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Date: Mon, 11 Nov 1996 17:25:18 GMTServer: NCSA/1.5Content-type: text/htmlLast-modified: Tue, 27 Feb 1996 15:51:55 GMTContent-length: 3318<html><head><title>CS 537 - Quiz #3</title></head><body><table border=0 width=100% align=center><tr><td width=25%><td width=50% align=center><b>UNIVERSITY OF WISCONSIN-MADISON<br>Computer Sciences Department</b><td width=25%><tr><tr><td><b>CS 537<br>Spring 1996 </b><td><td align=right><b>Bart Miller</b><tr><td><td align=center><b>Quiz #3</b><br>Wednesday, February 21<td></table><h2>Using Monitors to Save Your Life</h2>You are to write the code to provide the synchronization to controlaccess to a couple of pedestrian bridges.The bridges are used to cross a dangerous, shark-infested river.The picture below illustrates the problem.<CENTER><BR><IMG ALIGN=CENTER SRC="quiz3.river.gif" ALT="Bridges"><BR></CENTER><p>You have<b>provided to you</b> (already written)a procedure called<tt>CrossBridge(bridgenum)</tt>.You don't know how long this takes to execute.This procedure causes the person process to cross bridge<tt>bridgenum</tt> (which can have valid values of 1 or 2).A person starting on the east shore will end up on the west shore.A person starting on the west shore will end up on the east shore.This procedure returns when the action is completed.<p>Use<b>monitors</b>as the synchronization mechanism for your solution.Use the syntax that was presented in lecture: monitors are simply C++ classeswith the extra "monitor" keyword.<p>Each person is a process and these processes arrive randomly.You will write a procedure called<tt>Person()</tt>that will be called by people wanting to cross the river.You will write any additional procedures that you need.<p>Your solution must obey the following rules:<ol><li>Each bridge is strong enough to hold only 1 person at a time.Additional people will break the bridge and they all will fall in theriver and be eaten.<li>If there is more than one person wanting to cross the river, both bridgesshould be in use at the same time.<li>People should get to use the bridge in approximately the same order in whichthey arrived.<li>Initially, both bridges are unoccupied.</ol><p>Hint: the <tt>Person()</tt> procedure is probably <i>not</i> in the monitor.You might consider having this procedure call procedures in a "BridgeControl"monitor; this is similar to how we did the readers/writers problem.<p><table width=100% border=1 align=center><tr><td><pre>BridgeControl river;void Person(){ int b = river.GetBridge(); CrossBridge(b); river.Done(b);}</pre><tr><td><pre>monitor class BridgeControl {public: BridgeControl(); int GetBridge(); void Done(int);private: int busy[2]; cond waitList;};</pre><tr><td><pre>BridgeControl::BridgeControl() { busy[0] = busy[1] = 0;}</pre><tr><td><pre>intBridgeControl::GetBridge();{ while (1) { /* Note: this is NOT busy waiting. */ for (int i = 0; i < 2; i++) { if (!busy[i]) { busy[i] = 1; return (i+1); } } wait(waitList); }}</pre><tr><td><pre>void BridgeControl::Done(int bridge){ busy[bridge-1] = 0; wakeup(waitList);}</pre></table><hr><H4>Last modified:Wed Feb 21 10:59:47 CST 1996by<a href="http://www.cs.wisc.edu/~bart">bart</a></b></H4></body>?? 快捷鍵說明
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