使用 malloc 声明变量如何导致丢失位？

Question

首先，我有运行 valgrind 来确保（在默认设置下）错误为零。然后，我决定用类似的东西检查是否有泄漏：--leak-check=full

我的代码看起来像 char* variable=malloc(sizeof(char)*(strlen(in)+1)); 并且 valgrind 报告内存为 "definitely lost."

我唯一可以访问的其他代码行（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（虽然在这种情况下我希望我们可以安全地假设该值是空终止的。）

Answer 1

有

#include <stdlib.h>

char * G;

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

   G = malloc(20);
}

valgrind下的执行给出：

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，因为在执行结束时没有办法访问它（这里是main）

malloc(20) 没有丢失，因为仍然可以通过 G

到达

使用 malloc 声明变量如何导致丢失位？

How declaring a variable with malloc lead to lost bits?

c

debugging

valgrind