【问题标题】:pthread_cond_broadcast broken with dlsym?pthread_cond_broadcast 被 dlsym 破坏?
【发布时间】:2012-09-02 18:08:06
【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 函数只是调用了原始的 pthread_cond_broadcast。但是,对于一个同时调用 pthread_cond_wait 和 pthread_cond_broadcast 的非常简单的 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?

插入代码(编译为共享库):

#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:

#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;
}

编辑:

对于 glibc 2.11.1,gdb bt 给出:

(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 ()

编辑 2:

(已解决) 正如 R.. 所建议的(谢谢!),问题是在我的平台上 pthread_cond_broadcast 是一个版本符号,而 dlsym 给出了错误的版本。这篇博客非常详细地解释了这种情况:http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/

【问题讨论】:

  • 如果你在调试器下运行它,回溯是什么样的?您是否尝试过单步执行代码?
  • 我注意到这是一个版本符号。 dlsym 会不会给你错误的版本?

标签: c pthreads segmentation-fault glibc dlsym


【解决方案1】:

通过您的函数的调用似乎以不同版本的函数结束:

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

所以你的情况类似于这个问题,即你得到不兼容的 pthread 函数版本:symbol versioning and dlsym

这个页面提供了一种解决问题的方法,虽然有点复杂:http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/

【讨论】:

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