如何从标准输入中读取多行并将其存储在链表中

Question

我正试图编写一个程序，从用户（从STDIN）获取行并将它们存储在一个链表中。

现在我只能获取一行并终止程序。如何更改代码以不断从stdin获取行？

另外，如果有人告诉我是否正在按照应该分配和释放内存的方式进行操作，将非常有帮助。

谢谢。

#include <stdio.h>
#include <stdlib.h>

int BUFF_SIZE = 128;

struct Node {
    char* data;
    struct Node* next;
};

struct Node* head = NULL;
struct Node* tail = NULL;

void free_list(struct Node* head)
{
    if (head != NULL)
    {
        free_list(head->next);
        free(head);
    }
}

int main()
{
    int curr_size = 0;

    char* pStr = malloc(BUFF_SIZE);
    curr_size = BUFF_SIZE;

    printf("%s", "please print multiple lines\n");
    if (pStr != NULL)
    {
        char c;
        int i = 0;

        while ((c = getchar()) != '\n' && c != EOF)
        {
            pStr[i++] = c;

            if (i == curr_size)
            {
                curr_size = i + BUFF_SIZE;
                pStr = realloc(pStr, curr_size);
                if (pStr == NULL) return;
            }
        }
        pStr[i] = '\0';

        struct Node* new_node = malloc(sizeof(struct Node*));
        char* new_data = malloc(sizeof(pStr));
        new_data = pStr;
        new_node->data = new_data;

        if (head == NULL)
        {
            head = new_node;
            tail = new_node;
        }

        else
        {
            tail->next = new_node;
        }
    }

    free_list(head);
}

- beginnerCoder

Answer 1

一些问题：

1. As of now you are terminating the reading upon reacieving the \n.

   if (pStr == NULL) return; //error
   
   int c;
   int i = 0;
   
   while ((c = getchar()) != EOF)
   {
      /*New word, insert into linked list*/
      if (c == '\n'){
          pStr[i] = '\0';
   
          struct Node* new_node = malloc(sizeof(*new_node));
          char* new_data = malloc(i+1);
          strcpy(new_data, pStr);
          new_node->data = new_data;
   
          if (head == NULL)
          {
               head = new_node;
               tail = new_node;
          }
          else
          {
               tail->next = new_node;
               tail = new_node;
          }
          i = 0; //Reset the index
      }
      else {
   
          pStr[i++] = c;
          if (i == curr_size)
          {
              curr_size = i + BUFF_SIZE;
              pStr = realloc(pStr, curr_size);
              if (pStr == NULL) return;
          }
      }
   }
   

2. Memory leaks and node data will be always pointing to latest content of pStr.

   char* new_data = malloc(sizeof(pStr)); 
   new_data = pStr;   //Memory leak here
   new_node->data = new_data;
   

   change it to

   char* new_data = malloc(i+1);
   strcpy(new_data, pStr);
   new_node->data = new_data;
   

   sizeof(pStr) is size of pointer not the string length.

3. You need to update tail after each node is inserted to list.

    else
    {
        tail->next = new_node;
    }
   

   to

    else
   {
       tail->next = new_node;
       tail = new_node;
    }
   

- kiran Biradar

---

抱歉，我是新来的，我该怎么做？ - beginnerCoder

在答案旁边的点赞计数器下面，有一个复选标记。点击它，它应该变成绿色。另外，需要注意的是，这段代码仍然存在问题。c 应该是 int，而不是 char。 - WhozCraig

@WhozCraig 谢谢，为什么它应该是 int 类型？ - beginnerCoder

因为这就是 getchar() 返回的内容，而这非常重要，尤其是在与 EOF 进行比较时（正如您所做的那样）。 - WhozCraig

struct Node* new_node = malloc(sizeof(struct Node*)); (copied from the original source) is wrong. Must be sizeof(struct Node) or sizeof *new_node - Ingo Leonhardt