C++11 “overloaded lambda” with variadic template and variable capture

Question

I\'m investigating a C++11 idiom which might be called \"overloaded lambda\":

• http://cpptruths.blogspot.com/2014/05/fun-with-lambdas-c14-style-part-2.html

Answer 1

Overload resolution works only for functions that exist in a common scope. This means that the second implementation fails to find the second overload because you don't import function call operators from overload<Frest...> into overload<F0, Frest...>.

However, a non-capturing lambda type defines a conversion operator to a function pointer with the same signature as the lambda's function call operator. This conversion operator can be found by name lookup, and this is what gets invoked when you remove the capturing part.

The correct implementation, that works for both capturing and non-capturing lambdas, and that always calls operator() instead of a conversion operator, should look as follows:

template <class... Fs>
struct overload;

template <class F0, class... Frest>
struct overload<F0, Frest...> : F0, overload<Frest...>
{
    overload(F0 f0, Frest... rest) : F0(f0), overload<Frest...>(rest...) {}

    using F0::operator();
    using overload<Frest...>::operator();
};

template <class F0>
struct overload<F0> : F0
{
    overload(F0 f0) : F0(f0) {}

    using F0::operator();
};

template <class... Fs>
auto make_overload(Fs... fs)
{
    return overload<Fs...>(fs...);
}

DEMO

In c++17, with class template argument deduction and pack expansion of using declarations in place, the above implementation can be simplified to:

template <typename... Ts> 
struct overload : Ts... { using Ts::operator()...; };

template <typename... Ts>
overload(Ts...) -> overload<Ts...>;

DEMO 2