // 1

  constexpr auto fp = static_cast<void(*)()>([]{});
  A<fp>{}; // 2
}

Now, this compiles in clang (since 5.0.0) but gcc(>=7.2) complains:

error: lambda-expression in template-argument
   A<static_cast<void(*)()>([]{ /*whatever*/ })>{}; // 1
                            ^
error: 'main()::<lambda()>::_FUN' is not a valid template argument for type 'void (*)()' because 'static constexpr void main()::<lambda()>::_FUN()' has no linkage
   A<fp>{}; // 2

The question is, who's right?

A simpler example using constexpr local variables should exist, no? — Yakk - Adam Nevraumont, Dec 02 '17 at 11:07
@Yakk, done; this is not entirely equivalent to the original code though ... or not ? — Massimiliano Janes, Dec 02 '17 at 11:25
I cannot help wondering what you are trying to do exactly and why this is needed. — Germán Diago, Dec 02 '17 at 12:36
@germandiago this is a language lawyer question... why one would use it is irrilevant. Anyway, the ability of portably casting a lambda to, say, a member pointer may have use cases I think ... — Massimiliano Janes, Dec 02 '17 at 13:12
Pointer to function with no linkage should be allowed to be a non-type template argument in C++17. See [N4198](http://wg21.link/N4198). — cpplearner, Dec 02 '17 at 13:22
@cpplearner ah I see, so clang's right; may you write an answer ? — Massimiliano Janes, Dec 02 '17 at 14:04
@massim not quite the same, but 3 times simpler. And I can see the use; fastest possible delegates the portable version. — Yakk - Adam Nevraumont, Dec 02 '17 at 14:55
@Klaus, indeed the [gcc standards support page](https://gcc.gnu.org/projects/cxx-status.html) claims support for [n4268](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4268.html) (AFAIK formerly known as N4198), so I'll file a bug report (unless anybody else has something to say ) ... — Massimiliano Janes, Dec 02 '17 at 16:02

score 7 · Accepted Answer · answered Dec 02 '17 at 22:23

This is a gcc bug, filed 83258.

In C++14, we used to have a linkage requirement for non-type template parameters of pointer type. But in C++17 (as a result of N4268), the parameter just needs to be a converted constant expression of the correct type, with a few other restrictions (none of which are relevant here). Once we can construct fp, we should be able to use it as a template parameter.

Can I use the result of a C++17 captureless lambda constexpr conversion operator as a function pointer template non-type argument?

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