这是我的代码。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char buf1[100];
char buf2[100];
int main()
{
char **p = (char**)(buf1+sizeof(long));
char **q = (char**)(buf2+1);
*p = (char*)malloc(100);
*q = (char*)malloc(100);
strcpy(*p, "xxxx");
strcpy(*q, "zzzz");
printf("p:%s q:%s\n", *p, *q);
return 0;
}
我使用gcc编译代码,并像这样运行valgrind-3.6.1
valgrind --leak-check=full --log-file=test.log --show-reachable=yes ~/a.out
valgrind给了我下面的日志
==20768== Memcheck, a memory error detector
==20768== Copyright (C) 2002-2010, and GNU GPL'd, by Julian Seward et al.
==20768== Using Valgrind-3.6.1 and LibVEX; rerun with -h for copyright info
==20768== Command: /home/zxin11/a.out
==20768== Parent PID: 12686
==20768==
==20768==
==20768== HEAP SUMMARY:
==20768== in use at exit: 200 bytes in 2 blocks
==20768== total heap usage: 2 allocs, 0 frees, 200 bytes allocated
==20768==
==20768== 100 bytes in 1 blocks are still reachable in loss record 1 of 2
==20768== at 0x4C2488B: malloc (vg_replace_malloc.c:236)
==20768== by 0x4005FD: main (test2.c:12)
==20768==
==20768== 100 bytes in 1 blocks are definitely lost in loss record 2 of 2
==20768== at 0x4C2488B: malloc (vg_replace_malloc.c:236)
==20768== by 0x400611: main (test2.c:13)
==20768==
==20768== LEAK SUMMARY:
==20768== definitely lost: 100 bytes in 1 blocks
==20768== indirectly lost: 0 bytes in 0 blocks
==20768== possibly lost: 0 bytes in 0 blocks
==20768== still reachable: 100 bytes in 1 blocks
==20768== suppressed: 0 bytes in 0 blocks
==20768==
==20768== For counts of detected and suppressed errors, rerun with: -v
==20768== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 3 from 3)
为什么第一个malloc仍然可以访问而第二个malloc肯定丢失了? 也许是关于对齐,你不能将malloced内存的地址放入一个未对齐的变量,如果是这样,我怎么能抑制这种积极的报告呢? 非常想你。
答案 0 :(得分:5)
来自memcheck manual(强调我的):
如果
--leak-check
设置正确,则对于每个剩余的块,Memcheck确定是否可以从根集内的指针访问该块。根集包括(a)所有线程的通用寄存器,以及(b)可访问客户端内存中的初始化,对齐,指针大小的数据字,包括堆栈。
所以关于对齐的猜想是正确的。不幸的是,强有力地抑制这种警告的最好方法可能就是在退出程序之前将任何这样的已知值复制到对齐的位置(可能这个代码是你真实应用程序的模型,它对你来说有点意义存储未对齐的指针。)
您也可以--gen-suppressions=yes
尝试writing or generating a suppression file。但是,如果您的应用程序是非确定性的,或者您使用不同的输入数据运行它,这种方法会很快变得烦人。