pthread_cond_broadcast被dlsym破坏了? [英] pthread_cond_broadcast broken with dlsym?
问题描述
我正在尝试使用LD_PRELOAD机制插入对pthread_cond_broadcast的调用.我插入的pthread_cond_broadcast函数仅调用原始的pthread_cond_broadcast.但是,对于非常简单的同时调用pthread_cond_wait和pthread_cond_broadcast的pthread代码,我要么在glibc中出现段错误(对于glibc 2.11.1),要么程序挂起(对于glibc 2.15).有什么线索吗?
I am trying to interpose calls to pthread_cond_broadcast using LD_PRELOAD mechanism. My interposed pthread_cond_broadcast function just calls the original pthread_cond_broadcast. However, for a very simple pthread code where both pthread_cond_wait and pthread_cond_broadcast get invoked, I either end up with a segfault in glibc (for glibc 2.11.1) or the program hangs (for glibc 2.15). Any clues on that is going on?
插入代码(将其编译为共享库):
The interposition code (that gets compiled as a shared library):
#define _GNU_SOURCE
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <dlfcn.h>
static int (*orig_pthread_cond_broadcast)(pthread_cond_t *cond) = NULL;
__attribute__((constructor))
static void start() {
orig_pthread_cond_broadcast =
(int (*)()) dlsym(RTLD_NEXT, "pthread_cond_broadcast");
if (orig_pthread_cond_broadcast == NULL) {
printf("pthread_cond_broadcast not found!!!\n");
exit(1);
}
}
__attribute__((__visibility__("default")))
int pthread_cond_broadcast(pthread_cond_t *cond) {
return orig_pthread_cond_broadcast(cond);
}
简单的pthread程序:
The simple pthread program:
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>
pthread_mutex_t cond_mutex;
pthread_cond_t cond_var;
int condition;
void *thread0_work(void *arg) {
pthread_mutex_lock(&cond_mutex);
printf("Signal\n");
condition = 1;
pthread_cond_broadcast(&cond_var);
pthread_mutex_unlock(&cond_mutex);
return NULL;
}
void *thread1_work(void *arg) {
pthread_mutex_lock(&cond_mutex);
while (condition == 0) {
printf("Wait\n");
pthread_cond_wait(&cond_var, &cond_mutex);
printf("Done waiting\n");
}
pthread_mutex_unlock(&cond_mutex);
return NULL;
}
int main() {
pthread_t thread1;
pthread_mutex_init(&cond_mutex, NULL);
pthread_cond_init(&cond_var, NULL);
pthread_create(&thread1, NULL, thread1_work, NULL);
// Slowdown this thread, so the thread 1 does pthread_cond_wait.
usleep(1000);
thread0_work(NULL);
pthread_join(thread1, NULL);
return 0;
}
对于glibc 2.11.1,gdb bt给出:
For glibc 2.11.1, gdb bt gives:
(gdb) set environment LD_PRELOAD=./libintercept.so
(gdb) run
Starting program: /home/seguljac/intercept/main
[Thread debugging using libthread_db enabled]
[New Thread 0x7ffff7436700 (LWP 19165)]
Wait
Signal
Before pthread_cond_broadcast
Program received signal SIGSEGV, Segmentation fault.
0x00007ffff79ca0e7 in pthread_cond_broadcast@@GLIBC_2.3.2 () from /lib/libpthread.so.0
(gdb) bt
#0 0x00007ffff79ca0e7 in pthread_cond_broadcast@@GLIBC_2.3.2 () from /lib/libpthread.so.0
#1 0x00007ffff7bdb769 in pthread_cond_broadcast () from ./libintercept.so
#2 0x00000000004008e8 in thread0_work ()
#3 0x00000000004009a4 in main ()
(已解决) 如R ..的建议(谢谢!),问题是在我的平台上pthread_cond_broadcast是版本化的符号,而dlsym提供了错误的版本.该博客详细介绍了这种情况: http ://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/
(Solved) As suggested by R.. (thanks!), the issue is that on my platform pthread_cond_broadcast is a versioned symbol, and dlsym gives the wrong version. This blog explains this situation in great detail: http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/
推荐答案
通过函数进行的调用似乎结束于该函数的另一个版本:
The call through your function seems to end up in a different version of the function:
With LD_PRELOAD: __pthread_cond_broadcast_2_0 (cond=0x804a060) at old_pthread_cond_broadcast.c:37
Without LD_PRELOAD: pthread_cond_broadcast@@GLIBC_2.3.2 () at ../nptl/sysdeps/unix/sysv/linux/i386/i686/../i486/pthread_cond_broadcast.S:39
因此您的情况类似于此问题,即您正在获取不兼容的pthread函数版本:符号版本控制和dlsym
So your situation is similar to this question, i.e. you are getting incompatible versions of pthread functions: symbol versioning and dlsym
This page gives one way to solve the problem, though a bit complex: http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/
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