使用指针在链表中进行结构构造时出现C分段错误

Question

I write a very basic linked list in C supporting only two operations - insert of a node to the top of the list and iterate over the list in order to print the value of each node. The problem I am facing is that I have a segmentation fault when executing. Here is my code:

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct linkedList
{
    int data;
    struct linkedList *next;
};
struct linkedList* createNode(int value)
{
    struct linkedList *node;
    node = malloc(sizeof(struct linkedList));
    node->next = malloc(sizeof(struct linkedList));
    node->next = NULL;
    node = NULL;
    return node;
}
//insert a node at the top of the linked list
struct linkedList* insertTop(struct linkedList* top,struct linkedList* node)
{
    if(top == NULL)//the element we insert is the 1st element for the linked list
    {
        node->next = NULL;
        //the first element points to NULL since it has no successors
    }
    else//there is already an element in the list
    {
        node->next = top;
    }
    return node;
}
void iterate(struct linkedList* top)
{
    while(top->next != NULL)
    {
        printf("Data = %d\n", top->data);
        top = top->next;
    }
}
int main()
{
    struct linkedList *a,*b,*c,*root;
    a = createNode(2);
    b = createNode(10);
    c = createNode(23);
    root = insertTop(NULL,a);//these 3 lines provoke a segmentation fault
    root = insertTop(root,b);
    root = insertTop(root,c);
    iterate(root);//the result I expect here is 23,10,2 
    return 0;
}

I know this question has been asked many times on stackoverflow and yet I still can't figure out why my code does not work as expected. Can you please explain to me where is the problem and how can I fix it? Thank you

Answer 1

This:

node = NULL;

doesn't look so good in the createNode() function. In fact, that entire function doesn't look so good. It should be something like this:

struct linkedList* createNode(int value)
{
    struct linkedList *node = malloc(sizeof *node);
    node->next = NULL;
    node->data = value;
    return node;
}

It really shouldn't have been allocating more than one node, that's not sane.

Answer 2

There are two main issues. First createNode :

You allocate space for a struct linkedList and assign it to node , which is good. But then you do the same for node->next . So you're actually creating two nodes, but then you set node->next to NULL ,loosing the reference to the second bit of memory you allocated, creating a memory leak. The same happens when you do the same for node . The end result is that you always return NULL for your new node. Also, you don't assign value to anything.

Just do the first allocation, assign value to data , and initialize next to NULL:

struct linkedList* createNode(int value)
{
    struct linkedList *node;
    node = malloc(sizeof(struct linkedList));
    node->data = value;
    node->next = NULL;
    return node;
}

The next issue is in iterate :

while(top->next != NULL)

This causes the iteration to stop when the current node is the last one, so you don't print the last node. Also, as a result of how createNode was originally implemented, your root pointer is NULL, so you end up dereferencing a NULL pointer, which causes a core dump.

You want to test if top is NULL instead:

while(top != NULL)

A third issue is cleanup. You need to define a function to free the memory you allocated before your program exits. You can do that like this:

void cleanup(struct linkedList *top)
{
    struct linkedList *temp;

    while (top != NULL) {
        temp = top;
        top = top->next;
        free(temp);
     }
}

Answer 3

If you like to create a node you only have to allocate mamory for one node. Adapt your code like this:

struct linkedList* createNode(int value)
{
    struct linkedList *node = malloc(sizeof(struct linkedList));
    node->next = NULL;
    node->data = value;
    return node;
}

Apart from this you have to change the termination condition of the while loop in your function iterate :

void iterate( struct linkedList *top )
{
    while( top != NULL) // if to is not NULL you can print its data
        // ^^^ 
    {
        printf("Data = %d\n", top->data);
        top = top->next;
    } 
}

And you can simplify your function insertTop :

struct linkedList* insertTop(struct linkedList* top,struct linkedList* node)
{
    if( node == NULL ) // test if node is not NULL
        return top;
    node->next = top;  // successor of new node is top (top possibly is NULL)
    return node;
}

Dont't forget to free your list at the end of main :

int main()
{
    struct linkedList *root = NULL;
    root = insertTop( root, createNode(2) );
    root = insertTop( root, createNode(10) );
    root = insertTop( root, createNode(23) );

    iterate(root);

    while ( root != NULL )
    {
        struct linkedList *temp = root;
        root = root->next; 
        free( temp );
    }
    return 0; 
}

Apart from this you can combine createNode and insertTop to one function createTop :

struct linkedList* createTop( struct linkedList* top, int value )
{
    struct linkedList *node = malloc(sizeof(struct linkedList));
    node->next = top;
    node->data = value;
    return node;
}

Answer 4

You have quite a few problems there..

In the function createNode() you are allocatting memory to node and then throwing it away by doing node = NULL .

Loose those statements, as they create a memory leak.

struct linkedList* createNode(int value)
{
    struct linkedList *node;
    node = malloc(sizeof(struct linkedList));
    node->data = value;
    node->next = NULL;
    return node;
}