如何在不将内部类的定义置于父类的情况下创建内部类? [英] How to make an inner class without putting the definition of inner class to parent class?

问题描述

我要写一个头文件，它很长.由于它太复杂了，所以我不想将内部类的定义放到根类中.我的意思是我如何在不编写内部类的情况下将其内部化根类.

I'll write an header file,and it's very long.Since it will be too complicated,i don't want to put inner class definition in root class.I mean how can i make a class inner without writing it in root class.

class outer
{

}

class inner
{

}

如果我可以那样使用，我认为头文件会更清楚.

If i can use like that, The header file will be clearer i think.

推荐答案

像这样:

// foo.hpp

class Foo
{
public:
  class Inner;
  Foo();
  void bar();
  Inner zoo();
};

// foo_inner.hpp

#include "foo.hpp"

class Foo::Inner
{
  void func();
};

然后，在实现中:

#include "foo.hpp"
#include "foo_inner.hpp"

void Foo::bar() { /* ... */ }
void Foo::Inner::func() { /* ... */ }

请注意，您可以在Foo的类定义(即foo.hpp中)内使用不完整类型Foo::Inner，但要遵守对不完整类型的通常限制，例如Inner可能显示为函数返回类型，函数参数，引用或指针.只要类Foo的成员函数实现可以看到Foo::Inner的类定义(通过包含foo_inner.hpp)，一切都很好.

Note that you can use the incomplete type Foo::Inner inside the class definition of Foo (i.e. in foo.hpp) subject to the usual restrictions for incomplete types, e.g. Inner may appear as a function return type, function argument, reference, or pointer. As long as the member function implementations for the class Foo can see the class definition of Foo::Inner (by including foo_inner.hpp), all is well.

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