C++11 转换复合体<;int16_t>;要复杂<;双倍>;

C++11 转换复合体<;int16_t>;要复杂<;双倍>;,c++11,casting,complex-numbers,C++11,Casting,Complex Numbers,C++11中是否有这样做的方法: std::complex<int16_t> integer(42,42); std::complex<double> doub(25.5,25.5); std::complex<double> answer = integer*doub; std::复整数(42,42); 标准:复杂双盲(25.5,25.5); std::复杂答案=整数*doub; 错误是 error: no match for ‘operator*

C++11中是否有这样做的方法:

 std::complex<int16_t> integer(42,42);
 std::complex<double> doub(25.5,25.5);
 std::complex<double> answer = integer*doub;
std::复整数(42,42);
标准:复杂双盲(25.5,25.5);
std::复杂答案=整数*doub;
错误是

error: no match for ‘operator*’ (operand types are    
‘std::complex<short     int>’ and ‘std::complex<double>’)
std::complex<double> answer = integer*doub;
错误:“运算符*”不匹配(操作数类型为
“std::complex”和“std::complex”)
std::复杂答案=整数*doub;
我试过静态的施法

std::complex<double> answer = static_cast<std::complex<double>>(integer)*doub;
std::复杂答案=静态(整数)*doub;

没有从
复合体
复合体
的预定义转换,反之亦然

您可以定义自己的:

template <typename D, typename S> std::complex<D> cast(const std::complex<S> s)
{
    return std::complex<D>(s.real(), s.imag());
}

int main()
{
    std::complex<int16_t> integer(42, 42);
    std::complex<double> doub(25.5, 25.5);
    std::complex<double> answer = cast<double, int16_t>(integer)*doub;
}
template std::complex cast(const std::complex s)
{
返回std::complex(s.real(),s.imag());
}
int main()
{
std::复整数(42,42);
标准:复杂双盲(25.5,25.5);
标准::复杂答案=强制转换(整数)*doub;
}

请注意,本标准不要求
std::complex
以任何有意义的方式工作(只有三种浮动类型定义了行为)