我在C中创建了一个链表数据结构。但是,我在addLast函数的实现中遇到了一些奇怪的行为。似乎添加的元素在我下次调用addLast之前不会出现。我的代码(我将通过内联注释解释我认为我的代码是如何工作的):
帮助程序代码:
typedef struct LinkedList linkedlist;
typedef int ListElement;
struct LinkedList{
ListElement data;
linkedlist *next;
};
//Initializes a list;
void CreateList(linkedlist *list, ListElement contents){
list->data = contents;
list->next = NULL;
}
//Prints the items of the list, head first.
void displayList(linkedlist *list){
printf("(%d", list->data);
linkedlist *node = list->next;
if(node == NULL){
}
else{
while(node->next != NULL){
printf(" %d", node->data);
node = node->next;
}
}
printf(")");
}
有问题的代码:
//Adds an element at the tail of the list
void addLast(linkedlist *list, ListElement forAdding){
linkedlist *node = list;
linkedlist *NewNode = (linkedlist *) malloc(sizeof(linkedlist));
//Go to the last element in the list
while(node->next != NULL){
node = node->next;
}
//Prepare the node we will add
NewNode->data = forAdding;
NewNode->next = NULL;
//Since node is pointing to the tail element, set its
//next to the NewNode---the new tail
node->next = NewNode;
}
//Special attention to this function!
void List(ListElement items[], linkedlist *list, int numItems){
int i = 0;
while(i < numItems){
addLast(list, items[i]);
printf("Before ");
displayList(list);
printf("\n");
printf("Added %d", items[i]);
displayList(list);
printf("\n");
i++;
}
}
主要功能:
int main(){
linkedlist *l= (linkedlist *) malloc(sizeof(linkedlist));
CreateList(l, 0);
int a_list[5] = {1, 2, 3, 5, 6};
List(a_list, l, sizeof(a_list)/sizeof(a_list[0]));
printf("A list of five elements: %d", sizeof(a_list)/sizeof(a_list[0]));
displayList(l);
removeLast(l);
addLast(l, 7);
printf("\nAdded something at last position: ");
displayList(l);
printf("\n");
}
我得到了输出:
Before (0)
Added 1(0)
Before (0)
Added 2(0 1)
Before (0 1)
Added 3(0 1 2)
Before (0 1 2)
Added 5(0 1 2 3)
Before (0 1 2 3)
Added 6(0 1 2 3 5)
A list of five elements: 5(0 1 2 3 5)
Added something at last position: (0 1 2 3 5 6)
如您所见,似乎添加的项目只会出现在我下次调用addLast时。
我到目前为止已经发现它实际上是 虽然由于某种原因它不会被打印出来。例如,如果我在关闭函数List之前执行另一个addLast(list, 6);
调用(当然在循环之外!),输出行Added something at last position...
(在调用{{1}之后发生)实际上会显示addLast(l, 7);
。
那么,我做错了什么?
谢谢!
答案 0 :(得分:4)
问题不在AddLast()
只是你的displayList()
功能:)。您在最后一个元素之前停止打印1个元素。
更改displayList()
功能:
void displayList(linkedlist *list){
printf("(");
linkedlist *node = list;
while(node != NULL){
printf(" %d", node->data);
node = node->next;
}
printf(")");
}
此功能也可以打印空列表。
答案 1 :(得分:1)
您只是不打印列表的最后一项。
当node->data
为NULL时,您知道当前节点之后没有更多节点。但你需要在停止之前显示最后一个!!
您可以将支票修改为node != NULL
来执行此操作。
答案 2 :(得分:1)
您的显示功能是错误的:在您的循环中,对于列表的最后一个元素,您有
node->next = NULL
node->data != NULL
你的循环不会显示它:
while(node->next != NULL){
printf(" %d", node->data);
node = node->next;
}
您可以像while (node != NULL) {
答案 3 :(得分:1)
您的问题出在displayList ::
中...
while(node->next != NULL){
printf(" %d", node->data);
node = node->next;
...
当node-&gt; next到达最后一项时,它将为null,因此不显示。