在链表函数"1 块中的 16 个字节肯定在丢失记录 1 of 1"中丢失内存泄漏"

Question

我正在学习C，并编写了一些函数来处理C中的链表。然而，我有内存泄漏(我想是因为第28行和第29行中节点*列表的重新分配(，并且正在寻找解决问题的方法。

我的代码：

#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
int number;
struct node *next;
}
node;
node *declare(int value);
int count(node *list);
node *lead(node *list, int value);
void trail(node *list, int value);
node *append(node *list, int value, int index);
node *fetch_node(node *list, int index);
int fetch(node *list, int index);
int *list_to_array(node *list);
node *array_to_list(int array[], int range);
void free_list(node *list);
node *delete(node *list, int index);
void print(node *list);
int main(void)
{
node *list = declare(1);
list = lead(list, -1);
trail(list, 2);
trail(list, 44);
list = delete(list, 0);
append(list, 232, 1);
int *array = list_to_array(list);
node *list_new = array_to_list(array, 4);
printf("%dn", fetch(list_new, 3));
print(list);
free_list(list);
free_list(list_new);
free(array);
}
node *declare(int value)
{
node *n = malloc(sizeof(node));
n->number = value;
n->next = NULL;
return n;
}
int count(node *list)
{
node *tmp = list;
int count = 1;
while (tmp->next != NULL)
{
tmp = tmp->next;
count++;
}
return count;
}
node *lead(node *list, int value)
{
node *n = declare(value);
node *tmp = list;
n->next = tmp;
return n;
}
void trail(node *list, int value)
{
node *n = declare(value);
node *tmp = list;
while (tmp->next != NULL)
{
tmp = tmp->next;
}
tmp->next = n;
}
node *append(node *list, int value, int index)
{
if (index == 0)
{
return lead(list, value);
}
else
{
int range = count(list);
node *n = declare(value);
node *prev = fetch_node(list, index - 1);
node *follow = fetch_node(list, index);
prev->next = n;
n->next = follow;
return list;
}
}
node *fetch_node(node *list, int index)
{
node *tmp = list;
if (index < count(list))
{
for (int i = 0; i < index; i++)
{
tmp = tmp->next;
}
}
else
{
fprintf(stderr, "VALUE IN FETCH OUT OF RANGE, RETURNING POINTER TO LISTn");
}
return tmp;
}
int fetch(node *list, int index)
{
node *fetched = fetch_node(list, index);
return fetched->number;
}
node *delete(node *list, int index)
{
if (index == 0)
{
node *follow = list->next;
list = list->next;
}
else
{
node *fetched = fetch_node(list, index);
node *prev, *follow;
if (index > 0)
{
prev = fetch_node(list, index - 1);
}
else
{
prev = list;
}
int range = count(list);
if (index != range - 1 && index < range)
{
follow = fetch_node(list, index + 1);
}
else
{
follow = NULL;
}
prev->next = follow;
}
return list;
}
int *list_to_array(node *list)
{
int range = count(list);
int *array = malloc(sizeof(int) * range);
for (int i = 0; i < range; i++)
{
array[i] = fetch(list, i);
}
return array;
}
node *array_to_list(int array[], int range)
{
node *list = declare(array[0]);
for (int i = 1; i < range; i++)
{
trail(list, array[i]);
}
return list;
}
void free_list(node *list)
{
int range = count(list);
node *tmp = list;
node *del;
for (int i = 0; i < range; i++)
{
del = tmp;
tmp = tmp->next;
free(del);
}
}
void print(node *list)
{
int range = count(list);
printf("[");
for (int i = 0; i < range; i++)
{
int value = fetch(list, i);
printf("%d", value);
if (i != range - 1)
{
printf(", ");
}
}
printf("]n");
}

这是Valgrind的输出：

==18184== 16 bytes in 1 blocks are definitely lost in loss record 1 of 1
==18184==    at 0x4C31B0F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==18184==    by 0x4007C4: declare (linked_lists.c:45)
==18184==    by 0x400806: lead (linked_lists.c:65)
==18184==    by 0x4006F3: main (linked_lists.c:29)

有什么想法吗？

Answer 1

您的delete函数正在创建内存泄漏，因为它无法释放已删除的节点。

添加代码以释放它。

node *delete(node *list, int index)
{
if (index == 0)
{
node *follow = list->next;
// list = list->next;
free(list); // free deleted node
list = follow;
}
else
{
node *fetched = fetch_node(list, index);
node *prev, *follow;
if (index > 0)
{
prev = fetch_node(list, index - 1);
}
else
{
prev = list;
}
int range = count(list);
if (index != range - 1 && index < range)
{
follow = fetch_node(list, index + 1);
}
else
{
follow = NULL;
}
free(prev->next); // free deleted node
prev->next = follow;
}
return list;
}