1 编译运行附件中的代码,提交运行结果截图,并说明程序功能
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#include
#include #include #include define NUM 5
int queue[NUM];
sem_t blank_number, product_number;void *producer ( void * arg )
{
static int p = 0;for ( ;; ) { sem_wait( &blank_number ); queue[p] = rand() % 1000; printf("Product %d \n", queue[p]); p = (p+1) % NUM; sleep ( rand() % 5); sem_post( &product_number ); }
}
void *consumer ( void * arg )
{static int c = 0; for( ;; ) { sem_wait( &product_number ); printf("Consume %d\n", queue[c]); c = (c+1) % NUM; sleep( rand() % 5 ); sem_post( &blank_number ); }
}
int main(int argc, char *argv[] )
{
pthread_t pid, cid;sem_init( &blank_number, 0, NUM ); sem_init( &product_number, 0, 0); pthread_create( &pid, NULL, producer, NULL); pthread_create( &cid, NULL, consumer, NULL); pthread_join( pid, NULL ); pthread_join( cid, NULL ); sem_destroy( &blank_number ); sem_destroy( &product_number ); return 0;
}
- 运行结果:
此代码实现了生产者—消费者问题的模拟,每一个生产者都要把自己生产的产品放入缓冲池,每个消费者从缓冲池中取走产品消费。在这种情况下,生产者消费者进程同步,因为只有通过互通消息才知道是否能存入产品或者取走产品。他们之间也存在互斥,即生产者消费者必须互斥访问缓冲池,即不能有两个以上的进程同时进行。
2 修改代码,把同步资源个数减少为3个,把使用资源的线程增加到 (你的学号%3 + 4)个,编译代码,提交修改后的代码和运行结果截图。
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#include
#include #include #include define NUM 3
int queue[NUM];
sem_t blank_number, product_number;void *producer ( void * arg )
{
static int p = 0;for ( ;; ) { sem_wait( &blank_number ); queue[p] = rand() % 1000; printf("Product %d \n", queue[p]); p = (p+1) % NUM; sleep ( rand() % 3); sem_post( &product_number ); }
}
void *consumer ( void * arg )
{static int c = 0; for( ;; ) { sem_wait( &product_number ); printf("Consume %d\n", queue[c]); c = (c+1) % NUM; sleep( rand() % 3 ); sem_post( &blank_number ); }
}
int main(int argc, char *argv[] )
{
pthread_t pid=3, cid=4;sem_init( &blank_number, 0, NUM ); sem_init( &product_number, 0, 0); pthread_create( &pid, NULL, producer, NULL); pthread_create( &cid, NULL, consumer, NULL); pthread_join( pid, NULL ); pthread_join( cid, NULL ); sem_destroy( &blank_number ); sem_destroy( &product_number ); return 0;
}
- 运行结果: