c - 本链表程序回顾

Question

这是链表上的一个程序，它打印给定的输入。但是是否可以在不使用 if(start==NULL) ...else .. 语句的情况下做到这一点？

#include <stdio.h>
#include <malloc.h>
struct node
{
    int data;
    struct node* next;
}* start = NULL;

void main()
{
    struct node* tmp;
    struct node *ptr, *tmp1;

    int n = 0;
    int choice = 2;

    while (1)
    {
        printf("Enter a number less than 3 to continue");
        scanf("%d", &choice);
        if (choice >= 3)
            break;

        printf("Enter the input");
        scanf("%d", &n);
        // n=n+10;
        if (start == NULL)
        {
            tmp = (struct node*)malloc(sizeof(struct node));
            start = tmp;
            start->data = n;
            ptr = start;
        }
        else
        {
            tmp = (struct node*)malloc(sizeof(struct node));
            ptr->next = tmp;
            tmp->data = n;
            ptr = tmp;
        }
    }
    tmp->next = NULL;


    printf("\nThe output of the linked list is\n");
    n = 0;
    ptr = start;
    while (ptr != NULL)
    {
        printf("\n  1 ptr =           %d", ptr->data);
        ptr = ptr->next;
    }
}

Answer 1

is it possible to do it Without using the if(start==NULL) ...else .. statement?

是

创建一个头节点并只使用它的 .next 成员。

#include <stdio.h>
#include <malloc.h>

struct node {
  int data;
  struct node* next;
};

int main(void) {
  struct node head = {.next = NULL};  // only need .next 
  struct node *ptr = &head;

  while (1) {
    // ...

    int n;
    printf("Enter the input ");
    fflush(stdout);
    if (scanf("%d", &n) != 1) {
      break;
    }
    ptr->next = malloc(sizeof *ptr->next);
    if (ptr->next == NULL) {
      break;
    }
    ptr = ptr->next;
    ptr->data = n;
    ptr->next = NULL;
  }

  printf("\nThe output of the linked list is\n");
  ptr = head.next; // The start of the list is here
  while (ptr) {
    printf("  1 ptr =           %d\n", ptr->data);
    ptr = ptr->next;
  }
}