类，函数的前向声明 [英] Forward Declaration of Class, Function

问题描述

当函数的向前声明在源文件（.cpp）中工作时，为什么同样不适用于类？

When forward declarations of functions work in a source file (.cpp), why would the same doesn't work for classes ?

感谢。

// main.cpp

void forwardDeclaredFunction() ; // This is correct 

class One ; // Why this would be wrong 

int One:: statVar = 10 ;

void
One :: anyAccess() {

 std::cout << "\n statVar:\t " << statVar ;
 std::cout << "\n classVar:\t" << classVar ;
}

class One {

 public:
  void anyAccess() ;
  static int statVar ;

 private:
  int  classVar ;

} ;


int main (int argc, char * const argv[]) {

 One *obj = new One ;

        return 0;
}

void forwardDeclaredFunction() {
}

推荐答案

转发声明也可以用于类：

Forward declaration can work for classes too:

class Foo;

class Bar {
public:
    Foo *myFoo; // This has to be a pointer, thanks for catching this!
};

class Foo {
public:
    int value;
};

上面的代码显示了Foo类的前向声明，使用Foo *类型的变量class（Bar），然后是Foo类的实际定义。 C ++不在乎如果你留下的东西未实现，只要你在使用它的代码之前实现它们。定义指向某个类型的对象的指针不是使用其代码。

The above code shows a forward declaration of the Foo class, using a variable of type Foo* in another class (Bar), then the actual definition of the Foo class. C++ doesn't care if you leave things unimplemented as long as you implement them before using its code. Defining pointers to objects of a certain type is not "using its code."

快速，脏的回复，但我希望它有帮助。

Quick, dirty reply but I hope it helps.

编辑：声明未实现的类的非指针变量不会编译为回复。这样做正是我所说的使用它的代码。在这种情况下，只要调用Bar构造函数，就会调用Foo构造函数，因为它具有类型为Foo的成员变量。由于编译器不知道你计划以后实现Foo，它会抛出一个错误。对不起，因为我的错误;）。

Declaring a non-pointer variable of a class thats unimplemented will NOT compile as the replies stated out. Doing so is exactly what I meant by "using its code." In this case, the Foo constructor would be called whenever the Bar constructor is called, given that it has a member variable of type Foo. Since the compiler doesn't know that you plan on implementing Foo later on, it will throw an error. Sorry for my mistake ;).

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