如何避免嵌套模板中的模板参数重复

你有几种可能:

• 越简单,我想你想要的是:

  template <typename T>
  class BST {
  public:
      Node<T>* m_root = nullptr;
  
  public:
      BST() = default;
  
      // ...
  };
  

  有用法

  BST<int> bst;
  

• 另一个是专业化:

  template <typename T> class BST;
  
  template <typename T>
  class BST<Node<T>> {
  public:
      Node<T>* m_root = nullptr;
  
  public:
      BST() = default;
  
      // ...
  };
  

  使用方法:

  BST<Node<int>> bst;
  

• 最后一个,模板参数:

  template <template <typename> class N, typename T>
  class BST {
  public:
      N<T>* m_root = nullptr;
  
  public:
      BST() = default;
  
      // ...
  };
  

  有用法

  BST<Node, int> bst;
  

template <typename T>
class BST;

template <template <typename> class C, typename T>
class BST<C<T>> {
    using Node = C<T>;
public:
    BST() {
        m_root = NULL;
    }

    bool find(T value, Node** parent, Node** location) {/* ... */}
};

然后你就可以像这样使用它

BST<Node<int>>* bst = new BST<Node<int>>(); // deduce C as Node, T as int

或者添加一些嵌套的typedef。

template <typename T>
struct Node {
    using value_type = T;
    ...
};

template <typename Node>
class BST {
    using T = typename Node::value_type;
    ...
};

为什么不干脆

template <typename T>
class BST {

public:

    Node< T >* m_root;

public:

    BST() {
        m_root = NULL;
    }

    bool find(T value, Node< T >** parent, Node< T >** location) {/* ... */}
};

Node 内部 BST

template <typename T>
class BST
{
public:
    struct Node
    {
        T value;
        Node* left;
        Node* right;
    };

    BST() { m_root = NULL; }

    bool find(T value, Node** parent, Node** location) {/* ... */}

private:
    Node* m_root;
};