dlopen / dlsym / dlclose（dlfcn.h）导致内存泄漏

当像这样使用dlfcn系列时：

#include  #include  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; } 

它会导致内存泄漏：

 ==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？

Answer 1

看起来像后者。 然而，这似乎不是什么大问题，因为如果你重复dlopen / dlsym / dlclose调用另一个例程，你会发现内存泄漏的大小相同，它不会随着dlopen / dlclose调用的数量而增长。