CODEWITHSUNDEEP
c++templates
David
David

Reputation: 28178

How can I find the number of arguments a template function takes?

I have the following type trait:

template <class T>
struct Arity : Arity<decltype(&T::operator())> {};

template <class T, class R, class... Args>
struct Arity<R(T::*)(Args...)> {
    static constexpr auto value = sizeof...(Args);
};

template <class T, class R, class... Args>
struct Arity<R(T::*)(Args...) const> {
    static constexpr auto value = sizeof...(Args);
};

template <class R, class... Args>
struct Arity<R(*)(Args...)>  {
    static constexpr auto value = sizeof...(Args);
};

Which works great to find the number of arguments a function takes for most use cases, but it fails for one common case:

auto l1 = [](int, double){};
Arity<decltype(l1)>::value; // works, 2

auto l2 = [](auto, auto){};
Arity<decltype(l2)>::value; // error: Reference to overloaded function could not be resolved; did you mean to call it?

I believe it's impossible to generally make this work for any templated function / operator() because depending on the types/values passed as template types, a different overload could be selected, or perhaps no overload may be available at all. Also, there's no way to know what valid types and values to pass as template arguments. But still, I want this to work for the common case of a lambda taking auto arguments. Is there any way to make this more robust and cover lambdas which take auto arguments?

Upvotes: 8

Views: 320

Answers (2)

Quentin
Quentin

Reputation: 63124

I guess I achieved half of a solution here. Only works up to a fixed number of parameters, but for most applications that shouldn't be an issue. Also, it's probably highly simplifiable but my brain is not into tricky SFINAE right now.

template <
    class, std::size_t N,
    class = std::make_index_sequence<N>,
    class = void_t<>
>
struct CanCall : std::false_type { };

template <class F, std::size_t N, std::size_t... Idx>
struct CanCall<
    F, N,
    std::index_sequence<Idx...>,
    void_t<decltype(std::declval<F>()((Idx, std::declval<Any const&&>())...))>
> : std::true_type { };

CanCall<F, N> will return whether F is callable with N parameters of arbitrary type. The Any helper type has templated implicit conversion operators that allows it to morph into any desired parameter type.

template <class F, std::size_t N = 0u, class = void>
struct Arity : Arity<F, N + 1u, void> { };

template <class F, std::size_t N>
struct Arity<F, N, std::enable_if_t<CanCall<F, N>::value>>
: std::integral_constant<std::size_t, N> { };

template <class F>
struct Arity<F, MAX_ARITY_PROBING, void>
: std::integral_constant<std::size_t, ARITY_VARIADIC> { };

Arity<F> just checks whether an F can be called with zero, one, two... parameters. First positive check wins. If we reach MAX_ARITY_PROBING parameters, Arity bails out and supposes that the function is either variadic, or is not a function at all.

See it live on Coliru

Upvotes: 4

R Sahu
R Sahu

Reputation: 206567

I don't think you can use lambda functions in your use case whose argument types are auto. The operator() functions of such lambda functions are most likely implemented using function templates.

Hence, decltype can't be used with:

auto l2 = [](auto, auto){};
Arity<decltype(l2)>::value;

See this answer to another SO question for more on the subject.

Upvotes: 2

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