当像这样使用dlfcn系列时:
#include <stdio.h>
#include <dlfcn.h>
typedef int(*timefunc_t)(void*);
int main()
{
timefunc_t fun;
void* handle;
handle = dlopen("libc.so.6", RTLD_LAZY);
fun = (timefunc_t)dlsym(handle, "time");
printf("time=%d\n", fun(NULL));
dlclose(handle);
return 0;
}
导致内存泄漏:
==28803== Memcheck, a memory error detector
==28803== Copyright (C) 2002-2010, and GNU GPL'd, by Julian Seward et al.
==28803== Using Valgrind-3.6.1 and LibVEX; rerun with -h for copyright info
==28803== Command: ./dl
==28803==
time=1309249569
==28803==
==28803== HEAP SUMMARY:
==28803== in use at exit: 20 bytes in 1 blocks
==28803== total heap usage: 1 allocs, 0 frees, 20 bytes allocated
==28803==
==28803== LEAK SUMMARY:
==28803== definitely lost: 0 bytes in 0 blocks
==28803== indirectly lost: 0 bytes in 0 blocks
==28803== possibly lost: 0 bytes in 0 blocks
==28803== still reachable: 20 bytes in 1 blocks
==28803== suppressed: 0 bytes in 0 blocks
==28803== Rerun with --leak-check=full to see details of leaked memory
==28803==
==28803== For counts of detected and suppressed errors, rerun with: -v
==28803== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 13 from 6)
我的问题是,这是编程错误,还是dlfcn / libdl.so中的错误?
答案 0 :(得分:3)
看起来像后者。然而,这似乎并不是什么大问题,因为如果重复调用另一个例程的dlopen / dlsym / dlclose,你会看到内存泄漏的大小相同,它不会随着dlopen / dlclose调用的数量而增长。