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3鱼币
- #include <stdio.h>
- # include <stdlib.h>
- struct list
- {
- int data;
- struct list * pnext;
- };
- struct list * createlist(void);
- void printlist(struct liat *);
- int main(void)
- {
- struct list * p = NULL;
- p = createlist();
- printlist(p); 这里错了
- return 0;
- }
- struct list * createlist(void) //创建链表
- {
- int i;
- int len;
- int val;
- struct list* tail = (struct list*)malloc(sizeof(struct list));
- struct list* first = tail;
- printf("请输入一个数:");
- scanf("%d",&len);
- for (i = 0; i < len; i++)
- {
- printf("请输入第%d个元素的值:",i + 1);
- scanf("%d",&val);
- struct list * New = (struct list*)malloc(sizeof(struct list));
- New->data = val;
- New->pnext = NULL;
- tail->pnext = New;
- tail = New;
- }
- return first;
- }
- void printlist(struct list * frist )
- {
- while(frist != NULL)
- {
- printf("%d",frist->data );
- frist = frist->pnext ;
- }
- return ;
- }
复制代码 E:\c语言源代码\链表.cpp(17) : error C2664: 'printlist' : cannot convert parameter 1 from 'struct list *' to 'struct liat *'
Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
Error executing cl.exe.
求大侠帮助
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最佳答案
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链表,我刚学完不久。 我发现英文不好也有优点了。我写的时候,是用拼音的。就没错过。struct list* 和 struct liat* 这应该是你说的这里错的原因。偷偷告诉你一下,后面学到文件的时候,会让你很有动力学的。
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