可以 C++赋值运算符是自由函数吗？
我正在尝试这样的事情：
Foo & operator=(Foo & to, const Bar &from);
但我收到此错误：
E2239 'operator =(Foo &, const Bar &)' must be a member function
哪些运算符可以/不能定义为自由函数是否有限制，如果有，为什么？
I'm trying something like this:
Foo & operator=(Foo & to, const Bar &from);
But I'm getting this error:
E2239 'operator =(Foo &, const Bar &)' must be a member function
Are there limitations on which operators can/cannot be defined as Free Functions, and if so, why?
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赋值运算符必须是非静态成员函数，并且必须只有一个参数：
operator()、operator[]和operator->也必须实现为非静态成员函数。特定于类的
operator new和operator delete（及其变体）必须实现为静态成员函数（请注意，它们是隐式静态的，即使它们没有使用static关键字）。The assignment operator must be a non-static member function and must have exactly one parameter:
operator(),operator[], andoperator->must also be implemented as non-static member functions.Class-specific
operator newandoperator delete(and variants thereof) must be implemented as static member functions (note that these are implicitly static, even if they are not declared with thestatickeyword).它不能。
我猜原因与复制构造函数有关。它们具有非常相似的语义，并且您不能像其他构造函数一样在类外部定义复制构造函数。所以，他们不想把双胞胎分开太远（以避免双胞胎悖论：）。
PS C++ 中的一个遗憾是您无法向现有类添加成员。这没有低级的原因。如果可能的话，您可以通过不在类定义标头中声明私有函数来解耦标头和 cpp 依赖关系。
It cannot.
The reason, I guess, has to do with copy constructor. They have very similar semantics, and, you cannot define a copy constructor outside of a class just like other constructor. So, they didn't want to separate the twins far apart (to avoid the twins paradox:).
P.S. What's a shame in C++, is that you cannot add a member to existing class. There's no low-level reason for that. If it would be possible, you could decouple header and cpp dependencies by not declaring private functions in the class definition header.