Question

C++中编程出现function does not take 1 parameters

#include "stdio.h"
#include "stdlib.h"
#define NULL 0

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

void main()
{
struct node *creat();
void print();
struct node *head;
head=NULL;
head=creat(head);
print(head);
}

struct node *creat(struct node *head)
{
struct node *p1,*p2;
p1=p2=(struct node *)malloc(sizeof(struct node));
scanf("%d",p1->num);
p1->next=NULL;
while(p1->num!=0)
{
if(head==NULL) p1=head;
else
p2->next=p1;
p2=p1;
p1=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p1->num);
}
p2->next=NULL;
return head;
}

void print(struct node *head)
{
struct node *temp;
temp=head;
while(temp!=NULL)
{
printf("%6d",temp->num);
temp=temp->next;
}
}

出现的问题是：
error C2660: 'creat' : function does not take 1 parameters
error C2660: 'print' : function does not take 1 parameters

Answer 1

struct node *creat();
void print();
函数声明和定义不一致，参数类型和个数是要一致的，改成这么声明
struct node *creat(struct node *head);
void print(struct node *head);

Answer 2

struct node *creat(); 这句是对create的声明，应该写成
struct node *creat(struct node*); 并且放到main函数前面

Answer 3

这两个函数 你都没跟参数