CPP 中的分段错误（核心转储）

Question

I am confused a bit. I narrowed down it to:

Why having this line:

Node *root, *rootSafe = NULL;

give error:

Segmentation fault(core dumped)

While just switching it like below:

Node *rootSafe, *root = NULL;

runs perfect.

Here is the code, you can test.

#include <iostream>
using namespace std;
struct Node{
    int data;
    Node *left, *right;
    Node(int d){
        this->data = d;
        this->left = this->right = NULL;
    }
};
Node *newNode(int d){
    Node *temp = (Node *)malloc(sizeof(Node));
    temp->data = d;
    temp->left = temp->right = NULL;
    return temp;
}
void printInorder(Node *root){
    if(root == NULL){
        return;
    }
    else{
        printInorder(root->left);
        cout << "--" << root->data;
        printInorder(root->right);
    }
}
int main()
{
    //cout << "Hello World";
    Node *rootSafe, *root = NULL;
    int arr[] = {5, 3, 1, 4, 6};
    int sizeArr = sizeof(arr)/sizeof(arr[0]);
    
    for(auto i = 0; i < sizeArr; i++){
        if(root == NULL){
            rootSafe = newNode(arr[i]);
            root = rootSafe;
        }
        else{
            while(root != NULL){
                if(arr[i] < root->data){//Move left
                    if(root->left == NULL){
                        root->left = newNode(arr[i]);
                        root = NULL;
                    }
                    else{
                        root = root->left;
                    }
                }
                else{//Move right
                    if(root->right == NULL){
                        root->right = newNode(arr[i]);
                        root = NULL;
                    }
                    else{
                        root = root->right;
                    }
                }
            }
        }
        root = rootSafe;
    }
    
    cout << "\n Print Inorder: ----"; printInorder(rootSafe);
    return 0;
}

Answer 1

Without fiddling with your code, I think this:

Node *root, *rootSafe = NULL;

Doesn't do what you think it does. Do you think it sets both to NULL? It doesn't. root gets some random value and rootSafe gets NULL.

This might be what you really want:

Node *root = NULL, *rootSafe = NULL;

Frankly, I personally hate (and it's against coding conventions at some work places) specifying multiple variables on the same line. You will not find this in my code. Instead, you will see:

Node * root = nullptr;
Node * rootSafe = nullptr;

Note also that in modern C++, NULL is not a pointer. Get in the habit of using nullptr.

Answer 2

You are using root here:

        if(root == NULL){

Your original declaration

Node *root, *rootSafe = NULL;

doesn't initialize root , leaving it with indeterminate value. Therefore, root with some random invalid value may be dereferenced and it may lead to Segmentation Fault.

To avoid this, you should initialize root before using that.

It can be done in declaration:

Node *root = NULL, *rootSafe = NULL;

Or before the loop:

    root = NULL;
    for(auto i = 0; i < sizeArr; i++){

Answer 3

Node *root, *rootSafe = NULL;
Node *rootSafe, *root = NULL;

The first line leaves root uninitialized. This causes undefined behaviour while access root.

The second line leaves rootSafe uninitialized but the var later initialized rootSafe = newNode(arr[i]); .