将复数<int16_t>转换为复数<double>

Question

C++11 中是否有这样的功能：

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

我试过static_cast喜欢；

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

Answer 1

没有预定义的从 complex<double> 到 complex<int16_t> 的转换，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;
}