Question

C语言单链表的逆序，出现了错误，

//创建链表和遍历链表都没有问题，当程序运行到单链表逆序也就是OPP函数时崩溃，求告诉原因，不胜感激

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 100
typedef struct student{
char name[MAX];
int numb;
struct student *pNext;
}NODE,*PNODE;
PNODE crateList();
PNODE opp(PNODE pList);
void traverseList(PNODE pList);
int main(void){
PNODE pList;
pList=crateList();
traverseList(pList);
pList=opp(pList);
traverseList(pList);
return(0);
}
PNODE crateList(){
PNODE pHead=NULL,pCurrent,pPre;
char input[MAX];
puts("enter the name");
while(gets(input)!=NULL&&input[0]!='\0'){
pCurrent=(PNODE)malloc(sizeof(NODE));
if(pCurrent==NULL)
return(0);
if(pHead==NULL)
pHead=pCurrent;
else
pPre->pNext=pCurrent;
pCurrent->pNext=NULL;
strcpy(pCurrent->name,input);
puts("enter the number:");
scanf("%d",&pCurrent->numb);
pPre=pCurrent;
puts("enter the next one or enter to leave");
while(getchar()!='\n')
continue;
}
return(pHead);
}
void traverseList(PNODE pList){
PNODE pCurrent;
pCurrent=pList;
if(pCurrent==NULL)
puts("no data enter!");
else
while(pCurrent!=NULL){
printf("the name is %s,and number is %d\n",pCurrent->name,pCurrent->numb);
pCurrent=pCurrent->pNext;
}
}

PNODE opp(PNODE pList){
PNODE pFirst,pMid,pLast,pCurrent;
if(!pList&&pList->pNext==NULL){
puts("no data to convert!");
return(0);
}
pFirst=pList;
pMid=pFirst->pNext;
pLast=pMid;
while(pMid){
pFirst->pNext=NULL;
pMid->pNext=pFirst;
pFirst=pMid;
pMid=pLast;
pLast=pMid->pNext;
}
}

Answer 1

//分析opp函数

PNODE opp(PNODE pList){
	PNODE pFirst,pMid,pLast,pCurrent;
	if(!pList&&pList->pNext==NULL){//应改为if(!pList||pList->pNext==NULL)
	                               //&&表示两个条件都要同时满足,||表示只
	                               //满足一个，关键是使用&&,当pList为NULL
	                               //时，仍然会判断pList->pNext这会导致
	                               //访问空指针。
		puts("no data to convert!");
		return(0);
	}
	pFirst=pList;
	pMid=pFirst->pNext;
	pLast=pMid;    //应该是pLast=pMid->pNext;这行可以放入循环开头
	while(pMid){
		pFirst->pNext=NULL;//每次循环都会导致pFirst->pNext置为空指针
		                    //这应该不是期望的操作。因pFirst指针
		                    //会随循环而向下跳转，这是链首节点
		                    //的处理，应该放到循环外
		pMid->pNext=pFirst;
		pFirst=pMid;    
		pMid=pLast;
		pLast=pMid->pNext;//在无法确认pMid是否为NULL时，直接使用
		                   //pMid->pNext是不安全的。可以放到循环
		                   //开头
	}
}

思路没问题，但存在代码输入错误，赋值的时机错误，调整后为：

PNODE opp(PNODE pList){
	PNODE pFirst,pMid,pLast,pCurrent;
	if(!pList||pList->pNext==NULL){
		puts("no data to convert!");
		return(0);
	}
	pFirst=pList;
	pMid=pFirst->pNext;
	pFirst->pNext=NULL;	
	while(pMid){
		pLast=pMid->pNext; 
		pMid->pNext=pFirst;
		pFirst=pMid;
		pCurrent=pMid;
		pMid=pLast;
	}
	return pCurrent;
}

Answer 2

PNODE opp(PNODE pList){
 PNODE pFirst,pMid,pLast,pCurrent;
 if(!pList&&pList->pNext==NULL){
//  puts("no data to convert!");  个人认为没必要输出这种信息
  return(pList);
 }
 pFirst=pList;
 pMid=pFirst->pNext;
 pFirst->pNext=NULL;        // 不能每次都赋值成空，只有原来第一个要这样
 pLast=pMid->pNext;
 pMid->pNext=pFirst;        // 要把第一个节点连接上
 while(pLast){              // 后面就顺次向后移动指针
  pFirst=pMid;
  pMid=pLast;
  pLast=pMid->pNext;
  pMid->pNext=pFirst;       // 并把以前的链接在这后面
 }
 return pMid;
}

这样就没问题了