how to correctly find out the return type of a lambda

Question

basically how to make following code compile?

I know it failed because compiler was trying to evaluate something like ([](int &i){})(0) but how to s

Answer 1

Two ways. First:

using TResult = decltype(f(_e));

or second:

using TResult = typename std::result_of<Lambda&(TElement&)>::type;

Your code implicity says that the TElement is a temprary/rvalue. The & above makes them lvalues.

Answer 2

You might resolve it with:

template <typename Lambda, typename T>
struct lambda_return_type {
    private:
    template<typename U>
    static constexpr auto check(U*) -> decltype(std::declval<Lambda>()(std::declval<U>()));

    template<typename U>
    static constexpr auto check(...) -> decltype(std::declval<Lambda>()(std::declval<U&>()));

    public:
    typedef decltype(check<T>(nullptr)) type;
};

and

void bar(Lambda f) {
    typedef typename lambda_return_type<Lambda, TElement>::type TResult;
}

Answer 3

You may use following traits:

template <typename T>
struct return_type : return_type<decltype(&T::operator())>
{};
// For generic types, directly use the result of the signature of its 'operator()'

template <typename ClassType, typename ReturnType, typename... Args>
struct return_type<ReturnType(ClassType::*)(Args...) const>
{
    using type = ReturnType;
};