我想找到std::integer_sequence
中第一次出现值的位置。
-
以下是我的尝试。它有效,但我发现它非常优雅;当值不存在时(代码由于编译而被注释掉),它也无法产生干净的错误("未找到值")。 (另外,必须在Find_in_integer_sequence
中指定整数类型感觉多余,但我认为没有办法绕过它。)
代码只是为了您的娱乐,它不应该是建议解决方案的起点。
# include <iostream>
# include <utility>
# include <type_traits>
namespace detail
{
template < int Idx, typename T, T Match, T ...Values >
struct Find;
template < int Idx, bool B, typename T, T Match, T ...Values >
struct Find_impl;
template < int Idx, typename T, T Match, T ...Values >
struct Find_impl<Idx, true, T, Match, Values...>
{
static const int value = Idx;
};
template < int Idx, typename T, T Match, T Value, T ...Other_values >
struct Find_impl<Idx, false, T, Match, Value, Other_values...>
: public Find<(Idx + 1), T, Match, Other_values...>
{
};
template < int Idx, typename T, T Match, T Value, T ...Other_values >
struct Find<Idx, T, Match, Value, Other_values...>
: public Find_impl<Idx, (Match == Value), T, Match, Value, Other_values...>
{
};
//template < int Idx, typename T, T Match >
//struct Find<Idx, T, Match>
//{
// static_assert(false, "value not found");
//};
}
template < typename T, T Match, T ...Values >
struct Find
: public detail::Find<0, T, Match, Values...>
{
};
template < typename T, T Match, typename TIS >
struct Find_in_integer_sequence;
template < typename T, T Match, T ...Values>
struct Find_in_integer_sequence<T, Match, std::integer_sequence<T, Values...>>
: public Find<T, Match, Values...>
{
};
int main()
{
using i1 = std::integer_sequence<int, 2, 3, 3, 2, 3, 2, 0>;
auto k = Find_in_integer_sequence<int, 0, i1>::value;
std::cout << k << std::endl; # prints "6"
return 0;
}
答案 0 :(得分:3)
C ++ 14 constexpr
很棒:
template< class U, class T, T...ts >
constexpr std::size_t find( U t, std::integer_sequence<T, ts...> s )
{
T s_arr[] = {ts...};
for (std::size_t i = 0; i != sizeof...(ts); ++i) {
if (s_arr[i] == t) return i;
}
return sizeof...(ts);
}
只是搜索。线性。
这是一个更为复杂的解决方案,涉及构建具有对数深度的二叉树。它基于你需要做的事情来做类型,这些类型不能存储在constexpr
函数中的平面数组中。这个版本可以修改为在C ++ 11中有很多工作:
template<std::size_t N>using index=std::integral_constant<std::size_t,N>;
template<class T, T...t0s, T...t1s>
constexpr std::integer_sequence<T, t0s..., t1s... >
join( std::integer_sequence<T,t0s...>, std::integer_sequence<T,t1s...>){
return {};
}
template<class T, T...ts>
constexpr auto split( index<0>, std::integer_sequence<T,ts...> s ){
return std::make_pair( std::integer_sequence<T>{}, s );
}
template<class T, T t0, T...ts>
constexpr auto split( index<1>, std::integer_sequence<T, t0, ts...> s ){
return std::make_pair( std::integer_sequence<T, t0>{}, std::integer_sequence<T,ts...>{} );
}
template<std::size_t N, class T, T...ts>
constexpr auto split( index<N>, std::integer_sequence<T, ts...> s ){
auto a = split(index<N/2>{}, s );
auto b = split(index<(N+1)/2>{}, a.second );
return std::make_pair( join(a.first, b.first), b.second );
}
template< class T, T t0, T t1 >
constexpr index<1> find( std::integral_constant<T,t0>, std::integer_sequence<T, t1> )
{ return {}; }
template< class T, T t0, T...ts >
constexpr index<0> find( std::integral_constant<T,t0>, std::integer_sequence<T, t0, ts...> )
{ return {}; }
template< class T, T t0, T...ts >
constexpr index<1> find( std::integral_constant<T,t0>, std::integer_sequence<T> )
{ return {}; }
template< class T, T t0, T...ts >
constexpr auto find( std::integral_constant<T,t0> t, std::integer_sequence<T, ts...> s )
{
index<sizeof...(ts)/2> half;
auto halves = split( half, s );
auto front = find( t, halves.first );
auto back = find( t, halves.second );
return index<(front < half)?front:(back+half)>{};
}
这解决了对数递归深度的问题,允许搜索百万条目的整数列表。
答案 1 :(得分:1)
另一种方法是使用constexpr
函数:
#include <iostream>
#include <utility>
#include <type_traits>
template<class T>
constexpr int find_(std::integer_sequence<T>, T, int Idx) {
return Idx;
}
template<class T, T Head, T... Tail>
constexpr int find_(std::integer_sequence<T, Head, Tail...>, T Value, int Idx) {
return Value == Head ? Idx : find_(std::integer_sequence<T, Tail...>{}, Value, Idx + 1);
}
template<class S, class T>
constexpr auto findT(S s, T Value) {
return find_(s, Value, 0);
}
int main() {
using i1 = std::integer_sequence<int, 2, 3, 3, 2, 3, 2, 0>;
constexpr auto k = findT(i1{}, 0);
static_assert(k == 6, "Compile time constant check.");
std::cout << k << std::endl; // prints "6"
}