C链表-C语言入门经典例题

struct student

{

long num;

float score;

struct student *next;

};

注意：只是定义了一个struct student类型，并未实际分配存储空间。只有定义了变量才分配内存单元。

#include<iostream>
using namespace std;
int main() {
    struct student a,b,c,*head,*p;
    a.num = 99101;
    a.score = 89.5;
    b.num = 99103;
    b.score = 90;
    c.num = 99107;
    c.score = 85;        /*对结点的num和score成员赋值*/
    head = &a;        /*将结点a的起始地址赋给头指针head*/
    a.next = &b;        /*将结点b的起始地址赋给a结点的next成员*/
    b.next = &c;        /*将结点c的起始地址赋给b结点的next成员*/
    c.next = NULL;    /*c结点的next成员不存放其他结点地址*/
    p = head;        /*使p指针指向a结点*/
    do {
        cout<<p->num<<" "<<p->score;    /*输出p指向的结点的数据*/
        p=p->next;    /*使p指向下一结点*/
    } while (p!=NULL);    /*输出完c结点后p的值为NULL*/
    return 0;
}

C语言入门经典 单链表！

/* Program 11.4   Daisy chaining the horses */
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>

int main(void)
{
  struct horse                        /* Structure declaration       */
  {
    int age;
    int height;
    char name[20];
    char father[20];
    char mother[20];
    struct horse *next;               /* Pointer to next structure   */
  };

  struct horse *first = NULL;         /* Pointer to first horse      */
  struct horse *current = NULL;       /* Pointer to current horse    */
  struct horse *previous = NULL;      /* Pointer to previous horse   */

  char test = '\0';                   /* Test value for ending input */

  for( ; ; )
  {
    printf("\nDo you want to enter details of a%s horse (Y or N)? ",
                                       first != NULL?"nother " : "" );
    scanf(" %c", &test );
    if(tolower(test) == 'n')
      break;

    /* Allocate memory for a structure */
    current = (struct horse*) malloc(sizeof(struct horse));

    if(first == NULL)
      first = current;                /* Set pointer to first horse  */

    if(previous != NULL)

      previous -> next = current; /* Set next pointer for previous horse */

    printf("\nEnter the name of the horse: ");
    scanf("%s", current -> name);     /* Read the horse's name       */

    printf("\nHow old is %s? ", current -> name);
    scanf("%d", &current -> age);     /* Read the horse's age        */

    printf("\nHow high is %s ( in hands )? ", current -> name );
    scanf("%d", &current -> height);  /* Read the horse's height     */

    printf("\nWho is %s's father? ", current -> name);
    scanf("%s", current -> father);   /* Get the father's name       */

    printf("\nWho is %s's mother? ", current -> name);
    scanf("%s", current -> mother);   /* Get the mother's name       */

    current->next = NULL;             /* In case it's the last...    */
    previous = current;               /* Save address of last horse  */
  }

  /* Now tell them what we know. */
  current = first;                    /* Start at the beginning      */

  while (current != NULL)      /* As long as we have a valid pointer */
  { /* Output the data*/
    printf("\n\n%s is %d years old, %d hands high,",
                       current->name, current->age, current->height);
    printf(" and has %s and %s as parents.", current->father,
                                               current->mother);
    previous = current;    /* Save the pointer so we can free memory */
    current = current->next;          /* Get the pointer to the next */
    free(previous);                   /* Free memory for the old one */
  }
  return 0;
}

双向链表：

/* Program 11.5 Daisy chaining the horses both ways */
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>

int main(void)
{
  struct horse                        /* Structure declaration       */
  {
    int age;
    int height;
    char name[20];
    char father[20];
    char mother[20];
    struct horse *next;             /* Pointer to next structure     */
    struct horse *previous;         /* Pointer to previous structure */
  };

  struct horse *first = NULL;         /* Pointer to first horse      */
  struct horse *current = NULL;       /* Pointer to current horse    */
  struct horse *last = NULL;          /* Pointer to previous horse   */

  char test = '\0';                   /* Test value for ending input */

  for( ; ; )
  {
    printf("\nDo you want to enter details of a%s horse (Y or N)? ",
                                        first == NULL?"nother " : "");
    scanf(" %c", &test );
    if(tolower(test) == 'n')
      break;

    /* Allocate memory for each new horse structure */
    current = (struct horse*)malloc(sizeof(struct horse));

    if( first == NULL )
    {
      first = current;                /* Set pointer to first horse  */
     current->previous = NULL;
    }
    else
    {
      last->next = current;    /* Set next address for previous horse */
      current->previous = last; /* Previous address for current horse */
    }

    printf("\nEnter the name of the horse: ");
    scanf("%s", current -> name );    /* Read the horse's name   */

    printf("\nHow old is %s? ", current -> name);
    scanf("%d", &current -> age);     /* Read the horse's age    */

    printf("\nHow high is %s ( in hands )? ", current -> name);
    scanf("%d", &current -> height);  /* Read the horse's height */

    printf("\nWho is %s's father? ", current -> name);
    scanf("%s", current -> father);   /* Get the father's name   */

    printf("\nWho is %s's mother? ", current -> name);
    scanf("%s", current -> mother);   /* Get the mother's name   */

    current -> next = NULL;      /* In case it's the last horse..*/
    last = current;              /* Save address of last horse   */
  }

  /* Now tell them what we know. */
  while(current != NULL)         /* Output horse data in reverse order */
  {
    printf("\n\n%s is %d years old, %d hands high,",
               current->name, current->age, current->height);
    printf(" and has %s and %s as parents.", current->father,
                                            current->mother);
    last = current;       /* Save pointer to enable memory to be freed */
    current = current->previous; /* current points to previous in list */
    free(last);                 /* Free memory for the horse we output */
  }
  return 0;
}