獲取指向模板化 lambda 運算符 () 的指針，無需捕獲

Question

誰能告訴我一種有效的方法來獲取指向模板化的 lamda（無頭）運算符 () 的指針？ 已經嘗試了兩種選擇：

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
}

鏗鏘聲（-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()的地址類型是成員 function 指針，但是，您的fn_t的定義只是一個空閑的 function 指針。 您應該將fn_t定義為：

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

然后以下應該工作：

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

或者，為什么不呢？

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

演示。

Answer 2

它是 lambda 在某些情況下（此處不存在）可以轉換為 function 指針，而不是其成員operator() 。 並且您不能將成員指針轉換為 function 指針。

一種解決方法是使用另一個 lambda。

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

演示