C 为什么我不能从列表中删除节点？

每次我尝试这样做时，if语句中都会出现一个未经处理的异常？在这一点上，其他一切都很好

      void DeleteEmp(struct node* head, int tempID){
           struct node *curNode = head;
           struct node *prevNode = NULL;
           while (curNode != NULL) {
                if(curNode->empId == tempID) { // error here
                    free(curNode);
                    printf("Employee %d removed from database", tempID);
                }
                prevNode = curNode;
                curNode = curNode->next;
            }
        }

---

您正在使用

curNode

在

释放它之后

   while (curNode != NULL) {
        if(curNode->empId == tempID) { // error here
            free(curNode);
            printf("Employee %d removed from database", tempID);
        }
        prevNode = curNode;
        curNode = curNode->next; // PROBLEM HERE.
    }

也许你的意思是在呼叫
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后立即返回

void DeleteEmp(struct node* head, int tempID){
   struct node *curNode = head;
   struct node *prevNode = NULL;
   while (curNode != NULL) {
      if(curNode->empId == tempID) {
         if ( prevNode == NULL )
         {
            // This means curNode is the head.
            // Since curNode is about to be deleted,
            // make the next node the head.
            head = curNode->next;
         }
         else
         {
            // Since curNode is about to be deleted,
            // change the next node of prevNode.
            prevNode->next = curNode->next;
         }

         free(curNode);
         printf("Employee %d removed from database", tempID);
         return;
      }
      prevNode = curNode;
      curNode = curNode->next;
   }
}

释放节点之前，应将其从列表中删除

       while (curNode != NULL) {
            if(curNode->empId == tempID) {  
                if (prevNode){// remove the node from the list
                    prevNode->next = curNode->next;   
                }
                else{
                    head = curNode->next;
                } 
                free(curNode);
                printf("Employee %d removed from database", tempID);
                return;//assuming empId is unique return after you found it.
            }
            prevNode = curNode;
            curNode = curNode->next;
        }

在释放节点的位置，在释放节点之前，需要进行一些指针交换。
考虑这个

a->b->c->NULL

删除b应在释放前执行以下操作

a->c->NULL 

请注意，特殊情况包括删除c、删除导致新头的a以及删除单例列表中唯一的节点

注意GList类型和handler函数在library GObject中。使用此列表代码而不是编写和调试此列表代码可能是有意义的
 
free
在未将节点从列表中删除的情况下删除节点是错误的