从std :: complex< MyType>进行类型转换到std :: complex< double> [英] Type conversion from std::complex<MyType> to std::complex<double>

问题描述

我有一个实现用户定义的算术类型的类 MyType 。此类提供以下转换运算符

  struct MyType 
 {... 
运算符double（）
 {
 return to_double（）; //将我的类型转换为双精度值
} 
 ...}; 
  

按如下所示使用此类可以正常工作：

  double d = MyType（1）; 
  

但是，使用此类作为std :: complex中的类型，例如

  #include< complex> 
 std :: complex< double> c = std :: complex< MyType>（1,1）; 
  

失败，并出现以下编译器错误：

 错误：从'std :: complex< MyType>'转换为非标量类型'std :: complex< double>'要求
  

感谢您为解决此问题提供的帮助。

解决方案

专业化 std :: complex< float> ， std :: complex< double> 和 std :: complex< long double> 是用于表示和处理复数的LiteralType。 / p>

未指定实例化其他任何类型的模板复合体的效果。

所以 std :: complex< MyType> 是有问题的 ...

忽略该部分，

而< a href = https://en.cppreference.com/w/cpp/numeric/complex/complex rel = nofollow noreferrer> std :: complex< T> 具有通用骗局转换构造函数，专门化 std :: complex< double> 仅提供其他浮动复杂版本的转换。

但是 operator = 允许通用所有版本的转换（尽管只有Msvc接受代码）。

您必须提供自己的（显式/命名）转换功能：

  std :: complex< double> to_complex_double（std :: complex< MyType>& c）
 {
 #if 0 
 std :: complex< double> RES; 
 res = c; // gcc / clang不接受。 
美元的回报； 
 #else 
 return {c.real（），c.imag（）}; 
 #endif 
} 
  

I have a class MyType that implements a user-defined arithmetic type. This class provides the following conversion operator

struct MyType 
{ ...
  operator double()
  { 
    return to_double(); // This converts my type to a double value
  }
... };

Using this class as follows works fine:

double d = MyType(1);

However, using this class as type within std::complex, e.g.

#include <complex>
std::complex<double> c = std::complex<MyType>(1,1);

fails with the following compiler error:

error: conversion from 'std::complex<MyType>' to non-scalar type 'std::complex<double>' requested

Any help to solve this problem is appreciated.

Matthias

解决方案

The specializations std::complex<float>, std::complex<double>, and std::complex<long double> are LiteralTypes for representing and manipulating complex numbers.

The effect of instantiating the template complex for any other type is unspecified.

So std::complex<MyType> is "problematic"...

Ignoring that part,

whereas std::complex<T> has generic converting constructor, specialization std::complex<double> only provide conversions from other floating complex versions.

But operator= allows generic conversion for all versions (Only Msvc accepts the code though).

You have to provide your own (explicit/named) conversion function:

std::complex<double> to_complex_double(std::complex<MyType>& c)
{
#if 0
     std::complex<double> res;
     res = c; // gcc/clang doesn't accept that.
     return res;
#else
     return {c.real(), c.imag()};
#endif
}

这篇关于从std :: complex&lt; MyType&gt;进行类型转换到std :: complex&lt; double&gt;的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！