complex<long double>

项目
06/09/2015

存储描述了排序的对象对两种类型 long double，则第一个表示第二部分表示复数实部和虚部的对象。

template<>
   class complex<long double> {
public:
   complex(
      long double _RealVal = 0, 
      long double _ImagVal = 0
   );
complex(
      const complex<long double>& _ComplexNum
   );
   // rest same as template class complex
};

参数

_RealVal
复数的重载构造值的 long double 类型。
_ImagVal
复数加法的值的小数类型 long double 构造。
_ComplexNum
double 类型复数物理和虚部用于初始化类型 long double 构造的复数或浮动类型。

返回值

复数类型 long double。

备注

模板类的显式专用化复杂到类型 long double 复杂的类与它定义的模板仅类不同于构造函数。从 long double 的转换。浮动允许是隐式的，但是，要求 long double 到从 double 中转换为 explicit。使用分配语法，使用 explicit 排除与类型转换的启动。

有关类模板 complex的更多信息，请参见 complex 类。有关 complex类模板的成员列表，请参见复杂成员。

示例

// complex_comp_ld.cpp
// compile with: /EHsc
#include <complex>
#include <iostream>

int main( )
{
   using namespace std;
   double pi = 3.14159265359;

   // The first constructor specifies real & imaginary parts
   complex <long double> c1 ( 4.0 , 5.0 );
   cout << "Specifying initial real & imaginary parts,\n"
        << " as type float gives c1 = " << c1 << endl;

   // The second constructor initializes values of the real &
   // imaginary parts using those of complex number of type float
   complex <float> c2float ( 1.0 , 3.0 );
   complex <long double> c2longdouble ( c2float );
   cout << "Implicit conversion from type float to type long double,"
        << "\n gives c2longdouble = " << c2longdouble << endl;

   // The third constructor initializes values of the real &
   // imaginary parts using those of a complex number
   // of type double
   complex <double> c3double ( 3.0 , 4.0 );
   complex <long double> c3longdouble ( c3double );
   cout << "Implicit conversion from type long double to type float,"
        << "\n gives c3longdouble = " << c3longdouble << endl;

   // The modulus and argument of a complex number can be recovered
   double absc3 = abs ( c3longdouble );
   double argc3 = arg ( c3longdouble );
   cout << "The modulus of c3 is recovered from c3 using: abs ( c3 ) = "
        << absc3 << endl;
   cout << "Argument of c3 is recovered from c3 using:\n arg ( c3 ) = "
        << argc3 << " radians, which is " << argc3 * 180 / pi
        << " degrees." << endl;
}