Lambda函數，在編譯時確定參數的數量

Question

我想聲明一個具有N個參數的lambda函數，其中N是模板參數。 就像是...

template <int N>
class A {
    std::function<void (double, ..., double)> func;
                        // exactly n inputs
};

我想不出用元函數范式來做這個的方法。

Answer 1

您可以使用嵌套的typedef type編寫模板n_ary_function 。 此類型可以使用如下：

template <int N>
class A {
    typename n_ary_function<N, double>::type func;
};

以下代碼片段包含n_ary_function的定義：

template <std::size_t N, typename Type, typename ...Types>
struct n_ary_function
{
    using type = typename n_ary_function<N - 1, Type, Type, Types...>::type;
};

template <typename Type, typename ...Types>
struct n_ary_function<0, Type, Types...>
{
    using type = std::function<void(Types...)>;
};

Answer 2

獲取模板，計數和類型的元template ，並使用N個類型的副本調用模板：

template<template<class...>class target, unsigned N, class T, class... Ts>
struct repeat_type_N: repeat_type_N<target, N-1, T, T, Ts...> {};
template<template<class...>class target, class T, class... Ts>
struct repeat_type_N<target, 0, T, Ts...> {
  typedef target<Ts...> type;
};
template<template<class...>class target, unsigned N, class T>
using repeat_type_N_times = typename repeat_type_N<target, N, T>::type;

現在，我們使用它：

template<typename... Ts> using operation=void(Ts...);
template<unsigned N, class T> using N_ary_op = repeat_type_N_times< operation, N, T >;
template<unsigned N> using N_double_func = N_ary_op<N,double>;

我們測試它：

void three_doubles(double, double, double) {}

int main() {
  N_double_func<3>* ptr = three_doubles;
  std::function< N_double_func<3> > f = three_doubles;
}

贏了。

在上述系統中double, double, double您使用double, double, double究竟是什么完全取決於您。 例如，您可以使用初始化std::function的lambda。

你可以將double, double, double打包成template<class...>struct type_list{}; 所以你可以將它作為一個參數傳遞給另一個template ，然后專門解壓縮它。

對於大N遞歸較少的repeat_type ：

// package for types.  The typedef saves characters later, and is a common pattern in my packages:
template<class...>struct types{typedef types type;};

// Takes a target and a `types`, and applies it.  Note that the base has no implementation
// which leads to errors if you pass a non-`types<>` as the second argument:
template<template<class...>class target, class types> struct apply_types;
template<template<class...>class target, class... Ts>
struct apply_types<target, types<Ts...>>{
  typedef target<Ts...> type;
};
// alias boilerplate:
template<template<class...>class target, class types>
using apply_types_t=typename apply_types<target,types>::type;

// divide and conquer, recursively:
template<unsigned N, class T, class Types=types<>> struct make_types:make_types<
  (N+1)/2, T, typename make_types<N/2, T, Types>::type
> {};

// terminate recursion at 0 and 1:
template<class T, class... Types> struct make_types<1, T, types<Types...>>:types<T,Types...> {};
template<class T, class Types> struct make_types<0, T, Types>:Types{};

// alias boilerplate:
template<unsigned N, class T>
using make_types_t=typename make_types<N,T>::type;

// all of the above reduces `repeat_type_N_t` to a one-liner:    
template<template<class...>class target, unsigned N, class T>
using repeat_type_N_times = apply_types_t<target, make_types_t<N,T>>;

對於大N ，上面可以顯着減少編譯時間，並處理溢出template堆棧。

Answer 3

你不能直接這樣做。

你可以做這樣的事情

template <unsigned N> class UniformTuple;

template <>
class UniformTuple <0>
{
};

template <unsigned N>
class UniformTuple : public UniformTuple <N-1>
{
public:

    template <typename... Args>
    UniformTuple (double arg, Args... args)
    : UniformTuple <N-1> (args...)
    , m_value (arg)
    {
    }

private:

    double m_value;
};

template <int N>
class A
{
    std :: function <void (const UniformTuple <N> &)> func;
};

Answer 4

為了完整性，這里是一個沒有遞歸的解決方案：

template <class Ret, class Arg, class Idx>
struct n_ary_function_;

template <class Ret, class Arg, std::size_t... Idx>
struct n_ary_function_<Ret, Arg, std::index_sequence<Idx...>> {
    template <class T, std::size_t>
    using id = T;

    using type = std::function<Ret(id<Arg, Idx>...)>;
};

template <class Ret, class Arg, std::size_t N>
using n_ary_function = typename n_ary_function_<
    Ret, Arg, std::make_index_sequence<N>
>::type;

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