“不命名类型”。错误 [英] "does not name a type" error

问题描述

我有两个类声明如下：

class User
{
public:
  MyMessageBox dataMsgBox;
};

class MyMessageBox
{
public:
  void sendMessage(Message *msg, User *recvr);
  Message receiveMessage();
  vector<Message> *dataMessageList;
};

当我尝试使用gcc编译它时，会出现以下错误：

When I try to compile it using gcc, it gives the following error:

MyMessageBox未命名类型

MyMessageBox does not name a type

推荐答案

当编译器编译类 User 并访问 MyMessageBox 行时， MyMessageBox 尚未定义。它不知道它存在，所以不能理解你的类成员的意义。

When the compiler compiles the class User and gets to the MyMessageBox line, MyMessageBox has not yet been defined. It has no idea it exists, so cannot understand the meaning of your class member.

你需要确保 MyMessageBox 之前定义 之前，您将其用作成员。这是通过颠倒定义顺序来解决的。但是，你有一个循环依赖：如果你移动 MyMessageBox 上面 User ，那么在 MyMessageBox 不会定义 c>

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!

do is forward declare User ;即声明它，但不定义它。在编译期间，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;
}

通过向前声明 User ， MyMessageBox 仍然可以形成指针或对它的引用：

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;
};

您不能反过来这样做：成员需要有一个定义。 （原因是编译器需要知道有多少内存 User 占用，并知道它需要知道它的成员的大小。）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;
};

它不会工作，因为它不知道大小。

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

在旁注，此函数：

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

可能不应该使用指针。您不能发送没有消息的消息，也不能在没有用户发送消息的情况下发送消息。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示（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!)

而是使用引用：

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

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