使用Semaphores的程序在Linux上运行良好...在Mac OSX上出现意外结果

Question

我写了一个简单的程序,使用信号量解决了Readers-Writers问题.它在Linux操作系统上完美运行,但是当我在我的Mac OSX上运行时,我得到了意想不到的结果,我无法弄清楚原因.

我的计划:

#include <semaphore.h>
#include <sys/types.h>
#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

void* function1(void* val);
void* function2(void* val);

// shared values
volatile int X;
volatile int Y;

// declare semaphores
sem_t s1;
sem_t s2;

main()
{
void* status;

pthread_t thread1;
pthread_t thread2;
srand(time(NULL));

// initialize semaphores to zero
sem_init(&s1, 0, 0);
sem_init(&s2, 0, 0);

pthread_create(&thread1, NULL, function1, NULL);
pthread_create(&thread2, NULL, function2, NULL);

pthread_join(thread1, &status);
pthread_join(thread2, &status);

sem_destroy(&s1);
sem_destroy(&s2);

}

void* function1(void* val)
{
   while(1)
   {
   X = rand()%1000; // write 
   printf("After thread ID A writes to X, X = %d\n", X);
   sem_post(&s1); // signal
   sem_wait(&s2); // wait
   printf("After thread ID A reads from Y, Y = %d\n", Y); // read
   sleep(3);
   }   
}

void* function2(void* val)
{
   while(1)
   {
    sem_wait(&s1); // wait
    printf("After thread ID B reads from X, X = %d\n", X); // read
    Y = rand()%1000; // write
    printf("After thread ID B write to Y, Y = %d\n", Y);
    sem_post(&s2); // signal
    sleep(3);
   }
}

我在Linux上收到的输出(它应该是什么样子):

After thread ID A writes to X, X = 100
After thread ID B reads from X, X = 100
After thread ID B write to Y, Y = 234
After thread ID A reads from Y, Y = 234
...

Mac OSX上的输出(意外):

After thread ID A writes to X, X = 253
After thread ID A reads from Y, Y = 0
After thread ID B reads from X, X = 253
After thread ID B write to Y, Y = 728
...

Answer 1

检查sem_init调用的错误返回; 我打赌你会发现OS X版本返回"功能未实现"错误.

这是因为OS X上没有实现未命名的POSIX信号量.您需要使用命名信号量或pthread互斥/条件变量.