这是丑陋的，而且很可能不是一个常量表达式进行排序（因为需要实例化深度的）最好的方式..但瞧，合并排序：
助手类型，可回收阵列型与constexpr元素访问：
#include <cstddef>
#include <iterator>
#include <type_traits>
template<class T, std::size_t N>
struct c_array
{
T arr[N];
constexpr T const& operator[](std::size_t p) const
{ return arr[p]; }
constexpr T const* begin() const
{ return arr+0; }
constexpr T const* end() const
{ return arr+N; }
};
template<class T>
struct c_array<T, 0> {};
append函数，该函数数组类型：
template<std::size_t... Is>
struct seq {};
template<std::size_t N, std::size_t... Is>
struct gen_seq : gen_seq<N-1, N-1, Is...> {};
template<std::size_t... Is>
struct gen_seq<0, Is...> : seq<Is...> {};
template<class T, std::size_t N, class U, std::size_t... Is>
constexpr c_array<T, N+1> append_impl(c_array<T, N> const& p, U const& e,
seq<Is...>)
{
return {{p[Is]..., e}};
}
template<class T, std::size_t N, class U>
constexpr c_array<T, N+1> append(c_array<T, N> const& p, U const& e)
{
return append_impl(p, e, gen_seq<N>{});
}
合并排序：
template<std::size_t Res, class T, class It, std::size_t Accum,
class = typename std::enable_if<Res!=Accum, void>::type >
constexpr c_array<T, Res> c_merge(It beg0, It end0, It beg1, It end1,
c_array<T, Accum> const& accum)
{
return
beg0 == end0 ? c_merge<Res>(beg0 , end0, beg1+1, end1, append(accum, *beg1)) :
beg1 == end1 ? c_merge<Res>(beg0+1, end0, beg1 , end1, append(accum, *beg0)) :
*beg0 < *beg1 ? c_merge<Res>(beg0+1, end0, beg1 , end1, append(accum, *beg0))
: c_merge<Res>(beg0 , end0, beg1+1, end1, append(accum, *beg1));
}
template<std::size_t Res, class T, class It, class... Dummies>
constexpr c_array<T, Res> c_merge(It beg0, It end0, It beg1, It end1,
c_array<T, Res> const& accum, Dummies...)
{
return accum;
}
template<class T, std::size_t L, std::size_t R>
constexpr c_array<T, L+R> c_merge(c_array<T, L> const& l,
c_array<T, R> const& r)
{
return c_merge<L+R>(l.begin(), l.end(), r.begin(), r.end(),
c_array<T, 0>{});
}
template<class T>
using rem_ref = typename std::remove_reference<T>::type;
template<std::size_t dist>
struct helper
{
template < class It >
static constexpr auto merge_sort(It beg, It end)
-> c_array<rem_ref<decltype(*beg)>, dist>
{
return c_merge(helper<dist/2>::merge_sort(beg, beg+dist/2),
helper<dist-dist/2>::merge_sort(beg+dist/2, end));
}
};
template<>
struct helper<0>
{
template < class It >
static constexpr auto merge_sort(It beg, It end)
-> c_array<rem_ref<decltype(*beg)>, 0>
{
return {};
}
};
template<>
struct helper<1>
{
template < class It >
static constexpr auto merge_sort(It beg, It end)
-> c_array<rem_ref<decltype(*beg)>, 1>
{
return {*beg};
}
};
template < std::size_t dist, class It >
constexpr auto merge_sort(It beg, It end)
-> c_array<rem_ref<decltype(*beg)>, dist>
{
return helper<dist>::merge_sort(beg, end);
}
助手使用示例：
template<class T, std::size_t N>
constexpr std::size_t array_size(T (&arr)[N]) { return N; }
template<class T, std::size_t N>
constexpr T* c_begin(T (&arr)[N]) { return arr; }
template<class T, std::size_t N>
constexpr T* c_end(T (&arr)[N]) { return arr+N; }
用例：
constexpr int unsorted[] = {5,7,3,4,1,8,2,9,0,6,10}; // odd number of elements
constexpr auto sorted = merge_sort<array_size(unsorted)>(c_begin(unsorted),
c_end(unsorted));
#include <iostream>
int main()
{
std::cout << "unsorted: ";
for(auto const& e : unsorted) std::cout << e << ", ";
std::cout << '\n';
std::cout << "sorted: ";
for(auto const& e : sorted) std::cout << e << ", ";
std::cout << '\n';
}
输出：
unsorted: 5, 7, 3, 4, 1, 8, 2, 9, 0, 6, 10,
sorted: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
你真的可以从东西是一个'constexpr'（如您的构造函数）调用非'constexpr'功能（如排序）？能够做到这一点真的没有意义。 – masaers
@masaers很明显'std :: sort'不是constexpr;难题是写一些行为像'std :: sort'但是*是* constexpr的东西。 – Quuxplusone
我明白了，对不起。编译时排序元函数将非常酷...... – masaers