假设我们有以下结构:
typedef struct link {
char link[MAXLINK];
struct link *next;
} LinkT;
typedef struct node {
char keyword[MAXKEY];
LinkT *links;
struct node *next;
} NodeT;在主要功能中,我添加了很少的新关键字和链接到它。打印整个列表时得到的输出是:
Asteroid
Link-45
Link-23
Earth
Link-2
Moon
Link-3
Link-1我想添加一个新的链接-4到地球。所以输出将如下所示:
Asteroid
Link-45
Link-23
Earth
Link-2
Link-4
Moon
Link-3
Link-1我的想法是有一个findLink函数,遍历关键字列表,然后遍历关键字匹配,遍历链接列表,如果没有找到链接,我添加一个新的链接。
然而,问题是,当我运行下面的代码,它永远不会完成,没有输出任何东西(冻结?)我得逼着退出。
如果删除最后添加的关键字和链接,它将运行但不正确,并打印该链接不在列表中。类似的函数findKeyword做同样的-打印关键字没有在列表中。
我如何修正代码,使其运行于任何列表?
代码:
NodeT *addNewKeyword(NodeT **list, char *keyword) {
NodeT *new = malloc(sizeof(NodeT));
assert(new != NULL);
strcpy(new->keyword, keyword);
new->next = *list;
*list = new;
return 0;
}
LinkT *addNewLink(NodeT *list, char *link) {
LinkT *new = malloc(sizeof(LinkT));
assert(new != NULL);
strcpy(new->link, link);
new->next = list->links;
list->links = new;
return 0;
}
NodeT *findKeyword(NodeT *list, char *keyword) {
NodeT *current = list;
while (current != NULL) {
if (strcmp(current->keyword, keyword) == 0) {
return current;
}
current = current->next;
}
return NULL;
}
void *findLink (NodeT *list, char *keyword, char *link) {
NodeT *current = list;
while (current != NULL) {
if (strcmp(current->keyword, keyword) == 0) {
LinkT *currentLink = list->links;
while (currentLink != NULL) {
if (strcmp(currentLink->link, link) == 0) {
return currentLink;
}
currentLink = list->links->next;
}
return NULL;
}
current = current->next;
}
return NULL;
}
void *printList(NodeT *listNode) {
while (listNode != NULL) {
printf("%s\n", listNode->keyword);
LinkT *listLink = listNode->links;
if (listLink == NULL) {
printf("List is empty.\n");
} else {
while (listLink != NULL) {
printf(" %s\n", listLink->url);
listLink = listLink->next;
}
}
listNode = listNode->next;
}
return 0;
}
int main() {
NodeT *wordsList = NULL;
addNewKeyword(&wordsList, "Moon");
addNewLink(wordsList, "Link-1");
addNewLink(wordsList, "Link-3");
addNewKeyword(&wordsList, "Earth");
addNewLink(wordsList, "Link-2");
/* runs after removing following lines: */
// addNewKeyword(&wordsList, "Asteroid");
// addNewLink(wordsList, "Link-23");
// addNewLink(wordsList, "Link-45");
if (findKeyword(wordsList, "Moon") == 0) {
printf("Keyword is in list.\n");
} else {
printf("Keyword is not in list.\n");
}
if (findLink(wordsList, "Moon", "Link-1") == NULL) {
printf("Link is not in list.\n");
addNewLink(wordsList, "Link-1");
} else {
printf("Link is in list.\n");
}
printList(wordsList);
return 0;
}编辑1
我正在从一个函数类型返回一个节点,这可能是不按预期工作的原因。findLink函数需要的类型是什么?
编辑2
正如Sahin在“答案”中指出的那样,我更新了findLink函数,现在程序运行,但是向不正确的位置添加了一个元素。主要代码的更改:
int main() {
...
addNewKeyword(&wordsList, "Moon");
addNewLink(wordsList, "Link-1");
addNewLink(wordsList, "Link-3");
addNewKeyword(&wordsList, "Earth");
addNewLink(wordsList, "Link-2");
addNewKeyword(&wordsList, "Asteroid");
addNewLink(wordsList, "Link-23");
addNewLink(wordsList, "Link-45");
if (findLink(wordsList, "Earth", "Link-4") == NULL) {
printf("Link is not in list.\n");
addNewLink(wordsList, "Link-4");
} else {
printf("Link is in list.\n");
}
...
}输出:
Link is not in list.
Asteroid
Link-4
Link-45
Link-23
Earth
Link-2
Moon
Link-3
Link-1发布于 2021-01-18 20:40:36
void *findLink (NodeT *list, char *keyword, char *link) {
NodeT *current = list;
while (current != NULL) {
if (strcmp(current->keyword, keyword) == 0) {
LinkT *currentLink = current->links;//An error is here, you always look for the same list.
while (currentLink != NULL) {
if (strcmp(currentLink->link, link) == 0) {
return currentLink;
}
currentLink = currentLink->next;//The other error is here
}
return NULL;
}
current = current->next;
}
return NULL;
}我猜您的错误在这里,您需要得到下一个currentLink,因为相同节点的下一个节点总是相同的。您需要在进行这些更正之后测试代码,然后我们可以再次查看结果。
编辑:响应编辑2:
NodeT *addNewKeyword(NodeT **list, char *keyword) {
NodeT *new = malloc(sizeof(NodeT));
assert(new != NULL);
strcpy(new->keyword, keyword);
new->next = *list;
*list = new; //This row
return 0;
}在我放置注释的行中,您将更改参数内的值,列表变量中的值将更改。由于您使用"Asteroid“调用addNeKeyword,所以新的链接将被添加到其中。因此,请具体说明更多细节。你到底想要什么?
https://stackoverflow.com/questions/65772723
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