c - 使用 malloc 声明变量如何导致丢失位？

Question

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

我的代码看起来像 char*variable=malloc(sizeof(char)*(strlen(in)+1)); 并且 valgrind 报告内存“肯定丢失了”。

我有权访问的唯一其他代码行(这是在库回调函数中)是声明 in 的代码行。这是一个 void * 类型的函数参数(尽管在本例中我希望我们可以安全地假定该值以 null 终止。)

Answer 1

有

#include <stdlib.h>

char * G;

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

   G = malloc(20);
}

在valgrind下执行给出:

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，因为在执行结束时无法访问它(这里是在 main 之外)

malloc(20) 没有丢失，因为仍然可以通过 G

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