C++全局重载运算符=

发布于 2024-10-18 08:28:59 字数 596 浏览 3 评论 0原文

可能的重复：
什么是“operator = 必须是非-”静态成员”是什么意思？（C++）

嗨，

我有以下代码...

// Header file
  struct dataRecord{
     size_t id;
     char name[gcNameLength];
  };

  void operator=(dataRecord &adr, const dataRecord &bdr);

gcc 在编译时如何给出以下错误。

error: ‘void operator=(dataRecord&, const dataRecord&)’ must be a nonstatic member function

感谢您的帮助。

原文

Possible Duplicate:
What does “operator = must be a non-static member” mean? (C++)

Hi,

I have the following code...

// Header file
  struct dataRecord{
     size_t id;
     char name[gcNameLength];
  };

  void operator=(dataRecord &adr, const dataRecord &bdr);

How ever gcc gives me the following error when compiling.

error: ‘void operator=(dataRecord&, const dataRecord&)’ must be a nonstatic member function

Thanks for the help.

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反目相谮 2024-10-25 08:29:02

如 “operator = 必须是”中所述非静态成员”是什么意思？，运算符重载需要是成员函数。

请注意，当您重载运算符=时，您应该返回对左操作数的引用，这样它就不会破坏流程并且允许使用如下表达式：

dataRecord r1;
dataRecord r2;
...
dataRecord r3 = r2 = r1;

As stated in What does “operator = must be a non-static member” mean?, the operator overload needs to be a member function.

Note that when you overload the operator=, you should return a reference to the left operand, so it won't break the flow and will allow expressios like:

dataRecord r1;
dataRecord r2;
...
dataRecord r3 = r2 = r1;

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小嗷兮 2024-10-25 08:29:01

您需要在 struct dataRecord 本身上重载 = 操作。

像这样的东西：

struct dataRecord{
   size_t id;
   char name[gcNameLength];
   dataRecord& operator= (const dataRecord&) {
       // write overload code here
   }
};

You need to overload = operation on the struct dataRecord itself.

Something like:

struct dataRecord{
   size_t id;
   char name[gcNameLength];
   dataRecord& operator= (const dataRecord&) {
       // write overload code here
   }
};

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