面試官：Java 實現(xiàn)線程安全的三種方式

一個程序在運(yùn)行起來的時候會轉(zhuǎn)換成進(jìn)程，通常含有多個線程。

通常情況下，一個進(jìn)程中的比較耗時的操作（如長循環(huán)、文件上傳下載、網(wǎng)絡(luò)資源獲取等），往往會采用多線程來解決。

比如顯示生活中，銀行取錢問題、火車票多個售票窗口的問題，通常會涉及到并發(fā)的問題，從而需要多線程的技術(shù)。

當(dāng)進(jìn)程中有多個并發(fā)線程進(jìn)入一個重要數(shù)據(jù)的代碼塊時，在修改數(shù)據(jù)的過程中，很有可能引發(fā)線程安全問題，從而造成數(shù)據(jù)異常。例如，正常邏輯下，同一個編號的火車票只能售出一次，卻由于線程安全問題而被多次售出，從而引起實際業(yè)務(wù)異常。

現(xiàn)在我們就以售票問題來演示線程安全的問題

1， 在不對多線程數(shù)據(jù)進(jìn)行保護(hù)的情況下會引發(fā)的狀況

public?class?ThreadUnSecurity?{
????
????static?int?tickets?=?10;
????
????class?SellTickets?implements?Runnable{

????????@Override
????????public?void?run()?{
????????????//?未加同步時產(chǎn)生臟數(shù)據(jù)
????????????while(tickets?>?0)?{
????????????????
????????????????System.out.println(Thread.currentThread().getName()+"--->售出第：??"+tickets+"?票");
????????????????tickets--;
????????????????
????????????????try?{
????????????????????Thread.sleep(1000);
????????????????}?catch?(InterruptedException?e)?{
????????????????????e.printStackTrace();
????????????????}
????????????????
????????????}
????????????
????????????if?(tickets?<=?0)?{
????????????????
????????????????System.out.println(Thread.currentThread().getName()+"--->售票結(jié)束！");
????????????}
????????}
????}
????
????
????public?static?void?main(String[]?args)?{
????????
????????
????????SellTickets?sell?=?new?ThreadUnSecurity().new?SellTickets();
????????
????????Thread?thread1?=?new?Thread(sell,?"1號窗口");
????????Thread?thread2?=?new?Thread(sell,?"2號窗口");
????????Thread?thread3?=?new?Thread(sell,?"3號窗口");
????????Thread?thread4?=?new?Thread(sell,?"4號窗口");
????????
????????thread1.start();
????????thread2.start();
????????thread3.start();
????????thread4.start();
????????
????????
????}
????

}

上述代碼運(yùn)行的結(jié)果：

1號窗口--->售出第：? 10?票
3號窗口--->售出第：? 10?票
2號窗口--->售出第：? 10?票
4號窗口--->售出第：? 10?票
2號窗口--->售出第：? 6 票
1號窗口--->售出第：? 5 票
3號窗口--->售出第：? 4 票
4號窗口--->售出第：? 3 票
2號窗口--->售出第：? 2 票
4號窗口--->售出第：? 1 票
1號窗口--->售出第：? 1 票
3號窗口--->售票結(jié)束！
2號窗口--->售票結(jié)束！
1號窗口--->售票結(jié)束！
4號窗口--->售票結(jié)束！

我們可以看出同一張票在不對票數(shù)進(jìn)行保護(hù)時會出現(xiàn)同一張票會被出售多次！由于線程調(diào)度中的不確定性，讀者在演示上述代碼時，出現(xiàn)的運(yùn)行結(jié)果會有不同。

第一種方式：同步代碼塊

package?com.bpan.spring.beans.thread;

import?com.sun.org.apache.regexp.internal.recompile;

public?class?ThreadSynchronizedSecurity?{
????
????static?int?tickets?=?10;
????
????class?SellTickets?implements?Runnable{

????????@Override
????????public?void?run()?{
????????????//?同步代碼塊
????????????while(tickets?>?0)?{
????????????????
????????????????synchronized?(this)?{
????????????????????
//????????????????????System.out.println(this.getClass().getName().toString());
????????????????????
????????????????????if?(tickets?<=?0)?{
????????????????????????
????????????????????????return;
????????????????????}
????????????????????
????????????????????System.out.println(Thread.currentThread().getName()+"--->售出第：??"+tickets+"?票");
????????????????????tickets--;
????????????????????
????????????????????try?{
????????????????????????Thread.sleep(100);
????????????????????}?catch?(InterruptedException?e)?{
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????????
????????????????if?(tickets?<=?0)?{
????????????????????
????????????????????System.out.println(Thread.currentThread().getName()+"--->售票結(jié)束！");
????????????????}
????????????}
????????}
????}
????
????
????public?static?void?main(String[]?args)?{
????????
????????
????????SellTickets?sell?=?new?ThreadSynchronizedSecurity().new?SellTickets();
????????
????????Thread?thread1?=?new?Thread(sell,?"1號窗口");
????????Thread?thread2?=?new?Thread(sell,?"2號窗口");
????????Thread?thread3?=?new?Thread(sell,?"3號窗口");
????????Thread?thread4?=?new?Thread(sell,?"4號窗口");
????????
????????thread1.start();
????????thread2.start();
????????thread3.start();
????????thread4.start();
????????
????????
????}
????

}

輸出結(jié)果讀者可自行調(diào)試，不會出現(xiàn)同一張票被出售多次的情況。

第二種方式：同步方法

package?com.bpan.spring.beans.thread;

public?class?ThreadSynchroniazedMethodSecurity?{
????
????
????static?int?tickets?=?10;
????
????class?SellTickets?implements?Runnable{

????????@Override
????????public?void?run()?{
????????????//同步方法
????????????while?(tickets?>?0)?{
????????????????
????????????????synMethod();
????????????????
????????????????try?{
????????????????????Thread.sleep(100);
????????????????}?catch?(InterruptedException?e)?{
????????????????????//?TODO?Auto-generated?catch?block
????????????????????e.printStackTrace();
????????????????}
????????????????
????????????????if?(tickets<=0)?{
????????????????????
????????????????????System.out.println(Thread.currentThread().getName()+"--->售票結(jié)束");
????????????????}
????????????????
????????????}
????????????
????????????
????????}
????????
????????synchronized?void?synMethod()?{
????????????
????????????synchronized?(this)?{
????????????????if?(tickets?<=0)?{
????????????????????
????????????????????return;
????????????????}
????????????????
????????????????System.out.println(Thread.currentThread().getName()+"---->售出第?"+tickets+"?票?");
????????????????tickets--?;
????????????}
????????????
????????}
????????
????}
????public?static?void?main(String[]?args)?{
????????
????????
????????SellTickets?sell?=?new?ThreadSynchroniazedMethodSecurity().new?SellTickets();
????????
????????Thread?thread1?=?new?Thread(sell,?"1號窗口");
????????Thread?thread2?=?new?Thread(sell,?"2號窗口");
????????Thread?thread3?=?new?Thread(sell,?"3號窗口");
????????Thread?thread4?=?new?Thread(sell,?"4號窗口");
????????
????????thread1.start();
????????thread2.start();
????????thread3.start();
????????thread4.start();
????????
????}

}

讀者可自行調(diào)試上述代碼的運(yùn)行結(jié)果

第三種方式：Lock鎖機(jī)制

通過創(chuàng)建Lock對象，采用lock()加鎖，unlock()解鎖，來保護(hù)指定的代碼塊

package?com.bpan.spring.beans.thread;

import?java.util.concurrent.locks.Lock;
import?java.util.concurrent.locks.ReentrantLock;

public?class?ThreadLockSecurity?{
????
????static?int?tickets?=?10;
????
????class?SellTickets?implements?Runnable{
????????
????????Lock?lock?=?new?ReentrantLock();

????????@Override
????????public?void?run()?{
????????????//?Lock鎖機(jī)制
????????????while(tickets?>?0)?{
????????????????
????????????????try?{
????????????????????lock.lock();
????????????????????
????????????????????if?(tickets?<=?0)?{
????????????????????????
????????????????????????return;
????????????????????}
????????????????????????
????????????????????System.out.println(Thread.currentThread().getName()+"--->售出第：??"+tickets+"?票");
????????????????????tickets--;
????????????????}?catch?(Exception?e1)?{
????????????????????//?TODO?Auto-generated?catch?block
????????????????????e1.printStackTrace();
????????????????}finally?{
????????????????????
????????????????????lock.unlock();
????????????????????try?{
????????????????????????Thread.sleep(100);
????????????????????}?catch?(InterruptedException?e)?{
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????}
????????????????
????????????if?(tickets?<=?0)?{
????????????????
????????????????System.out.println(Thread.currentThread().getName()+"--->售票結(jié)束！");
????????????}
????????????
????????}
????}
????
????
????public?static?void?main(String[]?args)?{
????????
????????
????????SellTickets?sell?=?new?ThreadLockSecurity().new?SellTickets();
????????
????????Thread?thread1?=?new?Thread(sell,?"1號窗口");
????????Thread?thread2?=?new?Thread(sell,?"2號窗口");
????????Thread?thread3?=?new?Thread(sell,?"3號窗口");
????????Thread?thread4?=?new?Thread(sell,?"4號窗口");
????????
????????thread1.start();
????????thread2.start();
????????thread3.start();
????????thread4.start();
????????
????????
????}
????

