好的,我重寫了我的答案。這給了什麼TartanLlama和我之前建議的不同方法。這符合您的複雜性要求,並且不使用函數指針,因此所有內容都可以嵌套。
編輯:許多感謝Yakk的指出了意見相當顯著優化(所需的編譯時模板遞歸深度)
基本上我讓使用模板類型/功能處理器的二叉樹,並手動執行二進制搜索。
可能使用mpl或boost :: fusion更乾淨地做到這一點,但是這個實現是自包含的。
它絕對符合你的要求,函數是可以嵌套的,運行時查找的元組數量是O(log n)。
以下是完整列表:
#include <cassert>
#include <cstdint>
#include <tuple>
#include <iostream>
using std::size_t;
// Basic typelist object
template<typename... TL>
struct TypeList{
static const int size = sizeof...(TL);
};
// Metafunction Concat: Concatenate two typelists
template<typename L, typename R>
struct Concat;
template<typename... TL, typename... TR>
struct Concat <TypeList<TL...>, TypeList<TR...>> {
typedef TypeList<TL..., TR...> type;
};
template<typename L, typename R>
using Concat_t = typename Concat<L,R>::type;
// Metafunction First: Get first type from a typelist
template<typename T>
struct First;
template<typename T, typename... TL>
struct First <TypeList<T, TL...>> {
typedef T type;
};
template<typename T>
using First_t = typename First<T>::type;
// Metafunction Split: Split a typelist at a particular index
template<int i, typename TL>
struct Split;
template<int k, typename... TL>
struct Split<k, TypeList<TL...>> {
private:
typedef Split<k/2, TypeList<TL...>> FirstSplit;
typedef Split<k-k/2, typename FirstSplit::R> SecondSplit;
public:
typedef Concat_t<typename FirstSplit::L, typename SecondSplit::L> L;
typedef typename SecondSplit::R R;
};
template<typename T, typename... TL>
struct Split<0, TypeList<T, TL...>> {
typedef TypeList<> L;
typedef TypeList<T, TL...> R;
};
template<typename T, typename... TL>
struct Split<1, TypeList<T, TL...>> {
typedef TypeList<T> L;
typedef TypeList<TL...> R;
};
template<int k>
struct Split<k, TypeList<>> {
typedef TypeList<> L;
typedef TypeList<> R;
};
// Metafunction Subdivide: Split a typelist into two roughly equal typelists
template<typename TL>
struct Subdivide : Split<TL::size/2, TL> {};
// Metafunction MakeTree: Make a tree from a typelist
template<typename T>
struct MakeTree;
/*
template<>
struct MakeTree<TypeList<>> {
typedef TypeList<> L;
typedef TypeList<> R;
static const int size = 0;
};*/
template<typename T>
struct MakeTree<TypeList<T>> {
typedef TypeList<> L;
typedef TypeList<T> R;
static const int size = R::size;
};
template<typename T1, typename T2, typename... TL>
struct MakeTree<TypeList<T1, T2, TL...>> {
private:
typedef TypeList<T1, T2, TL...> MyList;
typedef Subdivide<MyList> MySubdivide;
public:
typedef MakeTree<typename MySubdivide::L> L;
typedef MakeTree<typename MySubdivide::R> R;
static const int size = L::size + R::size;
};
// Typehandler: What our lists will be made of
template<typename T>
struct type_handler_helper {
typedef int result_type;
typedef T input_type;
typedef result_type (*func_ptr_type)(const input_type &);
};
template<typename T, typename type_handler_helper<T>::func_ptr_type me>
struct type_handler {
typedef type_handler_helper<T> base;
typedef typename base::func_ptr_type func_ptr_type;
typedef typename base::result_type result_type;
typedef typename base::input_type input_type;
static constexpr func_ptr_type my_func = me;
static result_type apply(const input_type & t) {
return me(t);
}
};
// Binary search implementation
template <typename T, bool b = (T::L::size != 0)>
struct apply_helper;
template <typename T>
struct apply_helper<T, false> {
template<typename V>
static int apply(const V & v, size_t index) {
assert(index == 0);
return First_t<typename T::R>::apply(v);
}
};
template <typename T>
struct apply_helper<T, true> {
template<typename V>
static int apply(const V & v, size_t index) {
if(index >= T::L::size) {
return apply_helper<typename T::R>::apply(v, index - T::L::size);
} else {
return apply_helper<typename T::L>::apply(v, index);
}
}
};
// Original functions
inline int fun2(int x) {
return x;
}
inline int fun2(double x) {
return 0;
}
inline int fun2(float x) {
return -1;
}
// Adapted functions
typedef std::tuple<int, double, float> tup;
inline int g0(const tup & t) { return fun2(std::get<0>(t)); }
inline int g1(const tup & t) { return fun2(std::get<1>(t)); }
inline int g2(const tup & t) { return fun2(std::get<2>(t)); }
// Registry
typedef TypeList<
type_handler<tup, &g0>,
type_handler<tup, &g1>,
type_handler<tup, &g2>
> registry;
typedef MakeTree<registry> jump_table;
int apply(const tup & t, size_t index) {
return apply_helper<jump_table>::apply(t, index);
}
// Demo
int main() {
{
tup t{5, 1.5, 15.5f};
std::cout << apply(t, 0) << std::endl;
std::cout << apply(t, 1) << std::endl;
std::cout << apply(t, 2) << std::endl;
}
{
tup t{10, 1.5, 15.5f};
std::cout << apply(t, 0) << std::endl;
std::cout << apply(t, 1) << std::endl;
std::cout << apply(t, 2) << std::endl;
}
{
tup t{15, 1.5, 15.5f};
std::cout << apply(t, 0) << std::endl;
std::cout << apply(t, 1) << std::endl;
std::cout << apply(t, 2) << std::endl;
}
{
tup t{20, 1.5, 15.5f};
std::cout << apply(t, 0) << std::endl;
std::cout << apply(t, 1) << std::endl;
std::cout << apply(t, 2) << std::endl;
}
}
住在Coliru: http://coliru.stacked-crooked.com/a/3cfbd4d9ebd3bb3a
二叉搜索樹實際上比O(log(i))好得多。它是O(log(k)),其中k是交換機中不同i的數量。 – Borealid
https://gist.github.com/lichray/dd803a8bb3461fc842e5(不是我的代碼,它是C++ Now 2015閃電講座)。 –
@巴里我不認爲有一個。 –