Question

您好我需要帮助释放我在C中的链接列表。当我运行此代码时，我得到了一个段错误11.一切直到解放都能完美地运行。谢谢你的帮助！

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

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

int main(void){
    struct node *root, *next;

    root = malloc(sizeof(struct node));
    root -> x = 0;
    root -> next = NULL;

    next = root;

    for (int i = 0; i <=10; i++){
        next -> next = malloc(sizeof(struct node));
        next -> x = i;
        next = next -> next;
    }

    next = root;

    for (int j = 0; j <= 10; j ++){
        printf("%i", next -> x);
        next = next -> next;
    }
        next = root;

    while(next != NULL){
        next = root;
        root = root -> next;
        free(next);
    }
    free(root);
}

Answer 1

原始代码有一些问题：

1）while(next != NULL)循环尝试使用已释放的节点。

2）循环已经处理了释放root（next = root;），因此无需单独释放root。

3）为了使while循环正常工作，列表的tail / head必须正确NULL终止。（我在第一个for循环中添加了终止）

4）第二个循环应该打印所有x值。它没。柜台短了一个数字。

该计划的改进变化如下。请检查程序输出。

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

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

int main(void){
    struct node *root, *next, *to_free;

    root = malloc(sizeof(struct node));                   // 1 malloc

    root->x = 777;
    root->next = NULL;

    printf("allocating memory for root. root->x = %i\n\n", root-> x);

    next = root; // next points to "head" (root)

    for (int i = 0; i <= 10; i++){
        next->next = malloc(sizeof(struct node));      // 11 mallocs

        printf("allocating memory for next. next->x = %i\n", i+1 );   

        next->next->x = i+1;                           //

        next = next->next;
        next->next = NULL;                             // list termination is needed!
    }
    printf("\n");

    next = root; // next points to "head" (root)

    for (int j = 0; j <= 11; j ++){                   // 12 nodes has to be printed! 
        printf("Printing x = %i\n", next->x);
        next = next->next;
    }
    printf("\n");

    next = root;   //  next points to "head" (root)

    while(next != NULL)         
    {   
        to_free = next;         // we will free `next` as `to_free` soon

        next = next->next;      // move to the next node   

        printf("deallocating node with x = %i\n", to_free->x);

       free(to_free);          // now free the remembered `next`
    }
 return 0;
}

输出：

allocating memory for root. root->x = 777

allocating memory for next. next->x = 1
allocating memory for next. next->x = 2
allocating memory for next. next->x = 3
allocating memory for next. next->x = 4
allocating memory for next. next->x = 5
allocating memory for next. next->x = 6
allocating memory for next. next->x = 7
allocating memory for next. next->x = 8
allocating memory for next. next->x = 9
allocating memory for next. next->x = 10
allocating memory for next. next->x = 11

Printing x = 777
Printing x = 1
Printing x = 2
Printing x = 3
Printing x = 4
Printing x = 5
Printing x = 6
Printing x = 7
Printing x = 8
Printing x = 9
Printing x = 10
Printing x = 11

deallocating node with x = 777
deallocating node with x = 1
deallocating node with x = 2
deallocating node with x = 3
deallocating node with x = 4
deallocating node with x = 5
deallocating node with x = 6
deallocating node with x = 7
deallocating node with x = 8
deallocating node with x = 9
deallocating node with x = 10
deallocating node with x = 11

Answer 2

释放节点的循环检查{/ 1}} next 后是否调用了NULL。

Answer 3

正如另一个答案已经说明的那样，下面的free()位于错误的位置，因为它释放了在while循环条件下检查的内存。

while(next != NULL){
    root = root->next;
    next = root;
    free(next);
}

但是，如果您按照评论中的建议将该块的第一个语句移动到块的末尾，那么您的问题可能就是这个循环本身完全足以释放整个列表，因此声明free(root)循环之后可能是双重释放，这通常是一个错误。你可以删除它。

Answer 4

下面列出了代码中的几个问题，其中最重要的是数字1。

列表（根）的第一项中的指针next不会指向NULL，因为它将在第一个for循环中更改。因此，while(next != NULL)将不起作用，因为您的链接列表中没有指向NULL的项目。
malloc将返回类型为void *的指针，因此最好将其强制转换为您的next。
进行内存分配后，应检查结果是否成功分配了内存。另外，我建议您使用typedef来定义链接列表。

您的代码应如下所示：

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

typedef struct node{
    int x;
    struct node *next, *temp;
} node_t;

int main(int argc, char *argv[]){
  node_t *root, *next, *temp;

  root = (node_t *) malloc(sizeof(node_t));
  if(root == NULL){
    /* Do somthing */
    fprintf(stderr, "unable to allocate memory\n");
    exit(1);
  }
  root -> x = 100;
  root -> next = NULL;
  for (int i = 0; i <= 10; i++){
    next = (node_t *) malloc(sizeof(node_t));
    if(next == NULL){
      /* Do somthing */
      fprintf(stderr, "unable to allocate memory\n");
      exit(1);
    }
    next -> x = i;
    next -> next = root;
    root = next;
  }
  temp = next;
  while(temp != NULL){
    printf("x value is: %d\n", temp -> x);
    temp = temp -> next;
  }

  while(next != NULL){
    temp = next -> next;
    free(next);
    next = temp;
  }
}

C中的自由链接列表

4 个答案: