检查pthread互斥锁是锁定还是解锁(线程锁定后)

Question

我需要查看一个互斥锁是否在if语句中被锁定或解锁,所以我这样检查它...

if(mutex[id] != 2){
    /* do stuff */
}

但是当我检查它时,gcc给了我以下错误:

error: invalid operands to binary != (have 'ptherad_mutex_t' and 'int')

那么我该如何检查互斥锁是否被锁定？

编辑:

我的问题的一个关键组成部分是我的线程(按设计)在将控制传递给另一个线程后立即锁定自己.因此,当线程A将控制传递给线程B时,线程A被锁定,线程B执行某些操作,然后当线程B完成时,它将解锁线程A.

这样做的问题是,如果线程B尝试解锁线程A并且线程A尚未完成锁定,则解锁的调用将丢失,线程A保持锁定状态,从而导致死锁.

更新:

我根据caf的建议重新制作了我的程序,但我仍遇到问题.我已经将我的程序模拟到了尽可能最好的结构caf中,但我现在甚至无法分辨出导致死锁的原因...我在这里创建了一个新问题,寻求我的代码帮助.

下面是caf的建议的可运行版本.我在线程a的函数中做了一个小的重新排序,没有它,线程a和线程b都会在创建时被锁定,等待一个永远不会改变的条件.

#include <pthread.h>

int run_a = 0;
pthread_mutex_t lock_a = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_a = PTHREAD_COND_INITIALIZER;

int run_b = 0;
pthread_mutex_t lock_b = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_b = PTHREAD_COND_INITIALIZER;

void *a(void *);
void *b(void *);

int main(){
    int status;
    pthread_t thread_a;
    pthread_t thread_b;

    pthread_create(&thread_a, NULL, a, (void *)0);
    pthread_create(&thread_b, NULL, b, (void *)0);

    pthread_join(thread_a, (void **)&status);
    pthread_join(thread_b, (void **)&status);

}

/* thread A */
void *a(void *i){
    while (1) {
        printf("thread A is running\n");
        sleep(1);

        /* unlock thread B */
        pthread_mutex_lock(&lock_b);
            run_b = 1;
            pthread_cond_signal(&cond_b);
        pthread_mutex_unlock(&lock_b);

        /* wait for thread A to be runnable */
        pthread_mutex_lock(&lock_a);
            while (!run_a)
                pthread_cond_wait(&cond_a, &lock_a);
            run_a = 0;
        pthread_mutex_unlock(&lock_a);      
    }
}

/* thread B */
void *b(void *i){
    while (1) {
        /* wait for thread B to be runnable */
        pthread_mutex_lock(&lock_b);
            while (!run_b)
                pthread_cond_wait(&cond_b, &lock_b);
            run_b = 0;
        pthread_mutex_unlock(&lock_b);

        printf("thread B is running\n");
        sleep(1);

        /* unlock thread A */
        pthread_mutex_lock(&lock_a);
            run_a = 1;
            pthread_cond_signal(&cond_a);
        pthread_mutex_unlock(&lock_a);
    }
}

Answer 1

你可以用pthread_mutex_trylock.如果成功,互斥锁无人认领,你现在拥有它(所以你应该释放它并返回"unheld",在你的情况下).否则,有人持有它.

我必须强调,"检查互斥锁是否无人认领"是一个非常糟糕的主意.这种想法存在固有的竞争条件.如果这样的函数在时间告诉你t锁是未被保留的,那么对于其他一些线程是否获得锁定一无所知t+1.

如果使用代码示例更好地说明,请考虑:

bool held = is_lock_held();

if (!held)
{
  // What exactly can you conclude here?  Pretty much nothing.
  // It was unheld at some point in the past but it might be held
  // by the time you got to this point, or by the time you do your
  // next instruction...
}

Answer 2

互斥体不是您要实现的方案的正确原语.你应该使用条件变量:

int run_thread_a = 0;
pthread_mutex_t run_lock_a = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t run_cond_a = PTHREAD_COND_INITIALIZER;

int run_thread_b = 0;
pthread_mutex_t run_lock_b = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t run_cond_b = PTHREAD_COND_INITIALIZER;

/* thread A */
while (1) {
    /* Wait for Thread A to be runnable */
    pthread_mutex_lock(&run_lock_a);
    while (!run_thread_a)
        pthread_cond_wait(&run_cond_a, &run_lock_a);
    run_thread_a = 0;
    pthread_mutex_unlock(&run_lock_a);

    /* Do some work */

    /* Now wake thread B */
    pthread_mutex_lock(&run_lock_b);
    run_thread_b = 1;
    pthread_cond_signal(&run_cond_b);
    pthread_mutex_unlock(&run_lock_b);
}

/* thread B */
while (1) {
    /* Wait for Thread B to be runnable */
    pthread_mutex_lock(&run_lock_b);
    while (!run_thread_b)
        pthread_cond_wait(&run_cond_b, &run_lock_b);
    run_thread_b = 0;
    pthread_mutex_unlock(&run_lock_b);

    /* Do some work */

    /* Now wake thread A */
    pthread_mutex_lock(&run_lock_a);
    run_thread_a = 1;
    pthread_cond_signal(&run_cond_a);
    pthread_mutex_unlock(&run_lock_a);
}

每个线程都将阻塞,pthread_cond_wait()直到另一个线程发出信号唤醒它.这不会死锁.

它可以很容易地通过分配一个可以扩展到多个线程,int,pthread_cond_t和pthread_mutex_t每个线程.