c++ - 获取指向模板化 lambda 运算符 () 的指针，无需捕获

Question

谁能告诉我一种获取指向模板化 lamda(没有 caputes) operator () 的指针的有效方法吗？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
}

叮当(-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 的地址类型 operator()是成员函数指针，然而，你的fn_t的定义只是一个免费的函数指针。您应该定义您的 fn_t作为:

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

那么以下应该工作:

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

或者，为什么不呢？

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

Demo.