Question

更具体地说，这段代码应该是Unix函数dc的较小克隆。链表似乎工作正常。如果我尝试使用c清除内存，添加更多数字，然后再次打印f我得到段错误。它似乎是打印链表中应该是Null节点的内容。

Interaction:
$ ./test
1 2 3
f
3
2
1
c
4 5
f
5
4
0
Segmentation Fault

这里是代码本身：

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

struct Node{
  int val;
  struct Node *next;
};

void cons_node(struct Node **head, int num)
{
  struct Node *newNode = malloc(sizeof(struct Node));
  newNode->val = num;
  newNode->next = NULL;
  if (*head == NULL){
    *head = newNode;
  }
  else {
    newNode->next = *head;
    *head = newNode;
  }
}

我假设问题出现在display func：

中

void display_list(struct Node *head)
{
  struct Node *current = head;
  while(current != NULL)
    {
      printf("%d\n", current->val);
      current = current->next;}

}

void print_top(struct Node *head)
{
  printf("%d\n", head->val);
}

或在明确的函数中：

void clear_stack(struct Node *head)
{
  struct Node *current;
  while ((current = head)!= NULL) {
    head = head->next;
    free(current);
  }
}

void vorpal_func(struct Node *head)
{ 
  struct Node *current;
  current = head;
  free(current);
}
int main(){

  int input;
  int first = 1;
  char quit = 'n';
  char inputchar = ' ';

  struct Node *head = NULL;

  while (quit == 'n'){

    if (scanf("%d", &input) == 1){
      cons_node(&head, input);
      first = 0;
    }

    else{
      inputchar = getchar();

      if(first == 1)
    printf("List is empty\n");

      else{
    switch (inputchar){
    case 'q':
      quit = 'y';
      break;
    case 'E':
      quit = 'y';
      break;
    case 'f':
      display_list(head);
      break;
    case 'p':
      print_top(head);
      break;
    case 'c':
      clear_stack(head);
      first = 1;
      break;
    case 't':
      vorpal_func(head);
      break;
    }
      }
    }
  }

  return 0;
}

我现在一直试图找出问题几个小时。有什么提示吗？

堆栈清除后打印链表时的Segfault

0 个答案: