构造一个内部有结构的类 c++

发布于 2024-12-11 06:44:33 字数 592 浏览 0 评论 0原文

所以我有一个类，它在其私有变量内保存一个结构，在这个结构内我有一个数组，其中数组的大小仅在类构造后确定。

template <typename T> 
class btree {

  public:
    btree(size_t maxNodeElems);
    ~btree() {}

  private:
    // The details of your implementation go here
    size_t maxNodeElems;
    struct node {

      list <T> elements;
      node lvl[];

    };

};

首先，我是否必须将其设置为它的 node * lvl 以及如何调用该结构内的变量？它与私有变量相同吗？所以每当我在 btree 类的函数之一中使用它时，我都可以将其称为 btree.lvl 或者是 btree->node->lvl 还是有特殊的方法来做到这一点？

另外，我的数组必须是 maxNodeElems+1 如果有人可以帮助我，我将不胜感激！

原文

So I have a class which holds a struct inside its private variables and inside this struct I have an array where the size of the array is only determined after the construction of the class.

template <typename T> 
class btree {

  public:
    btree(size_t maxNodeElems);
    ~btree() {}

  private:
    // The details of your implementation go here
    size_t maxNodeElems;
    struct node {

      list <T> elements;
      node lvl[];

    };

};

Firstly, do I have to make it so its node * lvl and how do I call the variables inside this struct? Is it the same as a private variable, so whenever I use it inside one of the functions in btree class I can call it be btree.lvl or is it btree->node->lvl or is there a special way to do this?

Also, my array has to be of maxNodeElems+1 if someone can help me, that'd be much appreciated!

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奈何桥上唱咆哮 2024-12-18 06:44:33

您只是声明类型，而不是该类型的实际对象。您需要将结构声明设为公开，将对象设为私有：

template <typename T> 
class btree {

  public:
    btree(size_t maxNodeElems);
    ~btree() {}

    struct node {   // <- this is just a declaration of a private inner-class
      list <T> elements;
      node lvl[];
    };

  private:
    size_t maxNodeElems;
    node*  memberNode;   // <- this is the actual private member

};

您可以从外部创建该类型的对象：

btree<A>::node* n = new btree<A>::node;

为了访问成员，您可以使用公共 getter 和 getter。 btree 类中的设置器：

class btree {
public:
   node* getNode()
   {
      return memberNode;
   }
   //...........
   //...........
};

编辑：

以下内容对我有用（初始化成员）：

template <typename T> 
class btree {

  public:
    btree()
    {
       memberNode = new btree<T>::node;
    }
    ~btree() {}

    void init()
    {
       memberNode->lvl = new node[10];
    }

    struct node {   // <- this is just a declaration of a private inner-class
      list <T> elements;
      node* lvl;
    };

  private:
    size_t maxNodeElems;
    node*  memberNode;   // <- this is the actual private member

};

int _tmain(int argc, _TCHAR* argv[])
{
   btree<char> b;
   b.init();
}

You are just declaring the type, not an actual object of that type. You need to make your struct declaration public and the object private:

template <typename T> 
class btree {

  public:
    btree(size_t maxNodeElems);
    ~btree() {}

    struct node {   // <- this is just a declaration of a private inner-class
      list <T> elements;
      node lvl[];
    };

  private:
    size_t maxNodeElems;
    node*  memberNode;   // <- this is the actual private member

};

You can create objects of that type from outside:

btree<A>::node* n = new btree<A>::node;

For accessing members, you can have public getters & setters in your btree class:

class btree {
public:
   node* getNode()
   {
      return memberNode;
   }
   //...........
   //...........
};

EDIT:

The following works for me (initializing the member):

template <typename T> 
class btree {

  public:
    btree()
    {
       memberNode = new btree<T>::node;
    }
    ~btree() {}

    void init()
    {
       memberNode->lvl = new node[10];
    }

    struct node {   // <- this is just a declaration of a private inner-class
      list <T> elements;
      node* lvl;
    };

  private:
    size_t maxNodeElems;
    node*  memberNode;   // <- this is the actual private member

};

int _tmain(int argc, _TCHAR* argv[])
{
   btree<char> b;
   b.init();
}

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