王豔 201771010127《面向物件程式設計(java)》第十七週學習總結
實驗十七 執行緒同步控制
實驗時間 2018-12-10
一、理論部分
1.執行緒同步:多執行緒併發執行不確定性問題解決方案:引入執行緒同步機制,使得另一執行緒要使用該方法,就只能等待。
解決方案:
1)鎖物件與條件物件
有關鎖物件和條件物件的關鍵要點:
➢ 鎖用來保護程式碼片段,保證任何時刻只能有一個執行緒執行被保護的程式碼。
➢ 鎖管理試圖進入被保護程式碼段的執行緒。
➢ 鎖可擁有一個或多個相關條件物件。
➢ 每個條件物件管理那些已經進入被保護的程式碼段但還不能執行的執行緒。
2)synchronized關鍵字
某個類內方法用synchronized 修飾後,該方法被稱為同步方法;
只要某個執行緒正在訪問同步方法,其他執行緒欲要訪問同步方法就被阻塞,直至執行緒從同步方法返回前喚醒被阻塞執行緒,其他執行緒方可能進入同步方法。
一個執行緒在使用的同步方法中時,可能根據問題的需要,必須使用wait()方法使本執行緒等待,暫時讓出CPU的使用權,並允許其它執行緒使用這個同步方法。
執行緒如果用完同步方法,應當執行notifyAll()方法通知所有由於使用這個同步方法而處於等待的執行緒結束等待。
二、實驗部分
1、實驗目的與要求
(1) 掌握執行緒同步的概念及實現技術;
(2) 執行緒綜合程式設計練習
2、實驗內容和步驟
實驗1:測試程式並進行程式碼註釋。
測試程式
l 在Elipse環境下除錯教材651頁程式14-7,結合程式執行結果理解程式;
掌握利用鎖物件和條件物件實現的多執行緒同步技術。
程式如下:
package test2; /** * This program shows how multiple threads can safely access a data structure. * @version 1.31 2015-06-21 * @author Cay Horstmann */ public class SynchBankTest { public static final int NACCOUNTS = 100; public static final double INITIAL_BALANCE = 1000; public static final double MAX_AMOUNT = 1000; public static final int DELAY = 10; public static void main(String[] args) { Bank bank = new Bank(NACCOUNTS, INITIAL_BALANCE); for (int i = 0; i < NACCOUNTS; i++) { int fromAccount = i; Runnable r = () -> { try { while (true) { int toAccount = (int) (bank.size() * Math.random()); double amount = MAX_AMOUNT * Math.random(); bank.transfer(fromAccount, toAccount, amount); Thread.sleep((int) (DELAY * Math.random())); } } catch (InterruptedException e) { } }; Thread t = new Thread(r); t.start(); } } }
package test2; import java.util.*; import java.util.concurrent.locks.*; /** * A bank with a number of bank accounts that uses locks for serializing access. * @version 1.30 2004-08-01 * @author Cay Horstmann */ public class Bank { private final double[] accounts; private Lock bankLock; private Condition sufficientFunds; /** * Constructs the bank. * @param n the number of accounts * @param initialBalance the initial balance for each account */ public Bank(int n, double initialBalance) { accounts = new double[n]; Arrays.fill(accounts, initialBalance); bankLock = new ReentrantLock(); sufficientFunds = bankLock.newCondition(); } /** * Transfers money from one account to another. * @param from the account to transfer from * @param to the account to transfer to * @param amount the amount to transfer */ public void transfer(int from, int to, double amount) throws InterruptedException { bankLock.lock(); try { while (accounts[from] < amount) sufficientFunds.await(); System.out.print(Thread.currentThread()); accounts[from] -= amount; System.out.printf(" %10.2f from %d to %d", amount, from, to); accounts[to] += amount; System.out.printf(" Total Balance: %10.2f%n", getTotalBalance()); sufficientFunds.signalAll(); } finally { bankLock.unlock(); } } /** * Gets the sum of all account balances. * @return the total balance */ public double getTotalBalance() { bankLock.lock(); try { double sum = 0; for (double a : accounts) sum += a; return sum; } finally { bankLock.unlock(); } } /** * Gets the number of accounts in the bank. * @return the number of accounts */ public int size() { return accounts.length; } }
程式執行結果如下:
測試程式2:
l 在Elipse環境下除錯教材655頁程式14-8,結合程式執行結果理解程式;
l 掌握synchronized在多執行緒同步中的應用。
程式如下:
package test2; /** * This program shows how multiple threads can safely access a data structure, * using synchronized methods. * @version 1.31 2015-06-21 * @author Cay Horstmann */ public class SynchBankTest2 { public static final int NACCOUNTS = 100; public static final double INITIAL_BALANCE = 1000; public static final double MAX_AMOUNT = 1000; public static final int DELAY = 10; public static void main(String[] args) { Bank bank = new Bank(NACCOUNTS, INITIAL_BALANCE); for (int i = 0; i < NACCOUNTS; i++) { int fromAccount = i; Runnable r = () -> { try { while (true) { int toAccount = (int) (bank.size() * Math.random()); double amount = MAX_AMOUNT * Math.random(); bank.transfer(fromAccount, toAccount, amount); Thread.sleep((int) (DELAY * Math.random())); } } catch (InterruptedException e) { } }; Thread t = new Thread(r); t.start(); } } }
package test2; import java.util.*; import java.util.concurrent.locks.*; /** * A bank with a number of bank accounts that uses locks for serializing access. * @version 1.30 2004-08-01 * @author Cay Horstmann */ public class Bank { private final double[] accounts; private Lock bankLock; private Condition sufficientFunds; /** * Constructs the bank. * @param n the number of accounts * @param initialBalance the initial balance for each account */ public Bank(int n, double initialBalance) { accounts = new double[n]; Arrays.fill(accounts, initialBalance); bankLock = new ReentrantLock(); sufficientFunds = bankLock.newCondition(); } /** * Transfers money from one account to another. * @param from the account to transfer from * @param to the account to transfer to * @param amount the amount to transfer */ public void transfer(int from, int to, double amount) throws InterruptedException { bankLock.lock(); try { while (accounts[from] < amount) sufficientFunds.await(); System.out.print(Thread.currentThread()); accounts[from] -= amount; System.out.printf(" %10.2f from %d to %d", amount, from, to); accounts[to] += amount; System.out.printf(" Total Balance: %10.2f%n", getTotalBalance()); sufficientFunds.signalAll(); } finally { bankLock.unlock(); } } /** * Gets the sum of all account balances. * @return the total balance */ public double getTotalBalance() { bankLock.lock(); try { double sum = 0; for (double a : accounts) sum += a; return sum; } finally { bankLock.unlock(); } } /** * Gets the number of accounts in the bank. * @return the number of accounts */ public int size() { return accounts.length; } }
程式執行結果:
測試程式3:
l 在Elipse環境下執行以下程式,結合程式執行結果分析程式存在問題;
l 嘗試解決程式中存在問題。
package test2; class Cbank { private static int s=2000; public static void sub(int m) { int temp=s; temp=temp-m; try { Thread.sleep((int)(1000*Math.random())); } catch (InterruptedException e) { } s=temp; System.out.println("s="+s); } } class Customer extends Thread { public void run() { for( int i=1; i<=4; i++) Cbank.sub(100); } } public class Thread3 { public static void main(String args[]) { Customer customer1 = new Customer(); Customer customer2 = new Customer(); customer1.start(); customer2.start(); } }
程式執行後如圖:
分析程式,輸出結果應為1900—1200,而控制檯上顯示結果有重複。因為沒有對程式加上鎖功能。修改後程式如下:
package test2; class Cbank { private static int s=2000; public synchronized static void sub(int m) { int temp=s; temp=temp-m; try { Thread.sleep((int)(1000*Math.random())); } catch (InterruptedException e) { } s=temp; System.out.println("s="+s); } } class Customer extends Thread { public void run() { for( int i=1; i<=4; i++) Cbank.sub(100); } } public class Thread3 { public static void main(String args[]) { Customer customer1 = new Customer(); Customer customer2 = new Customer(); customer1.start(); customer2.start(); } }
程式執行結果如下:
實驗2 程式設計練習
利用多執行緒及同步方法,編寫一個程式模擬火車票售票系統,共3個視窗,賣10張票,程式輸出結果類似(程式輸出不唯一,可以是其他類似結果)。
Thread-0視窗售:第1張票
Thread-0視窗售:第2張票
Thread-1視窗售:第3張票
Thread-2視窗售:第4張票
Thread-2視窗售:第5張票
Thread-1視窗售:第6張票
Thread-0視窗售:第7張票
Thread-2視窗售:第8張票
Thread-1視窗售:第9張票
Thread-0視窗售:第10張票
程式如下:
package test2; public class Demo { public static void main(String[] args) { Mythread mythread = new Mythread(); Thread ticket1 = new Thread(mythread); Thread ticket2 = new Thread(mythread); Thread ticket3 = new Thread(mythread); ticket1.start(); ticket2.start(); ticket3.start(); } } class Mythread implements Runnable { int ticket = 1; boolean flag = true; @Override public void run() { while (flag) { try { Thread.sleep(500); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } synchronized (this) { if (ticket <= 10) { System.out.println(Thread.currentThread().getName() + "視窗售:第" + ticket + "張票"); ticket++; } if (ticket > 10) { flag = false; } } } } }
程式執行結果如下:
三:實驗總結
這周繼續學習了有關執行緒的知識,主要學習了有關執行緒同步的問題。執行緒同步主要是為了解決多執行緒併發執行不確定性問題,使得多個執行緒中在一個執行緒使用某種方法時候,另一執行緒要使用該方法,就只能等待。實驗課上,老師和學長通過演示具體的例子給我們展現了多執行緒中在不加鎖時會出現的情況,讓我們對執行緒同步有了更加清晰地認識。雖然很多地方還是不太懂還是存在很大的問題,但是實驗課上講的內容聽得比較清晰,課後自己再執行試驗時也有了更深的體會。