c - 使用 malloc 声明变量如何导致丢失位?
问题描述
首先,我运行了 valgrind 以确保(在默认设置下)有零错误。然后,我决定通过以下方式检查泄漏:--leak-check=full
我的代码看起来像char* variable=malloc(sizeof(char)*(strlen(in)+1));
valgrind 报告内存“肯定丢失了”。
我可以访问的唯一另一行代码(这是在库回调函数中)in
是声明的代码行。这是一个类型的函数参数void *
(尽管在这种情况下,我希望我们可以安全地假设该值以空值终止。)
解决方案
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)
肯定会丢失,malloc(10)
因为在执行结束时无法访问它(这里不在main)
不会丢失,malloc(20)
因为仍然可以通过G到达
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