我正在尝试编写一个程序,该程序从用户(来自 STDIN)获取行并将它们存储在链接列表中。
现在我只得到一行并终止程序。
如何更改代码以继续从标准输入获取行?
另外,如果有人能告诉我我是否正在按应有的方式分配和释放内存,那将非常有帮助。
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);
}
最佳答案
几个问题:
\n
时终止阅读。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;
}
}
}
data
将始终指向 pStr
的最新内容。char* new_data = malloc(sizeof(pStr));
new_data = pStr; //Memory leak here
new_node->data = new_data;
将其更改为
char* new_data = malloc(i+1);
strcpy(new_data, pStr);
new_node->data = new_data;
tail
。 else
{
tail->next = new_node;
}
至
else
{
tail->next = new_node;
tail = new_node;
}
关于c - 如何从标准输入读取多行并将其存储在链表中,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/56442478/