找不到错误C.

Abd*_*ebi 4 c multithreading mutex semaphore ipc

我必须写两个线程.每个人打印5个偶数/奇数从1到100这样(奇数是impair法语,偶数pair).

even 2,4,6,8,10
odd 1,3,5,7,9
even 12,14,16,18,20
odd 13,15,17,19,21
etc...
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我写了这段代码:

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

#define maxi 100

pthread_mutex_t mutex;
sem_t p;
sem_t imp;
int tour = 0;

void *pair(void *arg);
void *impair(void *arg);

int main() {
  pthread_t tidp, tidimp;

  pthread_mutex_init(&mutex, NULL);
  sem_init(&p, 0, 1);
  sem_init(&imp, 0, 1);

  pthread_create(&tidp, NULL, pair, (void *)2);
  pthread_create(&tidimp, NULL, impair, (void *)1);

  pthread_join(tidp, NULL);
  pthread_join(tidp, NULL);

  sem_destroy(&imp);
  sem_destroy(&p);
  pthread_mutex_destroy(&mutex);

  return 0;
}

void *pair(void *arg) {
  int i = (int)arg;
  int j, l;

  // sleep(5);

  pthread_mutex_lock(&mutex);
  if (!tour) {
    tour = 1;
    pthread_mutex_unlock(&mutex);
    sem_wait(&imp);
  } else {
    pthread_mutex_unlock(&mutex);
  }

  for (l = 0; l < maxi; l += 10) {
    sem_wait(&p);
    printf(" Pair  ");

    pthread_mutex_lock(&mutex);
    for (j = 0; j < 10; j += 2) {
      printf(" %4d \t", j + i);
    }

    pthread_mutex_unlock(&mutex);

    printf("\n");
    sem_post(&imp);
    i += 10;
  }

  pthread_exit(NULL);
}

void *impair(void *arg) {
  int i = (int)arg;
  int j, l;

  pthread_mutex_lock(&mutex);
  if (!tour) {
    tour = 1;
    pthread_mutex_unlock(&mutex);
    sem_wait(&p);
  } else {
    pthread_mutex_unlock(&mutex);
  }

  for (l = 0; l < maxi; l += 10) {
    sem_wait(&imp);
    printf("Impair  ");

    pthread_mutex_lock(&mutex);
    for (j = 0; j < 10; j += 2) {
      printf(" %4d \t", j + i);
    }
    pthread_mutex_unlock(&mutex);

    printf("\n");
    sem_post(&p);
    i += 10;
  }

  pthread_exit(NULL);
}
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我不明白的是,当我运行代码时,有时它会开始odd,有时候是even.更具体地说,当它开始时odd一切正常并且我得到从1到100的所有数字,但是当它开始时even有时我只得到91,有时93,有时97.

谁能告诉我有什么问题?下面的屏幕截图可能有所帮助.

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Car*_*rum 6

你不是在等两个线程退出:

pthread_join(tidp, NULL);
pthread_join(tidp, NULL);
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其中之一应该是tidimp.

  • 有时只需要第二双眼睛. (3认同)