Get a pointer to a templated lambda operator () without captures

Question

Can anyone tell me a valid way to get a pointer to a templated lamda (without caputes) operator () ? Already tried two alternatives:

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
}

clang(-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 latest update:

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

The type of the address of lambda's operator() is a member function pointer, however, the definition of your fn_t is just a free function pointer. You should define your fn_t as:

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

Then the following should work:

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

Or, why not?

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

Demo.