Valgrind ...大小为8的块中的4个字节

Question

尝试释放列表后,我在Valgrind中收到此错误.print_list将列表转储到syslog.我非常有信心输出是正确的.

Valgrind的:

==7028== 1 errors in context 1 of 10:
==7028== Invalid read of size 4
==7028==    at 0x8049603: free_list (list.c:239)
==7028==    by 0x80488B5: m61_close_for_valgrind (m61.c:36)
==7028==    by 0x8048825: main (mytest.c:19)
==7028==  Address 0x420006c is 4 bytes inside a block of size 8 free'd
==7028==    at 0x4028F0F: free (vg_replace_malloc.c:446)
==7028==    by 0x804960C: free_list (list.c:239)
==7028==    by 0x80488B5: m61_close_for_valgrind (m61.c:36)
==7028==    by 0x8048825: main (mytest.c:19)
==7028== 

mytest.c:

15  char *temp = malloc(10);
16  char *temp2 = malloc(10);
17  free(temp);
18  free(temp2);
19  m61_close_for_valgrind();

list.h

typedef struct lnode {
    ACTIVE_ALLOCATION *value;
    struct lnode *next;
} lnode;

list.c(由m61_close_for_valgrind()调用

void free_list(LIST *s) {

    lnode **nptr = &s->head;

    print_list(s);
    while (*nptr) {
        lnode **tmp = nptr;
        tmp = nptr;

        if ((*tmp)->value) {
            syslog(LOG_NOTICE,"Freeing (*tmp)->value=%p\n", (*tmp)->value);
            //printf("%p\n",(*nptr)->value);
            free((*tmp)->value);    //Free active allocation metadata
        }

        nptr = &(*nptr)->next;
        syslog(LOG_NOTICE,"New *nptr value=%p\n", (*nptr));

        syslog(LOG_NOTICE,"Freeing (*tmp)=%p\n", (*tmp));
        free(*tmp);             //Free node

    }

}

系统日志

Sep 19 00:37:02 appliance mytest[7759]:   -- Start List Dump --
Sep 19 00:37:02 appliance mytest[7759]:   (*nptr)=0x903f220 (*nptr)->value=0x903f208   (*nptr)->next=0x903f260  (*nptr)->value->ptr=0x903f1f0
Sep 19 00:37:02 appliance mytest[7759]: (*nptr)->value->ptr=0x903f1f0
Sep 19 00:37:02 appliance mytest[7759]:   (*nptr)=0x903f260 (*nptr)->value=0x903f248   (*nptr)->next=(nil)  (*nptr)->value->ptr=0x903f230
Sep 19 00:37:02 appliance mytest[7759]: (*nptr)->value->ptr=0x903f230
Sep 19 00:37:02 appliance mytest[7759]:   -- End List Dump --
Sep 19 00:37:02 appliance mytest[7759]: Freeing (*tmp)->value=0x903f208
Sep 19 00:37:02 appliance mytest[7759]: New *nptr value=0x903f260
Sep 19 00:37:02 appliance mytest[7759]: Freeing (*tmp)=0x903f220
Sep 19 00:37:02 appliance mytest[7759]: Freeing (*tmp)->value=0x903f248
Sep 19 00:37:02 appliance mytest[7759]: New *nptr value=(nil)
Sep 19 00:37:02 appliance mytest[7759]: Freeing (*tmp)=0x903f260

Answer 1

正如caf已经写过的那样,你正在访问刚被释放的内存.

要解决这个问题,不要使用双指针,单指针在这里会做得很好.

所以更换

lnode **nptr = &s->head;

通过

lnode *nptr = s->head;

同样的

lnode **tmp = nptr;

在循环.做了

lnode *tmp = nptr;

当你在它的时候放下双重任务.

然后访问值和下一个

tmp->value

和

tmp->next

直

Answer 2

在除第一个迭代之外的每个迭代tmp中,next指向前一个节点的指针 - 但是您已经释放了该节点(在上一次迭代中),因此tmp指向一个已释放的块,您无法取消引用它.