我们应该使用带有信号量的互斥量来进行正确的同步并防止竞争条件吗？

Question

我试图看到竞争条件发生在消费者 - 生产者问题中，所以我创造了多个生产者和多个消费者。

据我所知，我需要提供信号量的互斥量：Mutex用于竞争条件，因为多个生产者可以同时访问缓冲区。然后数据可能已损坏。

并且信号量在生产者和消费者之间提供信号

这里的问题是，当我不使用互斥锁时，同步正确发生（我只使用信号量）。我的理解是正确的，或者在下面的代码中有什么不对的地方：

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

int buffer;
int loops = 0;

sem_t empty;
sem_t full;
sem_t mutex; //Adding MUTEX

void put(int value) {
    buffer = value;       
}

int get() {
    int b = buffer;      
    return b;
}



void *producer(void *arg) {
    int i;
    for (i = 0; i < loops; i++) {
        sem_wait(&empty);

        //sem_wait(&mutex);
        put(i);
        //printf("Data Set from %s, Data=%d
", (char*) arg, i);
        //sem_post(&mutex);

        sem_post(&full);
    }
}

void *consumer(void *arg) {
    int i;
    for (i = 0; i < loops; i++) {
        sem_wait(&full);

        //sem_wait(&mutex);
        int b = get();
        //printf("Data recieved from %s, %d
", (char*) arg, b);
        printf("%d
", b);
        //sem_post(&mutex);

        sem_post(&empty);

    }
}

int main(int argc, char *argv[])
{
    if(argc < 2 ){
        printf("Needs 2nd arg for loop count variable.
");
        return 1;
    }

    loops = atoi(argv[1]);

    sem_init(&empty, 0, 1); 
    sem_init(&full, 0, 0);    
    sem_init(&mutex, 0, 1);

    pthread_t pThreads[3];
    pthread_t cThreads[3];


    pthread_create(&cThreads[0], 0, consumer, (void*)"Consumer1");
    pthread_create(&cThreads[1], 0, consumer, (void*)"Consumer2");
    pthread_create(&cThreads[2], 0, consumer, (void*)"Consumer3");

    //Passing the name of the thread as paramter, Ignore attr
    pthread_create(&pThreads[0], 0, producer, (void*)"Producer1");
    pthread_create(&pThreads[1], 0, producer, (void*)"Producer2");
    pthread_create(&pThreads[2], 0, producer, (void*)"Producer3");


    pthread_join(pThreads[0], NULL);
    pthread_join(pThreads[1], NULL);
    pthread_join(pThreads[2], NULL);

    pthread_join(cThreads[0], NULL);
    pthread_join(cThreads[1], NULL);
    pthread_join(cThreads[2], NULL);

    return 0;
}