转换complex< int16_t>复杂到< double> [英] Convert complex<int16_t> to complex<double>

问题描述

在C ++ 11中仍然可以做到这一点：

Is there anyway in C++11 to do this:

 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;

我尝试过static_cast；

I have tried static_cast like;

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

推荐答案

没有预定义的转换复杂< double> 到 complex< int16_t> 或反之亦然。

您可以定义您自己的：

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;
}

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