Question

function does not take 1 parameters

#include<stdio.h>
#include<stdlib.h>
struct student
{
char name[10];
float score;
struct student *next;
};
void main()
{
struct student *create();
struct student *pdelete();
void print();
struct student *head;
int n;
printf("Input the number of nodes:\n");
scanf("%d",&n);
head=create(n);
printf("Output all nodes:\n");
print(head);
head=pdelete(head,60);
printf("Output fail-nodes:\n");
print(head);
}
struct student *create(int n)
{
struct student *head,*pnew,*ptail;
int i;
pnew=(struct student*)malloc(sizeof(struct student));
scanf("%s%f",pnew->name,&pnew->score);
head=ptail=pnew;
for(i=1;i<n;i++)
{
pnew=(struct student*)malloc(sizeof(struct student));
scanf("%s%f",pnew->name,&pnew->score);
ptail->next=pnew;
ptail=pnew;
}
ptail->next=NULL;
return head;
}
struct student *pdelete(struct student *head,int grade)
{
struct student *p,*pold;//删除满足指定条件的链表头部的连续若干节点
p=head;
while(head!=NULL&&head->score>=grade)
{
head=head->next;
free(p);
p=head;
}
if(head==NULL) return head;
p=head->next;
pold=head;
while(p!=NULL)
{
if(p->score>=grade)
{
pold->next=p->next;
free(p);
p=pold->next;
}
else{
pold=p;
p=p->next;
}
}

return head;

}
void print(struct student *head)
{
struct student *p=head;
while(p!=NULL)
{
printf("%s %.1f\n",p->name,p->score);
p=p->next;

}
}Cpp1.cpp(18) (20)(21)(23)function does not take 2 parameters

Answer 1

main函数开始时的函数声明有问题，函数声明就是事先通知编译器函数的调用方式以检查错误的啊，你注意一下它报错说“函数不需要参数”，就该知道函数声明、定义或调用形式有问题。
struct student *create();
struct student *pdelete();
void print();
错，改为
struct student *create(int);
struct student *pdelete(struct student* ,int);
void print(struct student *);

没问题

Answer 2

你的函数声明里 没有写参数，函数实现里有参数，不对应

Answer 3

你上面声明了void exit();是想自定义这个函数？声明的时候没有参数为什么调用的时候还带一个参数exit(0)，如果你是想用库函数就不要声明这个函数了，直接include stdlib.h 就好了