本文共 948 字，大约阅读时间需要 3 分钟。

Classical Problems of Synchronization

Bounded-Buffer Problem

semaphore full, empty, mutex;Initially:full = 0, empty = n, mutex = 1

do { … 	produce an item in nextp    …     wait(empty);    wait(mutex);    …     add nextp to buffer    …     signal(mutex);     signal(full);     } while (1);

do {	wait(full) 	wait(mutex);	… 	remove an item from buffer to nextc	…	signal(mutex);	signal(empty);	… 	consume the item in nextc …	} while (1);

Readers-Writers Problem

读者优先 写者优先

semaphore mutex, wrt;Initiallymutex = 1, wrt = 1, readcount = 0

wait(mutex);readcount++; if (readcount == 1) 	wait(wrt);signal(mutex);	…reading is performed	… wait(mutex);readcount--;if (readcount == 0) signal(wrt);signal(mutex):

Dining-Philosophers Problem

Philosopher i:	do { 		wait(chopstick[i])        wait(chopstick[(i+1) % 5])        …         eat        …         signal(chopstick[i]);        signal(chopstick[(i+1) % 5]);        …        think        …       } while (1);

转载地址：http://rlgwi.baihongyu.com/