Structure specialization when the class has an operator()

Question

I would like to specialize a structure when the passed type has an operator() (a functor or a lambda function). For the moment, I have this code:

#include <iostream>
#include <type_traits>

template <class...>
struct mystruct: std::false_type {};

template <class C> 
struct mystruct<C, decltype(&C::operator())>: std::true_type {};

int main() {
    int n = 42;
    auto lambda = [&n](int i){return i + n;};

    // Should return false: OK
    std::cout<<mystruct<decltype(x)>::value<<std::endl;

    // Should return true: problem
    std::cout<<mystruct<decltype(lambda)>::value<<std::endl; 

    return 0;
}

How to make it work?

Bonus question: how to make it work even when the operator() is protected or private?

score 2 · Accepted Answer · edited May 23 '17 at 12:30

2

This would be an appropriate use of the void_t technique:

template <class, class = std::void_t<>>
struct mystruct: std::false_type {};

template <class C> 
struct mystruct<C, std::void_t<decltype(&C::operator())>>: std::true_type {};

Although void_t is only in the library since C++17, it is easy to add to your own utility library.

For the bonus question, see Detect existence of private member.

edited May 23 '17 at 12:30

Community

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answered Dec 09 '15 at 18:48

ecatmur

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Caveat: That breaks for overloaded or templated `operator()`'s. – T.C. Dec 10 '15 at 00:01

Structure specialization when the class has an operator()

1 Answers1