读者写者问题.pptVIP

读者写者问题 Classical Problems of Synchronization: Readers-Writers Problem Readers-writers problem One data object is to be shared by two kinds of processes. Reader processes may want only to read the shared object. Writer processes may want to update (that is, to read and write) the shared object. Synchronization requirement for A writer and some other processes may not access the shared object simultaneously. More than one reader process can access the shared object simultaneously. 进程通信 Readers first读者优先 No reader will be kept waiting unless a writer has already obtained permission to use the shared object. 如果读者来： 无读者、写者，新读者可以读 有写者等待，但有其他读者正在读，新读者可以读 有写者写，新读者等 如果写者来： 无读者，新写者可以写 有读者，新写者等待 有其他写者，新写者等待 读者： While(true){ P(mutex); readcount++; if(readcount==1) P(w); V(mutex); 读 P(mutex); readcount--; if(readcount==0) V(w); V(mutex); } 写者： While(true){ P(w); 写 V(w); } 写者优先 第二类读者写者问题： 写者优先 条件： 1）多个读者可以同时进行读 2）写者必须互斥（只允许一个写者写，也不能读者写者同时进行） 3）写者优先于读者（一旦有写者，则后续读者必须等待，唤醒时优先考虑写者） readcount=0; writecount=0； S=mutex=w=wmutex=1 读者： while (true) { P(S); P(mutex); readcount ++; if (readcount==1) P (w); V(mutex); V(S); 读 P(mutex); readcount --; if (readcount==0) V(w); V(mutex); }; 写者： while (true) { P(wmutex); writecount++; if(writecount==1) P(S); V(wmutex); P(w); 写 V(w); P(wmutex); writecount--; if(writecount=0) V(S); V(wmutex); }; 分析 S的意义？ 上面代表的情形 能否继续对写者进行提前？ readcount=0; writecount=0；S=mutex=w=wmutex=m=1 读者： while (true) { P(m); P(S); P(mutex); readcount ++; if (readcount==1) P (w); V(mutex); V(S); V(m); 读 P(mutex); readcount --; if (readcount==0)