获取指向模板化 lambda 运算符 () 的指针，无需捕获

Question

谁能告诉我一种有效的方法来获取指向模板化的 lamda（无头）运算符 () 的指针？ 已经尝试了两种选择：

int main()
{
    static
    auto l = []<bool a, bool b>( unsigned c ) -> unsigned
    {
        return (unsigned)a + b + c;
    };
    using fn_t = unsigned (*)( unsigned );
    fn_t fnA = &l.operator ()<true, false>; // doesn't work
    fn_t fnB = &decltype(l)::operator ()<true, false>; // also doesn't work
}

铿锵声（-cl）12：

x.cpp(9,13): error: cannot create a non-constant pointer to member function
        fn_t fnA = &l.operator ()<true, false> ; // doesn't work
                   ^~~~~~~~~~~~~~~~~~~~~~~~~~~
x.cpp(10,14): error: address of overloaded function 'operator()' does not match required type
      'unsigned int (unsigned int)'
        fn_t fnB = &decltype(l)::operator ()<true, false> ; // also doesn't work
                    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
x.cpp(4,11): note: candidate function template has different qualifiers (expected unqualified but found 'const')
        auto l = []<bool a, bool b>(unsigned c) -> unsigned
                 ^

MSVC 2019 最新更新：

x.cpp(9): error C2276: '&': illegal operation on bound member function expression
x.cpp(10): error C2440: 'initializing': cannot convert from 'overloaded-function' to 'fn_t'
x.cpp(10): note: None of the functions with this name in scope match the target type

Answer 1

lambda 的operator()的地址类型是成员 function 指针，但是，您的fn_t的定义只是一个空闲的 function 指针。 您应该将fn_t定义为：

using fn_t = unsigned int (decltype(l)::*)(unsigned int) const;

然后以下应该工作：

fn_t fnA = &decltype(l)::operator ()<true, false>;

或者，为什么不呢？

auto fnB = &decltype(l)::operator ()<true, false>;

演示。

Answer 2

它是 lambda 在某些情况下（此处不存在）可以转换为 function 指针，而不是其成员operator() 。 并且您不能将成员指针转换为 function 指针。

一种解决方法是使用另一个 lambda。

fn_t fn = [](unsigned c) { return decltype(l){}.operator()<true, false>(c); };

演示