使用迭代插入函数在第二次插入后初始化二叉搜索树时出现分段错误

Question

I'm getting a segmentation fault after trying to insert a second node into the tree, a segmentation fault is thrown. Is the segmentation fault due to the initialize function or insert ? And how do I solve it by making changes in server.c?

client.c file:

int main()
{
    Tree my_tree;
    initialize(&my_tree);
    while(1)
    {
        scanf("%d", &choice);
        switch (choice)
        {
        case 1:
            scanf("%d", &element);
            insert(&my_tree, element);
            break;
        case 2:
               exit(0);
       }
    }
    return 0;
}

server.c file

typedef struct node
{
    int data;
    struct node *left;
    struct node *right;
} Node;

typedef struct tree
{
    Node *root;
} Tree;

void initialize(Tree *tree)
{
    tree = (Tree*)malloc(sizeof(Tree));
    tree->root = NULL;
}

void insert(Tree *tree, int data)
{
    Node *temp = (Node*)malloc(sizeof(Node));
    temp->left = temp->right = NULL;
    temp->data = data;
    if(tree->root == NULL) 
        tree->root = temp;    
    else
    {
        Node *prev,*curr;
        prev = NULL;
        curr = tree->root;
        while(curr != NULL)
        {
            if(temp->data < curr->data){
                prev = curr;
                curr = curr->left;
            }
            if(temp->data >curr->data){
                prev = curr;
                curr = curr->right;
            }
        }
        if(temp->data < prev->data)
            prev->left = temp;
        if(temp->data > prev->data)
            prev->right = temp;
    }
}

Answer 1

void initialize(Tree *tree)
{
    tree = (Tree*)malloc(sizeof(Tree));
    tree->root = NULL;
}

Change to

void initialize(Tree *tree)
{
    if (tree) {
     tree->root = NULL;
    }
}

When you allocate memory to tree you are making it to point new memory location, thus your original tree 's root node will not be initialized to NULL and also you introduce memory leaks.

Answer 2

Instead of initializing the Tree object you actually use in your program ( tree in main), you allocate a new Tree object and initialize that object. Effectively, initialize doesn't do anything but leak memory.

To fix this, replace

void initialize(Tree *tree)
{
    tree = (Tree*)malloc(sizeof(Tree));
    tree->root = NULL;
}

with

void initialize(Tree *tree)
{
    tree->root = NULL;
}

---

The OP points out there's a second error. This where I point out that one shouldn't be depending on the kindness of strangers, and learn how to debug such problems.

You could use a debugger.

$ gcc -Wall -Wextra -pedantic -g a.c -o a && gdb --args ./a
GNU gdb (Ubuntu 8.1-0ubuntu3.2) 8.1.0.20180409-git
...
(gdb) r
Starting program: /tmp/ikegami/a
1
3
1
2

Program received signal SIGSEGV, Segmentation fault.
0x00005555555547b1 in insert (tree=0x7fffffffdda0, data=2) at a.c:39
39                  if(temp->data >curr->data){
(gdb) print temp
$1 = (Node *) 0x555555756690
(gdb) print curr
$2 = (Node *) 0x0
(gdb) quit
A debugging session is active.

        Inferior 1 [process 1433] will be killed.

Quit anyway? (y or n) y

Another approach would be to use -fsanitize=address when compiling the program.

$ gcc -Wall -Wextra -pedantic -g -fsanitize=address a.c -o a && ./a
1
3
1
2
ASAN:DEADLYSIGNAL
=================================================================
==1370==ERROR: AddressSanitizer: SEGV on unknown address 0x000000000000 (pc 0x556253e76e87 bp 0x7ffe3a28e390 sp 0x7ffe3a28e360 T0)
==1370==The signal is caused by a READ memory access.
==1370==Hint: address points to the zero page.
    #0 0x556253e76e86 in insert /home/ikegami/tmp/a.c:39
    #1 0x556253e771af in main /home/ikegami/tmp/a.c:64
    #2 0x7fc3bdadcb96 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x21b96)
    #3 0x556253e76b29 in _start (/tmp/ikegami/a+0xb29)

AddressSanitizer can not provide additional info.
SUMMARY: AddressSanitizer: SEGV /home/ikegami/tmp/a.c:39 in insert
==1370==ABORTING

But approach show that temp->data >curr->data is the culprit. Using the debugger, we saw that curr is NULL , so curr->data is at issue. Now you know that you need figure out if why curr is NULL when you don't expect it to be.