我可以将C ++ 17不捕获lambda constexpr转换运算符的结果用作函数指针模板非类型参数吗?

问题描述

回答我如何写一个看起来像方法的lambda表达式?，我试图利用以下事实将无捕获的lambda转换为成员函数指针:自C ++ 17起，无捕获的lambda具有 constexpr 将转换运算符转换为其函数指针类型.

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.

所以我想到一个问题，直到:

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
}

现在，此代码以clang编译(自5.0.0版本开始)，但gcc(> = 7.2)抱怨:

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

问题是，谁是对的?

推荐答案

这是gcc错误，已提交 83258 .

This is a gcc bug, filed 83258.

在C ++ 14中，我们曾经有一个链接指针类型的非类型模板参数的必要条件.但是在C ++ 17中(由于 N4268 )，则该参数只需将转换即可正确类型的常量表达式，但有其他一些限制(此处没有任何限制).一旦构造了fp，就应该能够将其用作模板参数.

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.

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