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c++templatesvisual-c++lambdac++11
k06a
k06a

Reputation: 18745

C++ overloading by functor param count type

I am working on "LINQ to Objects" library for C++11. I would like to do smth like this:

// filtering elements by their value
arr.where( [](double d){ return d < 0; } )

// filtering elements by their value and position
arr.where( [](double d, int i){ return i%2==0; } )

I down want to write arr.where_i( ... ) - it's ugly. So i need function/method overloading by lambda-type...

This is my solution:

template<typename F>
auto my_magic_func(F f) -> decltype(f(1))
{
    return f(1);
}

template<typename F>
auto my_magic_func(F f, void * fake = NULL) -> decltype(f(2,3))
{
    return f(2,3);
}

int main()
{
    auto x1 = my_magic_func([](int a){ return a+100; });
    auto x2 = my_magic_func([](int a, int b){ return a*b; });
    // x1 == 1+100
    // x2 == 2*3
}

Is it SFINAE solution? What can you suggest me?

Upvotes: 3

Views: 473

Answers (2)

Luc Danton
Luc Danton

Reputation: 35449

You certainly want SFINAE in your solution. Generally speaking, the result would look something like:

template<
    typename Functor
    , typename std::enable_if<
        special_test<Functor>::value
        , int
    >::type = 0
>
return_type
my_magic_func(Functor f);

template<
    typename Functor
    , typename std::enable_if<
        !special_test<Functor>::value
        , int
    >::type = 0
>
return_type
my_magic_func(Functor f);

such that only one overload would be active at any one time -- all that remains now is carefully crafting that special_test to have the behaviour we want. This is a careful balancing act as you don't want the test to be too specific; otherwise we lose generality. Quite a shame when writing generic code. You haven't given too much information (e.g. are you strictly interested in support for lambdas? monomorphic functors? polymorphic functors?), but I will assume for now that we have access to a value_type alias which would correspond to double in your example.

As such, here's an example condition that will check that a given type is Callable (that's a Standard concept) with signature bool(value_type); i.e. that it's a predicate of sorts:

template<typename Functor, typename ValueType>
struct is_unary_predicate {
    typedef char (&accepted)[1];
    typedef char (&refused)[2];

    void consume(bool);

    template<
        typename X
        , typename Y
        , typename = decltype( consume(std::declval<X>()(std::declval<Y>())) )
    >
    accepted
    test(X&&, Y&&);

    refused test(...);

    static constexpr bool value =
        sizeof test(std::declval<Functor>(), std::declval<ValueType>())
        == sizeof(accepted);
};

Personally I have an is_callable<F, Signature> trait so that I would only need to write something like template<typename Functor, typename ValueType> using is_unary_predicate = is_callable<Functor, bool(ValueType)>; (and similarly I could have an is_binary_predicate alias instead of letting the second overload of my_magic_func be a catch-all). Perhaps you'd want to use a similar trait for future uses of SFINAE (although it may be somewhat painful to write without variadic templates).

Upvotes: 2

Kerrek SB
Kerrek SB

Reputation: 476970

Maybe something variadic:

#include <utility>

template <typename F, typename ...Args>
decltype(f(std::declval<Args>()...) my_magic_func(F f, Args &&... args)
{
    return f(std::forward<Args>(args)...);
}

Edit: You can also use typename std::result_of<F(Args...)>::type for the return type, which does the same thing.

Upvotes: 3

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