如何访问链表的 null 指针（节点）？

Question

This method is supposed to append a Node at the end of a linked list. The method loops until it reaches the end, which is the null pointer. But when I try to change the null pointer to a value, it crashes. How should I fix this? (The Node pointer has a integer data and another Node variable which the current Node points to).

void appendItem(LinkedList* list, int value)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp = list->head;

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

    temp->data = value;
    temp->next = NULL;
}

Answer 1

Dereferencing NULL is prohibited.

Instead of that, you should manage a pointer to what should be changed.

Also note that allocating some buffer via malloc() and overwriting the result with another value just after that causes a memory leak.

One more point is that casting results of malloc() family is considered as a bad practice .

Fixed code:

void appendItem(LinkedList* list, int value)
{
    Node** temp = &list->head;

    while(*temp != NULL)
    {
        temp = &(*temp)->next;
    }

    *temp = malloc(sizeof(Node));
    if (*temp != NULL)
    {
        (*temp)->data = value;
        (*temp)->next = NULL;
    }
}

Answer 2

These lines

Node* temp = (Node*)malloc(sizeof(Node));
temp = list->head;

produces a memory leak. At first a memory was allocated and its address was stored in the pointer temp and then the value of the pointer temp was overwritten by the value of the expression list->head . As a result the address of the allocated memory was lost.

After this loop

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

the pointer temp is equal to NULL . So a null pointer is used to access memory in these statements

temp->data = value;
temp->next = NULL;

that invokes undefined behavior.

The function can be defined for example the following way.

int appendItem( LinkedList *list, int value )
{
    Node *new_node = malloc( sizeof( Node ) );
    int success = new_node != NULL;

    if ( success )
    {
        new_node->data = value;
        new_node->next = NULL;

        if ( list->head == NULL )
        {
            list->head = new_node;
        }
        else
        {
            Node *current = list->head;
            while ( current->next != NULL ) current = current->next;
            current->next = new_node;
        }
    }

    return success;
}

Pay attention to that the memory allocation can fail. You need to process such a case in your function. And the caller of the function should be informed about such a situation.

Also as you allow to append new nodes to a singly-linked list then the list should be defined as a two-sided singly-linked list. That is the list should keep two pointers: one pointer to the head node and other pointer to the tail node. Otherwise appending a node to the list will be inefficient.