为什么我在C队列中遇到分段错误(核心转储)?

时间:2016-09-30 03:32:47

标签: c linked-list segmentation-fault queue coredump

所以我试图在C中实现Queue数据结构,但我不断收到此错误:分段错误(核心转储)。我认为它来自我的入队方法。有人可以帮我解决这个问题......

这是我的queue.c文件

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

void createQueue(LinkedList* list)
{
  list = (LinkedList*)calloc(1, sizeof(LinkedList));
}

int isEmpty(LinkedList* list)
{
  int empty = 0;
  if(list->head == NULL)
  {
    empty = 1;
  }
  return empty;
}

void enqueue(LinkedList* list, Person p)
{
  LinkedListNode* newNode = (LinkedListNode*)malloc(sizeof(LinkedListNode));
  newNode->data = p;
  if (isEmpty(list))
  {
    list->head = list->tail = newNode;
  } else
  {
    LinkedListNode* first = list->head;
    first->previous = newNode;
    newNode->next = list->head;
    list->head = newNode;
  }
  list->count ++;
}

Person dequeue(LinkedList* list)
{
  LinkedListNode* temp;
  temp = list->tail;
  list->tail = list->tail->previous;
  list->count --;

  return temp->data;
}

void viewQueue(LinkedList* list)
{
  LinkedListNode* current;
  current = list->head;
  while (current != NULL)
  {
    printf("Name: %s Birthdate: %s\n", current->data.name, current->data.birthdate);
    current = current->next;
  }
  printf("\nPeople: %d\n", list->count);
}

这是我的头文件......

#ifndef LINKED_LIST
#define LINKED_LIST

#define FALSE  0
#define TRUE !FALSE

typedef struct{
  char name[20];
  char birthdate[10];
} Person;

typedef struct LinkedListNode{
  Person data;
  struct LinkedListNode* next;
  struct LinkedListNode* previous;
} LinkedListNode;

typedef struct{
  LinkedListNode* head;
  LinkedListNode* tail;
  int count;
} LinkedList;

void createQueue(LinkedList* list);
int isEmpty(LinkedList* list);
void enqueue(LinkedList* list, Person p);
Person dequeue(LinkedList* list);
void viewQueue(LinkedList* list);

#endif

我也有这种方法来测试我的队列。当我运行它时,我得到一个分段错误。我的程序怎么了?

#include <stdio.h>
#include "queue.h"

int main(void)
{
  LinkedList* list;
  createQueue(list);
  Person p1 = {"J Cole", "28/01/1985"};
  Person p2 = {"Kendrick Lamar", "19/01/1994"};
  Person p3 = {"Slim Shady", "17/10/1972"};
  Person p4 = {"Tupac Shakur", "12/09/1971"};

  enqueue(list, p1);
  enqueue(list, p2);
  enqueue(list, p3);
  enqueue(list, p4);

  viewQueue(list);
  return 0;
}

2 个答案:

答案 0 :(得分:2)

您正在传递列表指针的值。您需要传递参考。

void createQueue(LinkedList** list)
{
  *list = (LinkedList*)calloc(1, sizeof(LinkedList));
}

从主

开始
createQueue(&list);

答案 1 :(得分:2)

将值(如指针)传递给例程时,为了使例程设置该值,您必须传递其地址。因此,这一行:

createQueue(list);

应该是:

createQueue(&list);

如果您想将createQueue设置为设置变量的值。你还需要:

void createQueue(LinkedList** list)
{
    *list = (LinkedList*) calloc(1, sizeof(LinkedList));
}

不要担心 - 每个C程序员都会在某些时候犯这个错误。有关更多信息,请搜索“按值传递或引用”和“指向指​​针的c指针”。