c - 使用传递给线程的 pthread_mutex_t 锁

Question

我正在学习 C 中的并发性。我知道有几种方法可以初始化 pthread_mutex_t。

这个简单的程序运行良好:

#include <stdio.h>
#include <pthread.h>

pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;

void* worker(void* arg) {

    pthread_mutex_lock(&lock); 
    printf("worker says hi\n");
    pthread_mutex_unlock(&lock);

    return NULL;
}

int main(int argc, char *argv[]) {

    pthread_t p;
    pthread_create(&p, NULL, worker, NULL);

    pthread_mutex_lock(&lock);
    printf("master says hi\n");
    pthread_mutex_unlock(&lock);

    pthread_join(p, NULL);

    return 0;
}

但是，如果我不想全局定义锁，我想我可以在我的主线程中定义它并将它传递给另一个线程，如下所示:

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

void* worker(void* arg) {
    pthread_mutex_t lock = * (pthread_mutex_t *) arg;

    pthread_mutex_lock(&lock);
    printf("worker says hi\n");
    pthread_mutex_unlock(&lock);

    return NULL;
}

int main(int argc, char *argv[]) {

    pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;

    pthread_t p;
    pthread_create(&p, NULL, worker, &lock);

    pthread_mutex_lock(&lock);
    printf("master says hi\n");
    pthread_mutex_unlock(&lock);

    pthread_join(p, NULL);

    return 0;
}

但是，这不有效。相反，它只打印 master says hi 并挂起。

我认为这可能是死锁，所以我在master中插入了一些 sleep 时间如下:

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

void* worker(void* arg) {
    pthread_mutex_t lock = * (pthread_mutex_t *) arg;

    pthread_mutex_lock(&lock);
    printf("worker says hi\n");
    pthread_mutex_unlock(&lock);

    return NULL;
}

int main(int argc, char *argv[]) {

    pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;

    pthread_t p;
    pthread_create(&p, NULL, worker, &lock);

    sleep(1);

    pthread_mutex_lock(&lock);
    printf("master says hi\n");
    pthread_mutex_unlock(&lock);

    pthread_join(p, NULL);

    return 0;
}

正如预期的那样，这会打印出 worker says hi，然后在一秒后打印出 master says hi。

但是，如果我将 sleep 移动到 worker 中，如下所示，它会再次挂起:

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

void* worker(void* arg) {
    pthread_mutex_t lock = * (pthread_mutex_t *) arg;

    sleep(1);

    pthread_mutex_lock(&lock);
    printf("worker says hi\n");
    pthread_mutex_unlock(&lock);

    return NULL;
}

int main(int argc, char *argv[]) {

    pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;

    pthread_t p;
    pthread_create(&p, NULL, worker, &lock);

    pthread_mutex_lock(&lock);
    printf("master says hi\n");
    pthread_mutex_unlock(&lock);

    pthread_join(p, NULL);

    return 0;
}

如果我将 sleep 函数放在上方 pthread_mutex_t lock = * (pthread_mutex_t *) arg; 行，那么它不会挂起。我不确定如何解释这种行为。

Answer 1

worker 函数正在复制互斥锁，而它应该只使用现有的互斥锁。修复方法是将 lock 变量声明为指向互斥量的指针:

void* worker(void* arg) {

    pthread_mutex_t *lock = arg;

    pthread_mutex_lock(lock);
    printf("worker says hi\n");
    pthread_mutex_unlock(lock);

    return NULL;
}