operator =模板类的重载

Question

我有一个项目，所有的类都将以int，double和float为模板，getCoordinate返回CCoordinate类型的对象。

            tempCoordinate = m_shapes.at(i)->getCoordinate(j);

在应用模板之前，它已经正常工作。 但是随后出现一些错误。

据我了解，我需要丢失和operator =重载以强制转换值，以防万一例如我有一个浮点数并且我收到一个整数，例如：

        CCoordinate<float> coorFloat;
        CCoordinate<int> coorInt = coorFloat

如何在课堂上创建这个？ 需要什么格式？ 。

我以为它应该看起来像这样，但显然我弄错了。

//CCoordinate.h
template<class T>
class CCoordinate {
 //Code
 public:
 template<class U> template <class U> CCoordinate<T>
            operator= (const CCoordinate<U>& c1);
}

//CCoordinate.cpp
template <class U >
CCoordinate<U> CCoordinate<T>::operator= (const CCoordinate<U>& c1)
{
    // some kind of casting ? 
}

我的错误：

19:06:43 **** Incremental Build of configuration Debug for project ShapesRefV2 ****
Info: Internal Builder is used for build
g++ -O0 -g3 -Wall -c -fmessage-length=0 -Werror=return-type -o "myCode\\CRectangle.o"      "..\\myCode\\CRectangle.cpp" 
g++ -O0 -g3 -Wall -c -fmessage-length=0 -Werror=return-type -o "myCode\\CPlane.o"    "..\\myCode\\CPlane.cpp" 
..\myCode\CPlane.cpp: In instantiation of 'GraSys::CRectangle<T>       GraSys::CPlane<T>::boundingBox(std::string, std::string) [with T = int; std::string =    std::basic_string<char>]':
..\myCode\CPlane.cpp:165:24:   required from here
..\myCode\CPlane.cpp:115:20: error: no match for 'operator=' (operand types are   'GraSys::CCoordinate<double>' and 'const GraSys::CCoordinate<int>')
  tempCoordinate = m_shapes.at(i)->getCoordinate(j);
                ^
..\myCode\CPlane.cpp:115:20: note: candidate is:
In file included from ..\myCode\CGraphicElement.h:14:0,
               from ..\myCode\CPlane.h:11,
               from ..\myCode\CPlane.cpp:9:
..\myCode\CCoordinate.h:17:7: note: GraSys::CCoordinate<double>&   GraSys::CCoordinate<double>::operator=(const GraSys::CCoordinate<double>&)
class CCoordinate
        ^
..\myCode\CCoordinate.h:17:7: note:   no known conversion for argument 1 from 'const    GraSys::CCoordinate<int>' to 'const GraSys::CCoordinate<double>&'
..\myCode\CPlane.cpp: In instantiation of 'GraSys::CRectangle<T> GraSys::CPlane<T>::boundingBox(std::string, std::string) [with T = float; std::string =   std::basic_string<char>]':
..\myCode\CPlane.cpp:166:24:   required from here
..\myCode\CPlane.cpp:115:20: error: no match for 'operator=' (operand types are 'GraSys::CCoordinate<double>' and 'const GraSys::CCoordinate<float>')
     tempCoordinate = m_shapes.at(i)->getCoordinate(j);
                    ^
..\myCode\CPlane.cpp:115:20: note: candidate is:
In file included from ..\myCode\CGraphicElement.h:14:0,
                 from ..\myCode\CPlane.h:11,
                 from ..\myCode\CPlane.cpp:9:
..\myCode\CCoordinate.h:17:7: note: GraSys::CCoordinate<double>&   GraSys::CCoordinate<double>::operator=(const GraSys::CCoordinate<double>&)
 class CCoordinate
       ^
..\myCode\CCoordinate.h:17:7: note:   no known conversion for argument 1 from 'const GraSys::CCoordinate<float>' to 'const GraSys::CCoordinate<double>&'

19:06:44 Build Finished (took 674ms)

Answer 1

在成员声明中，您的template <class U>太多了，成员应返回对*this的引用，因此它需要返回CCordinate & （如果省略，则意味着<T> ）：

// Remove this       vvvvvvvvvvvvvvvvvv
template<class U> /* template <class U> */
CCoordinate & operator= (const CCoordinate<U>& c1);
//          ^- Return type changed to be a reference.

由于成员是模板，而类是模板，因此有两个级别的模板。 实现成员时，您需要指定两个级别。

它还返回错误的类型（它返回CCoordinate<U>但是您已在类中声明它返回CCoordinate<T> ）。

// You need the T template as well.
// vvvvvvvvvvvvvvv
template <class T>
template <class U>
CCoordinate<T> & CCoordinate<T>::operator= (const CCoordinate<U>& c1)
//          ^  ^- Added reference as per above.
//          \---- Changed to T; U makes no sense here and conflicts with your member
//                declaration in the class.
{
    // Your logic to make the conversion.

    return *this;
}

Answer 2

您的尝试有两个问题。 更简单的问题是， operator=的声明语法具有额外的template <class U> 。 它看起来应该像这样：

template<class U> CCoordinate<T>
            operator= (const CCoordinate<U>& c1);

但是，即使是正确定义的operator=也不允许您编写

CCoordinate<float> coorFloat;
CCoordinate<int> coorInt = coorFloat;

这是因为copy上方的第二行初始化了 coorInt 。 在复制初始化时不考虑使用operator= －它仅查看用户定义的转换 ，在这种情况下，该转换仅包括非显式构造函数和非显式转换函数。