轉換複雜<int16_t>複雜<double>

Question

反正在C++ 11要做到這一點：

std::complex<int16_t> integer(42,42); 
std::complex<double> doub(25.5,25.5); 
std::complex<double> answer = integer*doub; 

的錯誤是

error: no match for ‘operator*’ (operand types are  
‘std::complex<short  int>’ and ‘std::complex<double>’) 
std::complex<double> answer = integer*doub; 

我已經試過的static_cast喜歡;

std::complex<double> answer = static_cast<std::complex<double>>(integer)*doub; 

Answer 1

有沒有預定義的皈依從complex<double>到complex<int16_t>或反之亦然。

您可以定義自己：

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; 
}