Question

我有一个包含4个线程的程序，最后我应该打印线程和管道（2个周期）阶段。类似的东西：

Thread 2: Stage 1 and 2
Thread 3: Stage 2 and 3
Thread 1: Stage 3 and 4
Thread 4: Stage 4 and 5

但我不知道如何为阶段做这个计数器，因为我正在做的事情除了1和2,2和3之外我不能为每个线程显示任何东西，而是第1阶段和第2阶段......

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

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;

void delay (int miliseconds){

    long pause;
    clock_t now,then;

    pause = miliseconds*(CLOCKS_PER_SEC/1000);
    now = then = clock();
    while( (now-then) < pause )
        now = clock();
}

int print(int n, int parar){

    if (n == 5) {
        printf("\nEstágio 4 e 5");

    } else {
        printf("Estágio %i e %i\n",n, n+1);
        if (parar == 2) {
            return (n+1, parar+0);
        }

        return print(n+1, parar+1);
    }
}

void thread_cont(void *arg){

    int *pvalor;
    pvalor=arg;

    pthread_mutex_lock(&mutex);
    printf ("\n---Thread %i--- \n", *pvalor);
    int a = print(1, 1);
    delay(2000);
    pthread_mutex_unlock(&mutex);

/*
    if (*pvalor == 1){
        printf ("Estágio 1 e 2");
    }

    if (*pvalor == 2){
    printf ("Estágio 2 e 3");
    }

    if (*pvalor == 3){
    printf ("Estágio 3 e 4");
    }

    if (*pvalor == 4){
    printf ("Estágio 4 e 5");
    }

*/

}

int main() {

    pthread_t id1;
    int offset1 = 1;
    pthread_create(&id1, NULL, thread_cont, &offset1);

    pthread_t id2;
    int offset2 = 2;
    pthread_create(&id2, NULL, thread_cont, &offset2);

    pthread_t id3;
    int offset3 = 3;
    pthread_create(&id3, NULL, thread_cont, &offset3);

    pthread_t id4;
    int offset4 = 4;
    pthread_create(&id4, NULL, thread_cont, &offset4);

    pthread_join(id1, NULL);
    pthread_join(id2, NULL);
    pthread_join(id3, NULL);
    pthread_join(id4, NULL);

    printf("\n");

    return 0;
}

Answer 1

我想我对我的问题做了错误的解释。对此抱歉，不过我想我现在就结束了。 4个主题。 4指令，管道（也许）并没有使用互斥，但信号量。顺便说一句语言障碍已经变得更加困难，甚至在这里问。

@edit final，也许是正确的代码

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

sem_t mutex;
pthread_t id1;
pthread_t id2;
pthread_t id3;
pthread_t id4;

void delay (int miliseconds){

    long pause;
    clock_t now,then;

    pause = miliseconds*(CLOCKS_PER_SEC/1000);
    now = then = clock();
    while( (now-then) < pause )
        now = clock();
}

pthread_mutex_t lock;

void *thread_cont(void *arg){

    int *pvalor;
    pvalor=arg;


    pthread_mutex_lock(&lock);
    sem_wait(&mutex);
    printf ("\n---Thread %i---", *pvalor);
    printf ("Stage 1");
    delay(2000);


    printf ("\n---Thread %i---", *pvalor);
    printf ("Stage 2");
    sem_post(&mutex);
    pthread_mutex_unlock(&lock);


    printf ("\n---Thread %i---", *pvalor);
    printf ("Stage 3");
    delay(2000);

    printf ("\n---Thread %i---", *pvalor);
    printf ("Stage 4");
    delay(2000);

    pthread_exit(0);

}


int main(){

    //create threads

    sem_init(&mutex, 0, 2);


//ONE FOR EACH CREATE

int offset1 = 1;
int offset2 = 2;
int offset3 = 3;
int offset4 = 4;


    pthread_create(&id1, NULL, thread_cont, &offset1);      
    delay(2000);
    pthread_create(&id2, NULL, thread_cont, &offset2);
    delay(2000);
    pthread_create(&id3, NULL, thread_cont, &offset3);
    delay(2000);
    pthread_create(&id3, NULL, thread_cont, &offset3);
    delay(2000);
    pthread_create(&id4, NULL, thread_cont, &offset4);
    delay(2000);


    pthread_join(id1, NULL);
    pthread_join(id2, NULL);
    pthread_join(id3, NULL);
    pthread_join(id4, NULL);


    sem_destroy(&mutex);

    printf("\n");

    return 0;
}

如何使4个线程相互交互？

1 个答案: