与对dlsym调用pthread_cond_broadcast坏了吗?(pthread_cond_b

2019-09-22 06:50发布

我试图干预调用使用LD_PRELOAD机制来调用pthread_cond_broadcast。 我插调用pthread_cond_broadcast函数只是调用原始调用pthread_cond_broadcast。 然而,对于一个非常简单的并行线程代码中调用pthread_cond_broadcast被调用都调用pthread_cond_wait和,我要么在glibc的(对于glibc的2.11.1)段错误或程序挂起(对于glibc的2.15)结束。 对任何线索是怎么回事?

插置的代码(即获取编译作为共享库):

#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);
}

简单的并行线程的程序:

#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的2.11.1,GDB BT得到:

(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:

(解决)作为由R建议。(感谢!),问题是,在我的平台上调用pthread_cond_broadcast是一个版本符号,和dlsym给出了错误的版本。 这个博客介绍的很详细这种情况: http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/

Answer 1:

通过你的函数调用似乎不同版本的功能来结束:

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,你的情况与此类似的问题,即你所得到的并行线程功能不兼容的版本: 符号版本和dlsym

该页面提供了解决问题的一种方式,虽然有点复杂: http://blog.fesnel.com/blog/2009/08/25/preloading-with-multiple-symbol-versions/



文章来源: pthread_cond_broadcast broken with dlsym?