多线程程序如何在 C 中工作? [英] How Multi-threaded program works in C?

问题描述

我是一个完全的新手，我一直在编写固定缓冲区大小的生产者-消费者问题，我观察到的缓冲区大小 1 与我预期的非常不同.我打印x-->"当我产生 x 和--> x"时当 x 被消耗时.我得到的输出是这样的:\

I am a complete newbie and I have been coding producer-consumer problem with fixed buffer size what I observed for buffer-size 1 is very different from what I expected. I print "x-->" when I produce x and "-->x" when x is consumed. The output I get is like this: \

0 --->
1 --->
2 --->
---> 0
---> 1
---> 2
3 --->
4 --->
5 --->
---> 3
---> 4
---> 5

我很困惑 1,2,3 是如何一次复制然后一次消耗 1,2,3 的，有人可以解释一下吗.

I am confused about how 1,2,3 reproduced at once and then 1,2,3 consumed at once can someone explain, please.

这是我的代码

#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#define BUFFER_SIZE 1
#define OVER (-1)


int filled = 0;

struct prodcons {
    int buffer[BUFFER_SIZE];      /* the actual data */
    pthread_mutex_t lock;         /* mutex ensuring exclusive access to buffer */
    int readpos, writepos;        /* positions for reading and writing */
    pthread_cond_t notempty;      /* signaled when buffer is not empty */
    pthread_cond_t notfull;       /* signaled when buffer is not full */
};

void init(struct prodcons * b)
{
    pthread_mutex_init(&b->lock, NULL);
    pthread_cond_init(&b->notempty, NULL);
    pthread_cond_init(&b->notfull, NULL);
    b->readpos = 0;
    b->writepos = 0;
}

void put(struct prodcons * b, int data)
{
    pthread_mutex_lock(&b->lock);
    if(BUFFER_SIZE == 1) {
        if(filled) {
            pthread_cond_wait(&b->notfull, &b->lock);
        }
        b->buffer[0] = data;
        filled = 1;
    }
    else {
        /* Wait until buffer is not full */
        while ((b->writepos + 1) % BUFFER_SIZE == b->readpos) {
            pthread_cond_wait(&b->notfull, &b->lock);
            /* pthread_cond_wait reacquired b->lock before returning */
        }
        /* Write the data and advance write pointer */
        b->buffer[b->writepos] = data;
        b->writepos++;
        if (b->writepos >= BUFFER_SIZE) b->writepos = 0;
        /* Signal that the buffer is now not empty */
    }

    pthread_cond_signal(&b->notempty);
    pthread_mutex_unlock(&b->lock);
}

int get(struct prodcons * b)
{
    int data = 0;
    pthread_mutex_lock(&b->lock);
    if(BUFFER_SIZE == 1) {
        if(!filled) {
            pthread_cond_wait(&b->notempty, &b->lock);
        }
        data = b->buffer[0];
        filled = 0;
    }
    else {
        /* Wait until buffer is not empty */
        while (b->writepos == b->readpos) {
            pthread_cond_wait(&b->notempty, &b->lock);
        }
        /* Read the data and advance read pointer */
        data = b->buffer[b->readpos];
        b->readpos++;
        if (b->readpos >= BUFFER_SIZE) b->readpos = 0;
        /* Signal that the buffer is now not full */
    }
    pthread_cond_signal(&b->notfull);
    pthread_mutex_unlock(&b->lock);
    return data;
}

struct prodcons buffer;

void * producer(void * data)
{
    int n;
    for (n = 0; n < 10000; n++) {
        printf("%d --->\n", n);
        put(&buffer, n);
    }
    put(&buffer, OVER);
    return NULL;
}

void * consumer(void * data)
{
    int d;
    while (1) {
        d = get(&buffer);
        if (d == OVER) break;
        printf("---> %d\n", d);
    }
    return NULL;
}
int main(void)
{
    pthread_t th_a, th_b;
    void * retval;
    init(&buffer);
    /* Create the threads */
    pthread_create(&th_a, NULL, producer, 0);
    pthread_create(&th_b, NULL, consumer, 0);
    /* Wait until producer and consumer finish. */
    pthread_join(th_a, &retval);
    pthread_join(th_b, &retval);
    return 0;
}

推荐答案

不能保证输出如您所愿.我实现了一个时间戳，它显示了 put 和 get 方法被调用的时间(时间是相对于 init 被调用的时间):

The output is not guaranteed as you expect it. I implemented a timestamp, which shows, when the put and get methods are called (time is relative to when init was called):

[ 44491 ns] 0 --->
[115427 ns] 1 --->
[153189 ns] ---> 0
[178376 ns] 2 --->
[182891 ns] 3 --->
[183518 ns] ---> 1
[188542 ns] ---> 2
[198661 ns] 4 --->
[203461 ns] ---> 3
[206636 ns] ---> 4
[204180 ns] 5 --->
[212146 ns] 6 --->
[221019 ns] ---> 5
[224146 ns] ---> 6
[221877 ns] 7 --->
[230976 ns] 8 --->
[232130 ns] 9 --->
[232347 ns] ---> 7
[237920 ns] ---> 8
[239051 ns] ---> 9
[239312 ns] 10 --->
[244770 ns] 11 --->
[245918 ns] 12 --->
[246106 ns] ---> 10
...

你应该知道，stdout 是两个线程共享的，那么锁是怎么实现的，谁知道呢.

You should know, that stdout is shared by the two threads, so how the locking is implemented there, who knows.

如果你想要同步输出，你必须把 printf 放入同步部分(在 put 和 get 方法中访问你的缓冲区之后).

If you want synchronized output, you have to put printf into the synchronized sections (after accessing your buffer in the put and get methods).

然后你得到输出:

[ 31900 ns] 0 --->
[ 77228 ns] ---> 0
[ 85722 ns] 1 --->
[ 90959 ns] ---> 1
[ 95490 ns] 2 --->
[ 99591 ns] ---> 2
[103447 ns] 3 --->
[107385 ns] ---> 3
[111325 ns] 4 --->
[115215 ns] ---> 4
[119143 ns] 5 --->
[123066 ns] ---> 5
[126962 ns] 6 --->
[130860 ns] ---> 6
[134561 ns] 7 --->
[138427 ns] ---> 7
[142188 ns] 8 --->
[146115 ns] ---> 8
[149797 ns] 9 --->
[153534 ns] ---> 9
[157856 ns] 10 --->
[161631 ns] ---> 10
...

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