Question

int main(int argc, char *argv[])
{
  printf("successfully started main\n");
  struct uf_list myList;
  uf_list_allocate(&myList);
  printf("successfully allocated myList\n");
  insert_node(&myList, 'c');
  printf("successfully inserted into myList\n");

  return 0;
}

...

void uf_list_allocate(struct uf_list *list)
{
  list = malloc(sizeof(struct uf_list));
  if(list == NULL)
    {fprintf(stderr, "no memory for allocate");}
  list->head = list->tail = NULL;
}
//--------------------------------------------------------------------------------------
void insert_node(struct uf_list *list, const char label)   
{
  struct uf_node *it = malloc(sizeof(struct uf_node));
  if(it == NULL)
    {fprintf(stderr, "no memory for insert");}

  it->c = label;
  it->next = NULL;                                     
  it->rep = NULL;

  if(list->head == NULL)                                //the list is empty
    { list->head = list->tail = it;}
  else
    { list->tail->next = it; list->tail = it; }

  it->rep = list->head;
}
/*----------------------------------------------------------------------------*/
struct uf_node
{
  char c;
  struct uf_node *next;
  struct uf_node *rep;
};
/*----------------------------------------------------------------------------*/
struct uf_list
{
  struct uf_node *head;
  struct uf_node *tail;
};

我尝试将一个元素从main插入到我的列表中时出现分段错误。是什么导致了分段错误？如果您需要更多信息，例如structs的定义，请告诉我们。

编辑：我意识到我做了什么。在allocate内，我更改了本地变量list.的地址。这意味着myList没有发生任何变化。但是，现在我有以下谜题：我将myList的声明置于main,之外，一切正常：

struct uf_list myList;

int main(int argc, char *argv[])
{
  printf("successfully started main\n");
  uf_list_allocate(&myList);
  printf("successfully allocated myList\n");
  insert_node(&myList, 'c');
  insert_node(&myList, 'd');
  insert_node(&myList, 'e');
  printf("successfully inserted into myList\n");
  print_uf_list(&myList); 


  return 0;
}

我无法弄明白为什么。似乎应该应用相同的逻辑，即，因为我将myList的地址传递给allocate但随后更改了局部变量list地址并对该地址进行操作，这是如何反映在{{{ 1}}其内存地址没有被操作？

Answer 1

在分配中，你不会返回任何东西。达是问题所在。在main中，您应该只有一个指针作为局部变量，并为其分配分配器函数返回的内容。

修改

更简单，因为它已经被分配（在main的堆栈上），你可以从该函数中删除分配代码，并具有初始化函数。这就是你所需要的：

Uf_list_init(struct uf_list *list) { list->head = list->tail = NULL; }

在原始代码中：

list = malloc(sizeof(struct uf_list));

你有一个指向te struct的指针，但你用一个全新的指针覆盖它。

Answer 2

C按值传递参数。 uf_list_allocate应该uf_list **listRef，以便可以对其进行修改。

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

struct uf_node
{
  char c;
  struct uf_node *next;
  struct uf_node *rep;
};

struct uf_list
{
  struct uf_node *head;
  struct uf_node *tail;
};

void uf_list_allocate(struct uf_list **listRef)
{
  *listRef = malloc(sizeof(struct uf_list));
  if(*listRef == NULL)
    {fprintf(stderr, "no memory for allocate"); exit(0);}
  (*listRef)->head = (*listRef)->tail = NULL;
}

void insert_node(struct uf_list *list, const char label)
{
  struct uf_node *it = malloc(sizeof(struct uf_node));
  if(it == NULL)
    {fprintf(stderr, "no memory for insert"); exit(0);}

  it->c = label;
  it->next = NULL;
  it->rep = NULL;

  if(list->head == NULL)                                //the list is empty
    { list->head = list->tail = it;}
  else
    { list->tail->next = it; list->tail = it; }

  it->rep = list->head;
}

int is_empty(const struct uf_list *list)
{
  return list->head == NULL;
}

void remove_node(struct uf_list *list)
{
  if (is_empty(list))
  {
    printf("List underflow\n"); exit(0);
  }
  else
  {
    struct uf_node *oldhead = list->head;
    list->head = list->head->next;
    if (list->tail == oldhead)
      list->tail = NULL;
    free(oldhead);
    printf("Node removed\n");
  }
}

void deallocate(struct uf_list **listRef)
{
  struct uf_list *list = *listRef;
  if(!is_empty(list))
  {
    while(!is_empty(list))
      remove_node(list);
  }
  free(list);
  list = NULL;
  printf("List deallocated\n");
}

void printList(const struct uf_list *myList)
{
  struct uf_node *cur = myList->head;
  while(cur!=NULL)
  {
    printf("%c -> ", cur->c);
    cur = cur->next;
  }
  printf("\n");
}

int main(int argc, char *argv[])
{
  printf("successfully started main\n");
  struct uf_list *myList;
  uf_list_allocate(&myList);
  printf("successfully allocated myList\n");

  insert_node(myList, 'c');
  printf("successfully inserted c into myList\n");

  insert_node(myList, 'd');
  printf("successfully inserted d into myList\n");
  printList(myList);

  insert_node(myList, 'e');
  printf("successfully inserted e into myList\n");
  printList(myList);

  remove_node(myList);
  printf("successfully removed c (head) from myList\n");
  printList(myList);

  deallocate(&myList);

  return 0;
}

我的链表有什么问题？

2 个答案: