c ++ variadic模板参数迭代

Question

我对这些事情缺乏经验,但我正在尝试创建一个模板函数,在"旋转"参数下评估n变量函数(参见下面的示例)并返回所有这些值的向量.

例如,对于具有函数f(x,y,z)的n = 3,返回的三元\向量应该是

< f(x,0,0),f(0,x,0),f(0,0,x)>

我需要的天真版本可能如下所示(不必正确\工作)

typedef FunctionSignature Function;

template<class Function, size_t Dimensions>
std::array<Function::Out,Dimensions> F(Function::InComponent x)
{
  std::array<Function::Out,Dimensions> Result;

  for (i=0; i<Dimensions; i++)    
     Result[i] = Function::f("rotate((x,0,...,0),i)");

  return Result;
}

但是如何做到这rotate一点.

我也希望运行时for可以以某种方式被消除,因为n在编译时众所周知.

Answer 1

template<class Function, size_t... Is, size_t... Js>
typename Function::Out call_f(typename Function::InComponent x, 
                              std::index_sequence<Is...>, 
                              std::index_sequence<Js...>) {
    return Function::f((void(Is), 0)..., x, (void(Js), 0)...);
}

template<class Function, size_t Dimensions, size_t... Is>
std::array<typename Function::Out, Dimensions> F(typename Function::InComponent x,
                                                 std::index_sequence<Is...>)
{
  return {{ call_f<Function>(x, std::make_index_sequence<Is>(),
                                std::make_index_sequence<Dimensions - Is - 1>())... }};
}

template<class Function, size_t Dimensions>
std::array<typename Function::Out,Dimensions> F(typename Function::InComponent x)
{
   return F<Function, Dimensions>(x, std::make_index_sequence<Dimensions>());
}

对于C++ 11,在SO上搜索实现make_index_sequence.

演示.