“X 未命名类型"C++ 中的错误

You need to make sure MyMessageBox is defined before you use it as a member. This is solved by reversing the definition order. However, you have a cyclic dependency: if you move MyMessageBox above User, then in the definition of MyMessageBox the name User won't be defined!

你能做的是forward declare User;也就是说，声明它但不定义它.在编译期间，已声明但未定义的类型称为不完整类型.考虑一个更简单的例子:

What you can do is forward declare User; that is, declare it but don't define it. During compilation, a type that is declared but not defined is called an incomplete type. Consider the simpler example:

struct foo; // foo is *declared* to be a struct, but that struct is not yet defined

struct bar
{
    // this is okay, it's just a pointer;
    // we can point to something without knowing how that something is defined
    foo* fp; 

    // likewise, we can form a reference to it
    void some_func(foo& fr);

    // but this would be an error, as before, because it requires a definition
    /* foo fooMember; */
};

struct foo // okay, now define foo!
{
    int fooInt;
    double fooDouble;
};

void bar::some_func(foo& fr)
{
    // now that foo is defined, we can read that reference:
    fr.fooInt = 111605;
    fr.foDouble = 123.456;
}

通过前向声明User，MyMessageBox仍然可以形成一个指针或对它的引用:

By forward declaring User, MyMessageBox can still form a pointer or reference to it:

class User; // let the compiler know such a class will be defined

class MyMessageBox
{
public:
    // this is ok, no definitions needed yet for User (or Message)
    void sendMessage(Message *msg, User *recvr); 

    Message receiveMessage();
    vector<Message>* dataMessageList;
};

class User
{
public:
    // also ok, since it's now defined
    MyMessageBox dataMsgBox;
};

你不能反过来这样做:如前所述，类成员需要有一个定义.(原因是编译器需要知道User占用了多少内存，并且知道它需要知道其成员的大小.)如果你说:

You cannot do this the other way around: as mentioned, a class member needs to have a definition. (The reason is that the compiler needs to know how much memory User takes up, and to know that it needs to know the size of its members.) If you were to say:

class MyMessageBox;

class User
{
public:
    // size not available! it's an incomplete type
    MyMessageBox dataMsgBox;
};

这行不通，因为它还不知道大小.

It wouldn't work, since it doesn't know the size yet.

附带说明，此功能:

 void sendMessage(Message *msg, User *recvr);

可能不应该通过指针获取其中任何一个.你不能在没有消息的情况下发送消息，也不能在没有用户的情况下发送消息.这两种情况都可以通过将 null 作为参数传递给任一参数来表达(null 是一个完全有效的指针值！)

Probably shouldn't take either of those by pointer. You can't send a message without a message, nor can you send a message without a user to send it to. And both of those situations are expressible by passing null as an argument to either parameter (null is a perfectly valid pointer value!)

相反，使用引用(可能是常量):

Rather, use a reference (possibly const):

 void sendMessage(const Message& msg, User& recvr);

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