Overload resolution for inherited functions

Question

I want to have a struct that takes an arbitrary number of lambdas and serve as a central calling point for all their call operators.

If the call operator is invoked with an argument list that doesn't match any of the lambdas given on construction, a default call operator should be called.

I thought the following code would accomplish exactly this. The call operator for every lambda is "lifted" into the Poc class via using .

template <typename ...Lambdas>
struct Poc : Lambdas...
{
    using Lambdas::operator() ...; // Lift the lambda operators into the class

    template <typename ...Ts>
    auto operator() (Ts...)
    {
        std::cout << "general call" << std::endl;
    }
};

// Deduction guide
template <typename ...Ts>
Poc(Ts... ts)->Poc<Ts...>;

int main()
{
    auto l_int = [](int) {std::cout << "int" << std::endl; };

    Poc poc{l_int};
    poc(1);//calls the default operator. why?

    return 0;
}

When I don't have the default call operator in the struct, everything works as expected (with valid argument lists). If I add it to the struct (as in the code above), the default operator gets called everytime, no matter which arguments I call it with.

In my understanding, the lambda-call-operators and the structs (default) call-operator are present in the same scope. Therefore, they should all be looked at for the overload resolution. Since the lamdba-operator is more specific than the generic default operator, it should be chosen.

Apparently that is not the case. Why is that?

I tested on Microsoft Visual C++ , Clang and GCC (all in their latest versions).

Edit:

• MSVC 15.8.9
• GCC 9.0.0 (via Wandbox )
• Clang 8.0.0 (via Wandbox )

Answer 1

It's simple when you spot it: your operator is not const -qualified, while the lambda's one is (unless you define the lambda as mutable ). Hence, it is a better match for your non-const instance of Poc .

Just add the missing const :

auto operator() (Ts...) const