面试题：C语言POSIX线程的复杂同步与数据共享

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <semaphore.h>
#include <time.h>

#define BUFFER_SIZE 10
#define PRODUCER_THREADS 3
#define CONSUMER_THREADS 2

int buffer[BUFFER_SIZE];
int in = 0;
int out = 0;

sem_t empty;
sem_t full;
pthread_mutex_t mutex;

// 生产者线程函数
void* producer(void* arg) {
    int id = *((int*)arg);
    while (1) {
        int num = rand() % 100; // 生成随机整数
        sem_wait(&empty); // 等待缓冲区有空闲位置
        pthread_mutex_lock(&mutex);
        buffer[in] = num;
        printf("Producer %d produced %d at position %d\n", id, num, in);
        in = (in + 1) % BUFFER_SIZE;
        pthread_mutex_unlock(&mutex);
        sem_post(&full); // 通知缓冲区有新数据
    }
    return NULL;
}

// 消费者线程函数
void* consumer(void* arg) {
    int id = *((int*)arg);
    while (1) {
        sem_wait(&full); // 等待缓冲区有数据
        pthread_mutex_lock(&mutex);
        int num = buffer[out];
        printf("Consumer %d consumed %d from position %d, squared value: %d\n", id, num, out, num * num);
        out = (out + 1) % BUFFER_SIZE;
        pthread_mutex_unlock(&mutex);
        sem_post(&empty); // 通知缓冲区有空闲位置
    }
    return NULL;
}

int main() {
    pthread_t producers[PRODUCER_THREADS];
    pthread_t consumers[CONSUMER_THREADS];
    int producer_ids[PRODUCER_THREADS];
    int consumer_ids[CONSUMER_THREADS];

    sem_init(&empty, 0, BUFFER_SIZE);
    sem_init(&full, 0, 0);
    pthread_mutex_init(&mutex, NULL);

    srand(time(NULL));

    for (int i = 0; i < PRODUCER_THREADS; i++) {
        producer_ids[i] = i;
        pthread_create(&producers[i], NULL, producer, &producer_ids[i]);
    }

    for (int i = 0; i < CONSUMER_THREADS; i++) {
        consumer_ids[i] = i;
        pthread_create(&consumers[i], NULL, consumer, &consumer_ids[i]);
    }

    for (int i = 0; i < PRODUCER_THREADS; i++) {
        pthread_join(producers[i], NULL);
    }

    for (int i = 0; i < CONSUMER_THREADS; i++) {
        pthread_join(consumers[i], NULL);
    }

    sem_destroy(&empty);
    sem_destroy(&full);
    pthread_mutex_destroy(&mutex);

    return 0;
}

如何避免死锁

顺序加锁：在上述代码中，生产者和消费者线程都是先获取empty或full信号量，再获取互斥锁mutex，解锁时顺序相反。这保证了所有线程加锁的顺序一致，避免死锁。
资源分配图算法：可以使用如银行家算法等资源分配图算法来检测和避免死锁，但在这种简单场景下，上述顺序加锁的方式足以避免死锁。
避免嵌套锁：尽量减少锁的嵌套使用。如果必须嵌套，严格按照相同顺序获取锁。在本程序中，只有一个互斥锁mutex，且获取和释放逻辑简单清晰，减少了死锁风险。

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面试题：C语言POSIX线程的复杂同步与数据共享

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如何避免死锁