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2.18 普通信号量案例

Giovanna

About 112 wordsLess than 1 minute

2024-08-31

Caution

大题

在两个进程中都要包含两个对Dmutex和Emutex的操作,即

// process A:
wait(Dmutex);
wait(Emutex);

// process B:
wait(Emutex);
wait(Dmutex);

若进程A和B按下述次序交替执行wait操作:

process A:wait(Dmutex);,于是Dmutex=0

process B:wait(Emutex);,于是Emutex=0

process A:wait(Emutex);,于是Emutex=-1 A阻塞

process B:wait(Dmutex);,于是Dmutex=-1 B阻塞