我正在尝试编写一个非常简单的函数,该函数会在过滤掉(并释放)某些元素的同时更新链表中的元素。尽管根据valgrind它包含无效的可用空间和内存泄漏,但我已经能够派生此实现。我不知道实施有什么问题。
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
typedef struct List {
int head;
struct List *tail;
} List;
List *cons(int h, List *tail)
{
List *list = malloc(sizeof(List));
list->head = h;
list->tail = (struct List*) tail;
return list;
}
bool is_odd(int val)
{
return val % 2 != 0;
}
int square(int val)
{
return val * val;
}
void print_list(List *l)
{
while (l) {
printf("item: %d, ", l->head);
l = (List*) l->tail;
}
printf("\n");
}
List *square_odd(List *list)
{
List *new_head = NULL;
List *prev_head = NULL;
while (list != NULL) {
List *next = (List *) list->tail;
if (is_odd(list->head)) {
if (new_head == NULL) new_head = list;
if (prev_head != NULL) prev_head->tail = (struct List*) list;
list->head = square(list->head);
prev_head = list;
} else {
if (next == NULL) {
prev_head->tail = NULL;
}
free(list);
}
list = next;
}
return new_head;
}
int main()
{
List *t = NULL;
List init = {100, NULL};
t = &init;
t = cons(1, t);
t = cons(2, t);
t = cons(3, t);
t = cons(4, t);
t = cons(5, t);
t = cons(6, t);
t = cons(7, t);
t = cons(8, t);
t = square_odd(t);
List *tmp = NULL;
print_list(t);
while(t->tail != NULL) {
tmp = t;
t = (List*) t->tail;
if (tmp != NULL) free(tmp);
}
return 0;
}
valgrind输出为:
==17692== Memcheck, a memory error detector
==17692== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==17692== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==17692== Command: ./main
==17692==
==17692== Invalid free() / delete / delete[] / realloc()
==17692== at 0x4835948: free (in /nix/store/wrj8cjkfqzi0qlwnigx8vxwyyfl01lqq-valgrind-3.15.0/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==17692== by 0x40127D: square_odd (in /tmp/linked/main)
==17692== by 0x401371: main (in /tmp/linked/main)
==17692== Address 0x1ffeffeac0 is on thread 1's stack
==17692== in frame #2, created by main (???:)
==17692==
item: 49, item: 25, item: 9, item: 1,
==17692==
==17692== HEAP SUMMARY:
==17692== in use at exit: 16 bytes in 1 blocks
==17692== total heap usage: 9 allocs, 9 frees, 1,152 bytes allocated
==17692==
==17692== LEAK SUMMARY:
==17692== definitely lost: 16 bytes in 1 blocks
==17692== indirectly lost: 0 bytes in 0 blocks
==17692== possibly lost: 0 bytes in 0 blocks
==17692== still reachable: 0 bytes in 0 blocks
==17692== suppressed: 0 bytes in 0 blocks
==17692== Rerun with --leak-check=full to see details of leaked memory
==17692==
==17692== For lists of detected and suppressed errors, rerun with: -s
==17692== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)
实现有什么问题。
您正在将init
的地址传递给free
。 init
是一个变量,其自动存储期限未分配给malloc
,因此您无法free
。
因为最初是t = &init
,然后在第一次调用cons
内执行new_t->tail = t
,所以实际上是执行new_t->tail = &init
。因此,在所有cons
调用之后,链中的最后一个元素都指向&init
。
t->tail-> ... ->tail->tail == &init
在循环内部,然后将&init
的地址传递给free
函数。
我会说实现实际上是好的,第一个元素的分配是错误的。
我建议,只需移除init
并使用malloc
创建第一个链:
int main() {
List *t = cons(100, NULL);
t = cons(1, t);
// the rest unchanged
free(NULL)
不执行任何操作。您可以仅用if (tmp != NULL) free(tmp);
替换free(tmp)
。