4

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;
}

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？

4

1 回答 1

4

看起来是后者。然而，这似乎没什么大不了的，因为如果您重复 dlopen/dlsym/dlclose 调用另一个例程，您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。

于 2011-06-28T09:49:17.833 回答