在二叉树中插入节点会产生错误 [英] Inserting a node in binary tree gives error

问题描述

当我使用指针在二叉树中创建一个节点时，给我一个分段错误，而当我使用双指针时，它工作得很好。



To我这两段代码看起来都很好但是为什么前者不起作用呢？



因为我们刚刚在后一种情况下取​​消引用指针，为什么当我只使用指针时它不起作用吗？





我尝试了什么：



这是有效的（双指针）

When i use pointers to create a node in a Binary tree is gives me a segmentation error, whereas when i use double pointers it works just fine.

To me both the pieces of code seem to be pretty fine but then why doesn't the former work?

Since we are just dereferencing the pointer in the later case, why doesn't it work when i use just a pointer?


What I have tried:

This is the one which works (Double Pointers)

void Insert(struct bnode **root, int n){
    if(*root == 0){
        temp = (struct bnode *) malloc (sizeof(struct bnode));
        temp -> data = n;
        temp -> right = temp -> left = 0;
        *root = temp;
        return;
    }else{
        temp = *root;
        while(temp != 0){
            if(n >= temp -> data){
                temp = temp -> right;
            }else{
                temp = temp -> left;
            }
        }
        temp = (struct bnode *) malloc (sizeof(struct bnode));
        temp -> data = n;
        temp -> right = temp -> left = 0;
    }
}

我想知道为什么下面的代码不起作用（指针）

I want to know why the code below doesn't work (Pointers)

void Insert(struct bnode *root, int n){
    if(root == 0){
        temp = (struct bnode *) malloc (sizeof(struct bnode));
        temp -> data = n;
        temp -> right = temp -> left = 0;
        root = temp;
        return;
    }else{
        temp = root;
        while(temp != 0){
            if(n >= temp -> data){
                temp = temp -> right;
            }else{
                temp = temp -> left;
            }
        }
        temp = (struct bnode *) malloc (sizeof(struct bnode));
        temp -> data = n;
        temp -> right = temp -> left = 0;
    }
}

推荐答案

编译并不代表你的代码是对的！ ：笑：

将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。



所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。



首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。

所以，你可以查看代码和很明显，它在某处：

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

private int Double(int value)
   {
   return value * value;
   }

一旦你知道可能出现的问题，就开始使用teh调试器找出原因。在你的行上设一个断点：

Once you have an idea what might be going wrong, start using teh debugger to find out why. Put a breakpoint on your line:

if(root == 0){

并运行你的应用程序。在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。

如果没有，为什么不呢？它有何不同？



这是一项非常值得开发的技能，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！



是的，我可能会告诉你问题是什么 - 但这并不难做到，并且你会同时学到一些非常值得的东西！

and run your app. Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?

This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

Yes, I could probably tell you what "the problem" is - but it's not difficult to do this yourself, and you will learn something really worthwhile at the same time!

你没有向我们展示完整的代码，但它应该像

You did not show us the complete code but it should be something like

typedef struct bnode {
    int data;
    bnode *left;
    bnode *right;
};

bnode *root = NULL;

然后使用 Insert（bnode ** root）版本因为只会更新全局 root 这里

Then use the Insert(bnode **root) version because only that will update the global root here

*root = temp;

单指针版本将 temp 分配给局部变量 root （传递的参数）但不更新全局变量。



如果没有看到完整的代码，为什么你的分段错误无法回答。但我猜你在调用 Insert之后没有检查你的全局 root 是否仍然是 NULL 。使用单指针版本就是这种情况。

The single pointer version will assign temp to the local variable root (the passed argument) but not update the global one.

Why you got a segmentation fault can't be answered without seeing the full code. But I guess that you did not check if your global root is still NULL after calling Insert. And that is the case when using the single pointer version.

简单地说， C 参数总是按值传递，因此，如果你需要更改函数内部的外部变量值，然后你必须将这个变量的地址传递给函数。

因为你需要改变 root的值在插入函数内，然后你必须传递它的地址，即& root 。这将建立插入函数的正确签名。

请尝试，例如：

Simply put, C parameters are always passed by value, so, if you need to change an external variable value inside a function then you have to pass the address of such variable to the function.
Since you need to change the value of root inside the Insert function then you have to pass the its address, namely &root. This establishes the correct signature of the Insert function.
Try, for instance:

#include <stdio.h>
#include <stdlib.h>

void allocate_1(int * p)
{
  p = malloc(sizeof(int));
  *p = 1;
}

void allocate_2(int ** pp)
{
  *pp = malloc(sizeof(int));
  **pp = 2;
}

int main()
{
  int * p;


  // allocate_1
  p = NULL;

  allocate_1( p );

  printf("after allocate_1: p = %p", p);

  if ( p )
  {
    printf(", *p = %d", *p);
    free(p);
  }
  printf("\n");


  // allocate_2
  p = NULL;

  allocate_2( &p );

  printf("after allocate_2: p = %p", p);

  if ( p )
  {
    printf(", *p = %d", *p);
    free(p);
  }
  printf("\n");


  return 0;
}

输出：

output:

after allocate_1: p = (nil)
after allocate_2: p = 0x13f7440, *p = 2

这篇关于在二叉树中插入节点会产生错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！