重载运算符=作为非成员
根据此线程的回复，operator=不能被重载作为非成员函数。那么，举个例子，下面的代码会让编译器非常生气：
class MyClass
{
// ...
};
MyClass& operator=(MyClass& Left, MyClass& Right)
{
// ...
}
这是为什么呢？我有一个带有 getter 和 setter 的容器类，因此成员函数是不必要的，并且会破坏封装。上述线程的答案之一说，这是为了确保“L 值作为其第一个操作数被接收”，但我不完全理解这意味着什么。有人可以澄清一下吗？
此外，还有 operator=、operator()、operator[] 和 operator-> “oddball” 情况...？或者我应该将所有重载运算符实现为成员函数......？ （我知道这样做是完全合法的，但我正在寻找更好的做法。）
According to replies to this thread, operator= cannot be overloaded as a non-member function. So, for example, the following makes the compiler very angry:
class MyClass
{
// ...
};
MyClass& operator=(MyClass& Left, MyClass& Right)
{
// ...
}
Why is this? I have a container class with getters and setters, so a member function is unnecessary and it would break encapsulation. One of the answers to the aforementioned thread said that it's to make sure the "L-value is received as its first operand," but I don't fully understand what that means. Could someone please clarify?
Additionally, are operator=, operator(), operator[] and operator-> "oddball" cases...? Or should I implement all overloaded operators as member functions...? (I know it's perfectly legal to do otherwise, but I'm looking for the better practice.)
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如果您的类没有赋值运算符（作为成员），编译器会默认生成一个赋值运算符，就像您不提供赋值运算符时它会生成一个复制构造函数一样。
因此，如果您稍后尝试定义非成员赋值运算符，它将“生气”。那么就会有两个！
If your class doesn't have an assignment operator (as a member), the compiler generates one by default, just like it generates a copy constructor if you don't provide one.
Therefore it will be "angry" if you try to define a non-member assignment operator later. There will then be two!
除非您声明一个，否则编译器会在您的类中声明一个
operator=，其签名为operator= (C&, C&)或operator= (C&, const C&)。如果允许您重载分配，则大多数用途将是不明确的。
然后，您可能会游说附加规则来执行以下任一操作：
operator=，则假装没有声明编译器，就像非成员operator=隐藏成员operator=这两种选择都会使本来就极其复杂的规则变得复杂，只为
operator=添加一个特殊情况。很少有人愿意去那里。
而且您还没有公开此功能的合法用途。
或者，任何合理的用途。
当可以展示一些合理的用例时，C++ 规则只会变得更加复杂。
Unless you declare one, the compiler declares an
operator=in your class with signatureoperator= (C&, C&)oroperator= (C&, const C&).If you were allowed to overload assignment, most uses would be ambiguous.
Then you would likely lobby for additional rules to either:
operator=if a user declared one is visible, as if the non-memberoperator=hides the memberoperator=Both choices would complicate rules that are already extremely complicated, adding a special case just for
operator=.Few people want to go there.
And also you haven't exposed a legitimate use for this feature.
Or, any reasonable use at all.
The C++ rules are only made even more complicated when some reasonable use-case can be showed.