C - 节点去除问题中的单链表

Question

I'm implementing a singly-linked list in C and got stuck with the remove node function. It removes the element, links two of its neighbours, but the following node gets the next node address set to NULL. Why? Can anybody help?

  struct node{
        struct node* next;
        int value;
    };

    struct list{
        struct node* head;
        struct node* tail;
    };

void remove_node(struct list* plist, int value){

    struct node* current;
    struct node* temp;
    current = plist->head;
    if (!(current)) return; 
    if ( current->value == value ){
        if (!(current->next)){
            plist->head = NULL; plist->tail = NULL;
        }
        else { 
            plist->head = current->next;
            free(current);
        }
    }
    else {
        while(current->next){
            if(current->next->value==value){
                if ((current->next)->next){ 
                    temp = current->next;
                    current->next = (current->next)->next;
                    free(temp);
                }
                else{
                    temp = current->next;
                    plist->tail = current;      
                    current->next = NULL;
                    free(temp);
                    break;
                }
            }
        current = current->next;    
        }
    }
} 


Node current current->next
0 0x9f39018 0x9f39028
1 0x9f39028 0x9f39038
2 0x9f39038 0x9f39048
3 0x9f39048 0x9f39058
4 0x9f39058 0x9f39068
5 0x9f39068 0x9f39078
6 0x9f39078 0x9f39088
7 0x9f39088 0x9f39098
8 0x9f39098 0x9f390a8
9 0x9f390a8 (nil)

after remove(5)

0 0x9f39018 0x9f39028
1 0x9f39028 0x9f39038
2 0x9f39038 0x9f39048
3 0x9f39048 0x9f39058
4 0x9f39058 0x9f39078
6 0x9f39078 (nil)

Answer 1

This code:

if ((current->next)->next){  
    current->next = (current->next)->next; 
    free(current->next); 

frees the next node after you have already removed it from the list. In other words, you are freeing the wrong node.

Answer 2

请注意，您更正后的代码也会在空列表中崩溃。