linux c语言链表的简单应用之创建链表

灰太狼 2021-11-16 12:34 714阅读 0赞

/\*  \*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*  
\* Filename: link.c  
\* Description:  
\* Version: 1.0  
\* Created: 2011骞?4鏈?9鏃?17鏃?2鍒?3绉?  
\* Revision: none  
\* Compiler: gcc  
\* Author: wen hao (WH), hnrain1004@gmail.com  
\* Company: sunplusapp  
\* \*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\* \*/   
  
  
\#include  < stdio.h >   
\#include  < stdlib.h >   
\#include  " link.h "   
  
 \#define  LEN sizeof(struct stu)   
 // 声明结构体   
 typedef  struct  stu  
\{  
 int  num;  
 char  name\[ 10 \];  
 struct  stu  \* next;  
\}TYPE;  
  
 // 链表创建函数,返回类型为结构体指针类型   
 TYPE  \*  create( int  n)  
\{  
TYPE  \* head, \* prev, \* curre;  
 int  i;  
  
 for (i  =  0 ; i  <  n; i \++ )  
\{  
curre  =  (TYPE  \* )malloc(LEN); // 申请空间   
  printf( " input number and name:\\n " );  
scanf( " %d %s " , & curre \-> num,curre \-> name); // 等待用户输入数据   
   
 if (i  ==  0 )  
prev = head = curre;  
 else   
prev \-> next = curre;  
prev = curre;  
\}  
curre \-> next  = NULL;  
 return  head;  
\}  
  
 // 打印输出函数,形参为链表头指针   
 void  print(TYPE  \* head)  
\{  
printf( " \\nthe link message is :\\n " );  
printf( " number \\t\\tname \\n " );  
 while (head != NULL) // 如果没有指向链表尾就一直打印   
  \{  
printf( " %d\\t\\t%s\\n " ,head \-> num,head \-> name);  
head = head \-> next;  
\}  
printf( " \\n " );  
\}  
  
 int  main( void )  
\{  
TYPE  \* head; // 定义结构体变量   
  head  =  create( 3 ); // 创建链表   
  print(head); // 打印链表   
   
 return  0 ;  
\}

转载于:https://www.cnblogs.com/hnrainll/archive/2011/04/29/2032708.html