Rakesh K Rakesh K - 27 days ago 19
C++ Question

"does not name a type" error

I have two classes declared as below:

class User
{
public:
MyMessageBox dataMsgBox;
};

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


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


MyMessageBox does not name a type

Answer

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.

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!

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;
}

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;
};

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;
};

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


On a side note, this function:

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

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!)

Rather, use a reference (possibly const):

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