如何使用malloc声明变量会导致位丢失?

时间:2019-02-17 14:55:08

标签: c debugging valgrind

首先,我运行valgrind以确保(在默认设置下)零错误。然后,我决定使用类似以下内容的检查泄漏:--leak-check=full

我有一些看起来像char* variable=malloc(sizeof(char)*(strlen(in)+1));的代码,并且valgrind报告说内存“肯定丢失了”。

我可以访问的唯一另一行代码(在库回调函数中)是声明in的代码行。这是类型void *的函数参数(尽管在这种情况下,我希望我们可以放心地假定该值以null终止。)

1 个答案:

答案 0 :(得分:2)

拥有

#include <stdlib.h>

char * G;

int main()
{
   char * l = malloc(10);

   G = malloc(20);
}

valgrind 下的执行结果为:

pi@raspberrypi:/tmp $ valgrind --leak-check=full ./a.out
==11087== Memcheck, a memory error detector
==11087== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==11087== Using Valgrind-3.13.0 and LibVEX; rerun with -h for copyright info
==11087== Command: ./a.out
==11087== 
==11087== 
==11087== HEAP SUMMARY:
==11087==     in use at exit: 30 bytes in 2 blocks
==11087==   total heap usage: 2 allocs, 0 frees, 30 bytes allocated
==11087== 
==11087== 10 bytes in 1 blocks are definitely lost in loss record 1 of 2
==11087==    at 0x4847568: malloc (vg_replace_malloc.c:299)
==11087==    by 0x10453: main (mm.c:7)
==11087== 
==11087== LEAK SUMMARY:
==11087==    definitely lost: 10 bytes in 1 blocks
==11087==    indirectly lost: 0 bytes in 0 blocks
==11087==      possibly lost: 0 bytes in 0 blocks
==11087==    still reachable: 20 bytes in 1 blocks
==11087==         suppressed: 0 bytes in 0 blocks
==11087== Reachable blocks (those to which a pointer was found) are not shown.
==11087== To see them, rerun with: --leak-check=full --show-leak-kinds=all
==11087== 
==11087== For counts of detected and suppressed errors, rerun with: -v
==11087== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 6 from 3)

malloc(10)肯定丢失了,因为在执行结束时(在 main 之外)无法访问它

malloc(20)不会丢失,因为仍然可以通过 G

到达