链表的总结

题目：将26个英文字母储存在链表中

#include <stdlib.h>
#include <stdio.h>
struct list
{
   char Ach;
   struct list* next;
};
void create( struct list* head , char* ch ) 而这个head是定义在局部函数的变量，当出局部函数的时候，head变量被销毁，而不会影响main函数中的head变量
{
   char arr = *ch;
   struct list* pf = NULL;
   struct list* tail = NULL;
   while( arr - 1 != 'z' )
   {
   struct list* pf = (struct list*)malloc(sizeof(struct list));
   pf->Ach = arr++;
   if( head == NULL )
   {
       head = pf;
   }
   else
   {
       tail->next = pf;
   }
   tail = pf;
   }

   tail->next = NULL;

}
void print( struct list* head )
{
   struct list* pf = head;
   while( pf != NULL )
   {
       printf("%c ",pf->Ach );
       pf = pf->next ;
   }
}
int main()
{
   struct list* head = NULL; head是定义在main函数中的变量
   char ch = 'a';
   create(head,&ch);

   print(head);
   return 0;
}

此时不会打印出结果

原因：请看上面的代码解析

此时我们思考一下应该怎么做

当交换二个数时，应该传地址吧

#include <stdlib.h>
#include <stdio.h>
struct list
{
   char Ach;
   struct list* next;
};
void create( struct list** head , char* ch )
{
   char arr = *ch;
   struct list* pf = NULL;
   struct list* tail = NULL;
   while( arr - 1 != 'z' )
   {
   struct list* pf = (struct list*)malloc(sizeof(struct list));
   pf->Ach = arr++;
   if( *head == NULL )
   {
       *head = pf;
   }
   else
   {
       tail->next = pf;
   }
   tail = pf;
   }

   tail->next = NULL;
   //return head;
}
void print( struct list* head )
{
   struct list* pf = head;
   while( pf != NULL )
   {
       printf("%c ",pf->Ach );
       pf = pf->next ;
   }
}
int main()
{
   struct list* head = NULL;
   char ch = 'a';
   create(&head,&ch);
   print(head);
   return 0;
}