如何使用两个condvar创建互斥体?

我想在Rust中构建一个单生产者多消费者示例,其中生产者注定有不超过10个未完成项目。我在C中使用互斥量和两个condvar建模了一个解决方案。一种condvar是在没有消耗的东西时等待消费者,而另一种condvar是在未消费的物品数大于10时等待生产者。C代码如下。

据我从Rust文档了解的那样,std::sync::Mutexstd::sync::Condvar之间必须存在1-1连接,因此我无法准确翻译我的C解决方案。

是否还有其他方法可以使用std::sync::Mutexstd::sync::Condvar在Rust中达到相同的目的(我看不到)。

#define _GNU_SOURCE
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>

//
// This is a simple example of using a mutex and 2 condition variables to
// sync a single writer and multiple readers interacting with a bounded (fixed max size) queue
//
// in this toy example a queue is simulated by an int counter n_resource
//

int n_resource;
pthread_cond_t rdr_cvar;
pthread_cond_t wrtr_cvar;
pthread_mutex_t mutex;

void reader(void* data)
{
    long id = (long)data;
    for(;;) {

        pthread_mutex_lock(&mutex);
        while (n_resource <= 0) {
            pthread_cond_wait(&rdr_cvar,&mutex);
        }
        printf("Reader %ld n_resource = %d\n",id,n_resource);
        --n_resource;
        // if there are still things to read - singla one reader
        if(n_resource > 0) {
            pthread_cond_signal(&rdr_cvar);
        }
        // if there is space for the writer to add another signal the writer
        if(n_resource < 10) {
            pthread_cond_signal(&wrtr_cvar);
        }
        pthread_mutex_unlock(&mutex);
    }
}
void writer(void* data)
{
    for(;;) {

        pthread_mutex_lock(&mutex);
        printf("Writer before while n_resource %d \n",n_resource);
        while (n_resource > 10) {
            pthread_cond_wait(&wrtr_cvar,&mutex);
        }
        printf("Writer after while n_resource %d \n",n_resource);

        ++n_resource;
        // if there is something for a reader to read signal one of the readers.
        if(n_resource > 0) {
            pthread_cond_signal(&rdr_cvar);
        }
        pthread_mutex_unlock(&mutex);
    }
}

int main()
{
    pthread_t rdr_thread_1;
    pthread_t rdr_thread_2;
    pthread_t wrtr_thread;
    pthread_mutex_init(&mutex,NULL);
    pthread_cond_init(&rdr_cvar,NULL);
    pthread_cond_init(&wrtr_cvar,NULL);
    pthread_create(&rdr_thread_1,NULL,&reader,(void*)1L);
    pthread_create(&rdr_thread_2,(void*)2L);
    pthread_create(&wrtr_thread,&writer,NULL);
    pthread_join(wrtr_thread,NULL);
    pthread_join(rdr_thread_1,NULL);
    pthread_join(rdr_thread_2,NULL);
}
mumu1514 回答:如何使用两个condvar创建互斥体?

虽然CondVar仅需要与一个Mutex关联,但是Mutex不必仅与一个CondVar关联。

例如,以下代码似乎可以正常工作-您可以在playground上运行它。

use std::sync::{Arc,Condvar,Mutex};
use std::thread;

struct Q {
    rdr_cvar: Condvar,wrtr_cvar: Condvar,mutex: Mutex<i32>,}

impl Q {
    pub fn new() -> Q {
        Q {
            rdr_cvar: Condvar::new(),wrtr_cvar: Condvar::new(),mutex: Mutex::new(0),}
    }
}

fn writer(id: i32,qq: Arc<Q>) {
    let q = &*qq;
    for i in 0..10 {
        let guard = q.mutex.lock().unwrap();
        let mut guard = q.wrtr_cvar.wait_while(guard,|n| *n > 3).unwrap();

        println!("{}: Writer {} n_resource = {}\n",i,id,*guard);
        *guard += 1;

        if *guard > 0 {
            q.rdr_cvar.notify_one();
        }
        if *guard < 10 {
            q.wrtr_cvar.notify_one();
        }
    }
}

fn reader(id: i32,qq: Arc<Q>) {
    let q = &*qq;
    for i in 0..10 {
        let guard = q.mutex.lock().unwrap();
        let mut guard = q.rdr_cvar.wait_while(guard,|n| *n <= 0).unwrap();

        println!("{} Reader {} n_resource = {}\n",*guard);
        *guard -= 1;

        if *guard > 0 {
            q.rdr_cvar.notify_one();
        }
        if *guard < 10 {
            q.wrtr_cvar.notify_one();
        }
    }
}

fn main() {
    let data = Arc::new(Q::new());
    let data2 = data.clone();

    let t1 = thread::spawn(move || writer(0,data2));
    let t2 = thread::spawn(move || reader(1,data));

    t1.join().unwrap();
    t2.join().unwrap();
}
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