C ++：没有返回可以结束非空函数吗？

Question

这是一个相当笼统的问题，但长话短说，我试图找出一种在不使用返回的情况下突破函数的方法。我有一个从模板构建的函数，所以我的编译器不喜欢我使用＆＃34; return;＆＃34;如我之前所见，建议尽早终止。以下是我尝试做的简要版本：

V & find (K key) {
    TreeNode<K,V> *traverseFind;
    //TreeNode<K,V> * treeNode = new TreeNode<K,V> (key,NULL);
    traverseFind = root;
    unsigned int compareTraverseToHeight = 0;
    //A bunch of stuff happens
    if(keyFinder == 0)  //this is what I have tried to do
        return;
    return traverseFind->value; //This is what I'm working with, the value can be NULL
}

一些背景：这是二叉树中的查找函数，它包含两个参数key和value。搜索密钥，然后返回其对应的值。作为项目的一部分，我的任务是抛出一个异常，如果输入了一个不在树中的密钥，它会被测试者函数捕获。这样可以正常工作，但是，测试程序的配置方式会在解析之后重新回到函数中，在这种情况下我有traverseFind-＆gt; value = NULL，当我尝试返回它时会崩溃我的程序。我无法控制测试人员，所以在我必须返回一个值之前，我想要退出find函数。

这大部分都是好奇的背景。我真正的问题是：而不是我试图使用的if（keyFinder == 0）函数（它没有编译），是否有另一个选项可以用来提前结束函数？ A＆＃34;休息＆＃34;功能声明可以这么说吗？

更多代码：

#ifndef __CSC116__BST_H__
#define __CSC116__BST_H__

//
// An implementation of a binary search tree.
//
// This tree stores both keys and values associated with those keys.
//
// More information about binary search trees can be found here:
//
// http://en.wikipedia.org/wiki/Binary_search_tree
//
// Note: Wikipedia is using a different definition of 
//       depth and height than we are using.  Be sure
//       to read the comments in this file for the
//   height function.
//

#include <list>
#include <iostream>
#include <utility>
#include <algorithm>
#include <string>

using namespace std;

unsigned int heightCount = 0;
unsigned int sizeCount = 0;
int keyFinder = 0;



//
// Each node in the tree stores
// both a key and a value
//
template <class K, class V> class TreeNode
{
public:
    TreeNode(K k, V v): key(k), value(v), left(0), right(0) {}

    K       key;
    V       value;
    TreeNode<K,V>   *left;
    TreeNode<K,V>   *right;
    template <class X, class Y> friend std::ostream & operator<< (std::ostream &s, const TreeNode<X,Y> &t);
};

// 
// TreeNodes can output themselves to streams
template <class K, class V> std::ostream & operator << (std::ostream &s, const TreeNode<K,V> &n)
{
    s << "\"" << n.key << ":" << n.value << "\"";
    return s;
}

//
// This exception is thrown if you try to find 
// a key in the tree that isn't present.
//

class key_not_found_exception {

};



template <class K, class V> class BinarySearchTree {
public:
    //
    // Constructor
    //
    BinarySearchTree ()
    {
        sizeCount = 0;
        heightCount = 0;
        root = NULL;
    }


    ~BinarySearchTree()
    {
    }


void insert (K k, V v) {
  TreeNode<K,V> * treeNode = new TreeNode<K,V> (k,v);
  TreeNode<K,V> *temp=NULL;
  TreeNode<K,V> *prev=NULL;
  temp = root;

  while(temp) { 
    prev = temp;
    if (temp->key < treeNode->key)
      temp = temp->right;
    else
      temp = temp->left;
  }

  if (prev==NULL)
    {root = treeNode;heightCount++;}
  else {
    if (prev->key<treeNode->key)
      {
        prev->right = treeNode;  
        if(prev->left ==NULL){heightCount++;}
      }
    else
      {
        prev->left = treeNode;
        if(prev->right ==NULL){heightCount++;}
      }
  } 

  sizeCount++;
  cout << "Height of Tree is:"<<"_" <<heightCount<<endl;
}






    V & find (K key)                     
    {

    TreeNode<K,V> *traverseFind;
    //TreeNode<K,V> * treeNode = new TreeNode<K,V> (key,NULL);

    traverseFind = root; 
    unsigned int compareTraverseToHeight = 0;

    while (compareTraverseToHeight < heightCount) 
    {
        if (traverseFind == NULL)  // We've fallen off the tree without finding item.
        {
            keyFinder = 0;
            cout<<"code1"<<endl;
            cout<<"keyFinder = 0"<<endl;
            break;

        }

        else if ( key == traverseFind->key )// We've found the item. 
        {
            keyFinder = 1;   

            cout<<"code2"<<endl;
            cout<<"keyFinder = 1"<<endl;
            break;
        }       

        else if ( key < traverseFind->key )                   // If the item occurs, it must be in the left subtree,
        {                                                 // So, advance the runner down one level to the left.
            traverseFind = traverseFind->left;

            cout<<"code3";
        }   


        else  // If the item occurs, it must be in the right subtree. // So, advance the runner down one level to the right.
        {
            traverseFind = traverseFind->right;

            cout<<"code4";  
        }  


         compareTraverseToHeight++;      

    }  // end while

    cout<<"At end of loop, keyFinder ="<<keyFinder<<endl;

        try
        {
            if(keyFinder ==0)
            {
                cout<<"Try has found that keyFinder = 0"<<endl;
                throw key_not_found_exception();

            }
            else{cout<<"keyFinder = 1"<<endl;}
        }catch(...){}

        cout<<"RETURNED";
        cout<<traverseFind->value<<endl;

            if(keyFinder == 0)
            {

            }
            return traverseFind->value;



    }

private:

    unsigned int doHeight (TreeNode<K,V> *t)
    {
        return -1;
    }

    void        doDelete (TreeNode<K,V> * n )
    {
    }

    TreeNode<K,V>   *root;
    unsigned int    count;
    unsigned int    sizeCount;

#endif

TESTER FUNCTION

void test_insert_find()
{
    BinarySearchTree<string,string> t;

    t.insert("bob", "bobdata");
    t.insert("joe", "joedata");
    t.insert("jane", "janedata");

    try
    {
        cout<<"Looking for Bob"<<endl;
        string s = t.find("bob");
        if (s != "bobdata")
            throw bst_tester_exception(__func__,__LINE__);
    }
    catch (key_not_found_exception &e)
    {
            throw bst_tester_exception(__func__,__LINE__);
    }

    try
    {
        cout<<"Looking for Sarah"<<endl;
        string s = t.find("sarah");
        throw bst_tester_exception(__func__,__LINE__);

    }


    catch (key_not_found_exception &e)
    {

        cout<<"Successfully caught"<<endl;
        // this is expected.
    }

测试的全部输出：

Looking for Bob
code2
keyFinder = 1
At end of loop, keyFinder = 1
keyFinder = 1
RETURNEDbobdata
Looking for Sarah
code4code4code1
keyFinder = 0
At end of loop, keyFinder = 0
Try has found that keyFinder = 0, throwing exception
RETURNED /*Program now crashes when it attempts to return NULL*/

Answer 1

通常，要从查找类型函数返回“失败”，有两个主要选择：

1. 返回找到的元素（或引用），否则抛出异常。

2. 将指针返回给元素（如果找到），或nullptr（如果没有）。

还有其他方法，例如返回表示成功的bool并返回在附加引用或指针参数中找到的元素。

在您的情况下，返回V &引用，结束函数而不返回值的唯一合理选项是throw异常。