二进制运算符在模板类上重载 [英] Binary operator overloading on a templated class

问题描述

我最近尝试衡量我的运算符重载/模板能力，并作为一个小测试，创建了下面的Container类。虽然这个代码编译精细，并在MSVC 2008（显示11）下正常工作，MinGW / GCC和Comeau阻塞在运算符+ 超载。因为我相信他们比MSVC更多，我想弄清楚我做错了什么。

I was recently trying to gauge my operator overloading/template abilities and as a small test, created the Container class below. While this code compiles fine and works correctly under MSVC 2008 (displays 11), both MinGW/GCC and Comeau choke on the operator+ overload. As I trust them more than MSVC, I'm trying to figure out what I'm doing wrong.

这里是代码：

#include <iostream>

using namespace std;

template <typename T>
class Container
{
      friend Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs);
   public: void setobj(T ob);
     T getobj();
      private: T obj;
};

template <typename T>
void Container<T>::setobj(T ob)
{
   obj = ob;
}

template <typename T>
T Container<T>::getobj()
{
   return obj;
}

template <typename T>
Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs)
{
      Container<T> temp;
      temp.obj = lhs.obj + rhs.obj;
      return temp;
}

int main()
{    
    Container<int> a, b;

 a.setobj(5);
    b.setobj(6);

 Container<int> c = a + b;

 cout << c.getobj() << endl;

    return 0;
}

这是Comeau给出的错误：

This is the error Comeau gives:

Comeau C/C++ 4.3.10.1 (Oct  6 2008 11:28:09) for ONLINE_EVALUATION_BETA2
Copyright 1988-2008 Comeau Computing.  All rights reserved.
MODE:strict errors C++ C++0x_extensions

"ComeauTest.c", line 27: error: an explicit template argument list is not allowed
          on this declaration
  Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs)
               ^

1 error detected in the compilation of "ComeauTest.c".

我很难尝试让Comeau / MingGW玩球，转向你们。这是很长时间，因为我的大脑已经融化了这很多在C ++语法的重量，所以我觉得有点尴尬。）

I'm having a hard time trying to get Comeau/MingGW to play ball, so that's where I turn to you guys. It's been a long time since my brain has melted this much under the weight of C++ syntax, so I feel a little embarrassed ;).

编辑

EDIT: Eliminated an (irrelevant) lvalue error listed in initial Comeau dump.

推荐答案

我发现解决方案感谢转到此论坛发帖。基本上，你需要有一个函数原型，然后才能在类中使用friend，但是您还需要声明该类才能正确定义函数原型。因此，解决方案是在顶部具有两个原型定义（函数和类）。以下代码在所有三个编译器下编译：

I found the solution thanks to this forum posting. Essentially, you need to have a function prototype before you can use 'friend' on it within the class, however you also need the class to be declared in order to properly define the function prototype. Therefore, the solution is to have two prototype definitons (of the function and the class) at the top. The following code compiles under all three compilers:

#include <iostream>

using namespace std;

//added lines below
template<typename T> class Container;
template<typename T> Container<T> operator+ (Container<T>& lhs, Container<T>& rhs); 

template <typename T>
class Container
{
      friend Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs);
      public: void setobj(T ob);
    		  T getobj();
      private: T obj;
};

template <typename T>
void Container<T>::setobj(T ob)
{
      obj = ob;
}

template <typename T>
T Container<T>::getobj()
{
      return obj;
}

template <typename T>
Container<T> operator+ (Container<T>& lhs, Container<T>& rhs)
{
      Container<T> temp;
      temp.obj = lhs.obj + rhs.obj;
      return temp;
}

int main()
{    
    Container<int> a, b;

    a.setobj(5);
    b.setobj(6);

    Container<int> c = a + b;

    cout << c.getobj() << endl;

    return 0;
}

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