单个链接列表按循环顺序打印

Question

我试图按照我在链接列表中创建每个节点的顺序打印出链接列表。例如，它应该打印出“0 1 2 3 4”，但我的代码是错误的，不会打印出任何东西。我认为问题出在我的for循环中。

#include <stdio.h> 
#include <stdlib.h> 
struct node 
{ 
    int data; 
    struct node *next; 
}; 

int main(void) 
{ 
    struct node *head = NULL; 
    struct node *tail = NULL; 
    struct node *current; 
    current = head; 
    int i; 
    for(i = 0; i <= 9; i++) 
    { 
     current = (struct node*)malloc(sizeof(struct node)); 
     current-> data = i; 
     current-> next = tail; 
     tail = current; 
     current = current->next; 
    } 

    current = head; 
    while(current) 
    { 
     printf("i: %d\n", current-> data); 
     current = current->next; 
    } 
} 

Answer 1

在构建列表时，您似乎被指针算术绊倒了。试试这个：

int main(void) 
{ 
    struct node *head = NULL; 
    struct node *tail = NULL; 
    struct node *current; 
    int i; 
    for (i=0; i <= 9; i++) 
    { 
     struct node *temp = (struct node*)malloc(sizeof(struct node)); 
     temp-> data = i; 
     temp-> next = NULL; 
     if (head == NULL)   // empty list: assign the head 
     { 
      head = temp; 
      tail = temp; 
      current = head; 
     } 
     else       // non-empty list: add new node 
     { 
      current-> next = temp; 
      tail = temp; 
      current = current->next; 
     } 
    } 

    // reset to head of list and print out all data 
    current = head; 

    while (current) 
    { 
     printf("i: %d\n", current-> data); 
     current = current->next; 
    } 
}