}

最后總結(jié)

由于synchronized是在JVM層面實現(xiàn)的，因此系統(tǒng)可以監(jiān)控鎖的釋放與否；而ReentrantLock是使用代碼實現(xiàn)的，系統(tǒng)無法自動釋放鎖，需要在代碼中的finally子句中顯式釋放鎖lock.unlock()。

另外，在并發(fā)量比較小的情況下，使用synchronized是個不錯的選擇；但是在并發(fā)量比較高的情況下，其性能下降會很嚴(yán)重，此時ReentrantLock是個不錯的方案。

補(bǔ)充

在使用synchronized 代碼塊時,可以與wait()、notify()、nitifyAll()一起使用，從而進(jìn)一步實現(xiàn)線程的通信。

其中，wait()方法會釋放占有的對象鎖，當(dāng)前線程進(jìn)入等待池，釋放cpu,而其他正在等待的線程即可搶占此鎖，獲得鎖的線程即可運(yùn)行程序；

線程的sleep()方法則表示，當(dāng)前線程會休眠一段時間，休眠期間，會暫時釋放cpu，但并不釋放對象鎖，也就是說，在休眠期間，其他線程依然無法進(jìn)入被同步保護(hù)的代碼內(nèi)部，當(dāng)前線程休眠結(jié)束時，會重新獲得cpu執(zhí)行權(quán),從而執(zhí)行被同步保護(hù)的代碼。

wait()和sleep()最大的不同在于wait()會釋放對象鎖，而sleep()不會釋放對象鎖。

notify()方法會喚醒因為調(diào)用對象的wait()而處于等待狀態(tài)的線程，從而使得該線程有機(jī)會獲取對象鎖。調(diào)用notify()后，當(dāng)前線程并不會立即釋放鎖，而是繼續(xù)執(zhí)行當(dāng)前代碼，直到synchronized中的代碼全部執(zhí)行完畢，才會釋放對象鎖。JVM會在等待的線程中調(diào)度一個線程去獲得對象鎖，執(zhí)行代碼。

需要注意的是，wait()和notify()必須在synchronized代碼塊中調(diào)用。

notifyAll()是喚醒所有等待的線程。

下面是示例代碼，

package?com.bpan.spring.beans.thread;

public?class?ThreadDemo?{
????
????static?final?Object?obj?=?new?Object();
????
????//第一個子線程
????static?class?ThreadA?implements?Runnable{

????????@Override
????????public?void?run()?{
????????????
????????????
????????????int?count?=?10;
????????????while(count?>?0)?{
????????????????
????????????????synchronized?(ThreadDemo.obj)?{
????????????????????
????????????????????System.out.println("A-----"+count);
????????????????????count--;
????????????????????
????????????????????synchronized?(ThreadDemo.obj)?{
????????????????????????
????????????????????????//notify()方法會喚醒因為調(diào)用對象的wait()而處于等待狀態(tài)的線程，從而使得該線程有機(jī)會獲取對象鎖。
????????????????????????//調(diào)用notify()后，當(dāng)前線程并不會立即釋放鎖，而是繼續(xù)執(zhí)行當(dāng)前代碼，直到synchronized中的代碼全部執(zhí)行完畢，
????????????????????????ThreadDemo.obj.notify();
????????????????????????
????????????????????????try?{
????????????????????????????ThreadDemo.obj.wait();
????????????????????????}?catch?(InterruptedException?e)?{
????????????????????????????//?TODO?Auto-generated?catch?block
????????????????????????????e.printStackTrace();
????????????????????????}
????????????????????}
????????????????}
????????????}
????????????
????????}
????????
????}
????
????static?class?ThreadB?implements?Runnable{
????????
????????
????????@Override
????????public?void?run()?{
????????????
????????????int?count?=?10;
????????????
????????????while(count?>?0)?{
????????????????
????????????????synchronized?(ThreadDemo.obj)?{
????????????????????System.out.println("B-----"+count);
????????????????????count--;
????????????????????
????????????????????synchronized?(ThreadDemo.obj)?{
????????????????????
????????????????????????//notify()方法會喚醒因為調(diào)用對象的wait()而處于等待狀態(tài)的線程，從而使得該線程有機(jī)會獲取對象鎖。
????????????????????????//調(diào)用notify()后，當(dāng)前線程并不會立即釋放鎖，而是繼續(xù)執(zhí)行當(dāng)前代碼，直到synchronized中的代碼全部執(zhí)行完畢，
????????????????????????ThreadDemo.obj.notify();
????????????????????????
????????????????????????try?{
????????????????????????????ThreadDemo.obj.wait();
????????????????????????}?catch?(InterruptedException?e)?{
????????????????????????????//?TODO?Auto-generated?catch?block
????????????????????????????e.printStackTrace();
????????????????????????}
????????????????????}
????????????????????
????????????????}
????????????????
????????????}
????????????
????????}
????????
????}
????
????public?static?void?main(String[]?args)?{
????????
????????
????????new?Thread(new?ThreadA()).start();
????????new?Thread(new?ThreadB()).start();
????????
????}

}

來源：cnblogs.com/revel171226/p/9411131.html

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