使用信号量的程序在Linux上运行正常… Mac OSX上的意外结果
发布时间:2020-12-13 19:45:46 所属栏目:Linux 来源:网络整理
导读:我写了一个简单的程序解决读者作家问题使用信号量.它在 Linux操作系统上运行完美,但是当我在Mac OSX上运行它时,我会得到意想不到的结果,我不知道为什么. 我的计划: #include semaphore.h#include sys/types.h#include stdio.h#include pthread.h#include un
我写了一个简单的程序解决读者作家问题使用信号量.它在
Linux操作系统上运行完美,但是当我在Mac OSX上运行它时,我会得到意想不到的结果,我不知道为什么.
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); sem_init(&s2,0); pthread_create(&thread1,NULL,function1,NULL); pthread_create(&thread2,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 = %dn",X); sem_post(&s1); // signal sem_wait(&s2); // wait printf("After thread ID A reads from Y,Y = %dn",Y); // read sleep(3); } } void* function2(void* val) { while(1) { sem_wait(&s1); // wait printf("After thread ID B reads from X,X); // read Y = rand()%1000; // write printf("After thread ID B write to Y,Y); sem_post(&s2); // signal sleep(3); } } 我在Linux上收到的输出(应该是什么样的): 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 ... 解决方法
检查sem_init调用的错误返回;我敢打赌你会发现OS X版本返回一个“功能未实现”的错误.
这是因为unnamed POSIX semaphores are not implemented on OS X.您需要使用命名信号量,或pthread互斥体/条件变量. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |