Question

我想知道在编译时获取可变参数模板类的第N个参数的最简单和更常用的方法（返回值必须是编译器的静态const）为了做一些优化）。这是我的模板类的形式：

template<unsigned int... T> MyClass
{
    // Compile-time function to get the N-th value of the variadic template ?
};

非常感谢。

编辑：由于MyClass将包含200多个函数，我无法专门化它。但我可以在MyClass中专门化一个结构或函数。

编辑：从经过验证的答案中得出的最终解决方案：

#include <iostream>

template<unsigned int... TN> class MyClass
{
    // Helper
    template<unsigned int index, unsigned int... remPack> struct getVal;
    template<unsigned int index, unsigned int In, unsigned int... remPack> struct getVal<index, In,remPack...>
    {
        static const unsigned int val = getVal<index-1, remPack...>::val;
    };
    template<unsigned int In, unsigned int...remPack> struct getVal<1,In,remPack...>
    {
        static const unsigned int val = In;
    };

    // Compile-time validation test
    public:
        template<unsigned int T> inline void f() {std::cout<<"Hello, my value is "<<T<<std::endl;}
        inline void ftest() {f<getVal<4,TN...>::val>();} // <- If this compile, all is OK at compile-time
};
int main()
{
    MyClass<10, 11, 12, 13, 14> x;
    x.ftest();
    return 0;
}

Answer 1

“感应设计”应该是这样的：

template<unsigned int N, unsigned int Head, unsigned int... Tail>
struct GetNthTemplateArgument : GetNthTemplateArgument<N-1,Tail...>
{
};


template<unsigned int Head, unsigned int... Tail>
struct GetNthTemplateArgument<0,Head,Tail...>
{
    static const unsigned int value = Head;
};

template<unsigned int... T> 
class MyClass
{
     static const unsigned int fifth = GetNthTemplateArgument<4,T...>::value;
};

Answer 2

这是另一种方法：

template<unsigned int index, unsigned int In, unsigned int... remPack> struct getVal
{
    static const unsigned int val = getVal<index-1, remPack...>::val;
};
template<unsigned int In, unsigned int...remPack> struct getVal<0,In,remPack...>
{
    static const unsigned int val = In;
};

template<unsigned int... T> struct MyClass
{
    //go to any arg by : getVal<Some_Unsigned_Index, T...>::val;
};

测试：http://liveworkspace.org/code/4a1a9ed4edcf931373e7ab0bf098c847

如果你因为＃34而受到刺痛，就无法扩展＆＃39; T ...＆＃39;进入固定长度的参数列表＆＃34; http://ideone.com/YF4UJ

Answer 3

这也是你可以做的事情

template<int N, typename T, T ... Ts>
struct GetN {
  constexpr T values[sizeof...(Ts)] = { Ts... };
  static const T value = values[N];
};

template<int N, typename T, T ... Ts>
constexpt T GetN<N, T, Ts...>::values[sizeof...(Ts)];

然后你可以简单地做

template<int N, unsigned int... T> struct MyClass {
  static const unsigned int value = GetN<N, unsigned int, T...>::value;
};

Answer 4

又一个简单的方法：

#include <array>

template<int... Args>
class Foo {
 public:
  static constexpr int Element(int index) {
    return std::array<int, sizeof...(Args)>{ Args... }[index];
  }

 int First = Element(0);
 int Second = Element(1);
};


int main() {
  return Foo<0, 1>::Element(0);  
}

// or just
int Nth = std::array<int, sizeof...(Args)>{ Args... }[N];

BTW，这是提取任何可变参数模板的第N个参数的一般方法：

#include <tuple>

template<typename... Args>
class Foo {
 public:
  template <int N>
  using Nth = typename std::remove_reference<decltype(std::get<N>(std::declval<std::tuple<Args...>>()))>::type;
};


int main() {
  Foo<int, float>::Nth<1> val = 3.14159f;
  return 0; 
}

获取可变参数模板类的第N个参数的最简单方法是什么？

4 个答案: