为什么是复杂的< double&gt ;?* int在C ++中未定义吗? [英] Why is complex<double> * int not defined in C++?

问题描述

C ++程序

#include <complex>
#include <iostream>

int main()
{
  std::complex<double> z(0,2);
  int n = 3;
  std::cout << z * n << std::endl;
}

产生一个错误:'z * n'中的'operator *'不匹配.为什么?

yields an error: no match for ‘operator*’ in ‘z * n’. Why?

我正在使用g ++ 4.4.1进行编译.也许编译器只是遵循C ++标准，在这种情况下，我的问题是:为什么标准不允许这样做?

I'm compiling with g++ 4.4.1. Perhaps the compiler is just following the C++ standard, in which case my question is: why does the standard not allow this?

推荐答案

这有效:

#include <complex>
#include <iostream>

int main()
{
    std::complex<double> z(0,2);
    double n = 3.0; // Note, double
    std::cout << z * n << std::endl;
}

因为复数由双精度数组成，所以它乘以双精度数.查看声明:

Because complex is composed of doubles, it multiplies with doubles. Looking at the declaration:

template <typename T>
inline complex<T> operator*(const complex<T>&, const T&);

(以下内容感谢 dribeas ).在模板推导过程中，不允许编译器进行隐式类型转换，因此通过传递 T 为 double 的 complex ，然后传递另一个 T 为 int 的 complex ，当尝试匹配将 T 视为 double 的函数时，会导致第二个参数不匹配，反之亦然.

(The following is thanks to dribeas) The compiler is not allowed to make implicit type conversions during template deduction, so by passing a complex with T being double, and then another T being int, when trying to match the function treating T as double causes the second argument to mis-match, and vice-versa.

对于您要工作的东西，必须定义一个类似于以下内容的函数:

For what you want to work, it would have to have a function defined similar to this:

template <typename T, typename U>
inline std::complex<T> operator*(std::complex<T> lhs, const U& rhs)
{
    return lhs *= rhs;
}

哪个允许该函数采用不同的类型，从而允许在调用 operator * = 时进行强制转换.

Which allows the function to take differing types, which allows the cast to be done when calling operator*=.

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