Question

我正在尝试使用模板元编程实现正弦函数。但是我得到错误'radtest'不是double＆amp;的有效模板参数。因为cv资格的冲突。这是代码：

#include <iostream>
using namespace std;

template <double&, int, int> struct 
Series;

template <double& rad> struct Sine
{
    enum
    {
        maxterms=10 
    };      

    static inline double sin()
    {
        return (rad) * 
  Series<(rad), 0, maxterms>::val(); 
    }
};

template <double& rad, int i, int 
maxterms> struct Series
{
   enum 
   {
         cont = i+1 != maxterms, 
         nxt1 = (i+1)*cont,
         nxtmax = maxterms*cont      
   };

   // uses recursive definition of 
   // Sine
   // sin(x)=x*term(0)
   // term(n)=1-
   // x*x/(2*n+2)/(2*n+3)*term(n+1)
   static inline double val()
   {
       return 1 - (rad)*
(rad)/(2.0*i+2.0)/(2.0*i+3.0)
      * Series<rad * cont, nxt1, 
nxtmax>::val(); 

   }
};

#define SineT(rad) Sine<rad>::sin()
constexpr double radtest=0.707;

int main() 
{
    cout << "Sine of " << radtest 
<< " is: " << SineT(radtest);
    return 0;
}

可能是什么问题？提前谢谢。

Answer 1

The problem is that radtest is const (implied by constexpr) so you cannot have double& parameter of sine bind to it.

If you try and make it double radtest (not constexpr) or if you try and make all template parameters const double& then you get into another problem: you cannot bind a temporary to a reference non-type template parameter. This is explicitly disallowed by the standard:

§ 14.3.2 Template non-type arguments

A template-argument for a non-type template-parameter shall be a converted constant expression (5.20) of the type of the template-parameter. For a non-type template-parameter of reference or pointer type, the value of the constant expression shall not refer to (or for a pointer type, shall not be the address of):

...

(1.2) a temporary object (12.2),

...

I personally don't see a way out of this mess. This (using references as template non-type parameters) is really pushing the boundaries of what the C++ template system can do.

What I can recommend is to just create a constexpr sin function.

模板元编程，cv资格错误中的冲突

1 个答案: