作者：ZimZz

鏈接：https://www.cnblogs.com/zemliu/p/3298685.html

簡(jiǎn)介

volatile關(guān)鍵字保證了在多線程環(huán)境下,被修飾的變量在別修改后會(huì)馬上同步到主存,這樣該線程對(duì)這個(gè)變量的修改就是對(duì)所有其他線程可見(jiàn)的,其他線程能夠馬上讀到這個(gè)修改后值.

Thread的本地內(nèi)存

• 每個(gè)Thread都擁有自己的線程存儲(chǔ)空間
• Thread何時(shí)同步本地存儲(chǔ)空間的數(shù)據(jù)到主存是不確定的

例子

借用Google JEREMY MANSON 的解釋,上圖表示兩個(gè)線程并發(fā)執(zhí)行,而且代碼順序上為Thread1->Thread2

1、不用 volatile

假如ready字段不使用volatile,那么Thread 1對(duì)ready做出的修改對(duì)于Thread2來(lái)說(shuō)未必是可見(jiàn)的,是否可見(jiàn)是不確定的.假如此時(shí)thread1 ready泄露了(leak through)了,那么Thread 2可以看見(jiàn)ready為true,但是有可能answer的改變并沒(méi)有泄露,則thread2有可能會(huì)輸出 0 (answer=42對(duì)thread2并不可見(jiàn))

2、使用 volatile

使用volatile以后,做了如下事情

• 每次修改volatile變量都會(huì)同步到主存中
• 每次讀取volatile變量的值都強(qiáng)制從主存讀取最新的值(強(qiáng)制JVM不可優(yōu)化volatile變量,如JVM優(yōu)化后變量讀取會(huì)使用cpu緩存而不從主存中讀取)
• 線程 A 中寫(xiě)入 volatile 變量之前可見(jiàn)的變量, 在線程 B 中讀取該 volatile 變量以后, 線程 B 對(duì)其他在 A 中的可見(jiàn)變量也可見(jiàn). 換句話說(shuō), 寫(xiě) volatile 類似于退出同步塊, 而讀取 volatile 類似于進(jìn)入同步塊

所以如果使用了volatile,那么Thread2讀取到的值為read=>true,answer=>42,當(dāng)然使用volatile的同時(shí)也會(huì)增加性能開(kāi)銷

注意

volatile并不能保證非源自性操作的多線程安全問(wèn)題得到解決,volatile解決的是多線程間共享變量的可見(jiàn)性問(wèn)題,而例如多線程的i++,++i,依然還是會(huì)存在多線程問(wèn)題,它是無(wú)法解決了.如下:使用一個(gè)線程i++,另一個(gè)i--,最終得到的結(jié)果不為0

public?class?VolatileTest?{

????private?static?volatile?int?count?=?0;
????private?static?final?int?times?=?Integer.MAX_VALUE;

????public?static?void?main(String[]?args)?{

????????long?curTime?=?System.nanoTime();

????????Thread?decThread?=?new?DecThread();
????????decThread.start();

????????//?使用run()來(lái)運(yùn)行結(jié)果為0,原因是單線程執(zhí)行不會(huì)有線程安全問(wèn)題
????????//?new?DecThread().run();

????????System.out.println("Start?thread:?"?+?Thread.currentThread()?+?"?i++");

????????for?(int?i?=?0;?i?????????????count++;
????????}

????????System.out.println("End?thread:?"?+?Thread.currentThread()?+?"?i--");

????????//?等待decThread結(jié)束
????????while?(decThread.isAlive());

????????long?duration?=?System.nanoTime()?-?curTime;
????????System.out.println("Result:?"?+?count);
????????System.out.format("Duration:?%.2fs\n",?duration?/?1.0e9);
????}

????private?static?class?DecThread?extends?Thread?{

????????@Override
????????public?void?run()?{
????????????System.out.println("Start?thread:?"?+?Thread.currentThread()?+?"?i--");
????????????for?(int?i?=?0;?i?????????????????count--;
????????????}
????????????System.out.println("End?thread:?"?+?Thread.currentThread()?+?"?i--");
????????}
????}
}

最后輸出的結(jié)果是

“

Start thread: Thread[main,5,main] i++Start thread: Thread[Thread-0,5,main] i--End thread: Thread[main,5,main] i--End thread: Thread[Thread-0,5,main] i--Result: -460370604Duration: 67.37s

原因是i++和++i并非原子操作,我們?nèi)舨榭醋止?jié)碼,會(huì)發(fā)現(xiàn)

void?f1()?{?i++;?}

的字節(jié)碼如下

void?f1();
Code:
0:?aload_0
1:?dup
2:?getfield?#2;?//Field?i:I
5:?iconst_1
6:?iadd
7:?putfield?#2;?//Field?i:I
10:?return

可見(jiàn)i++執(zhí)行了多部操作, 從變量i中讀取讀取i的值 -> 值+1 -> 將+1后的值寫(xiě)回i中,這樣在多線程的時(shí)候執(zhí)行情況就類似如下了

Thread1?????????????Thread2
r1?=?i;?????????????r3?=?i;
r2?=?r1?+?1;????????r4?=?r3?+?1;
i?=?r2;?????????????i?=?r4;

這樣會(huì)造成的問(wèn)題就是 r1, r3讀到的值都是 0, 最后兩個(gè)線程都將 1 寫(xiě)入 i, 最后 i 等于 1, 但是卻進(jìn)行了兩次自增操作

可知加了volatile和沒(méi)加volatile都無(wú)法解決非原子操作的線程同步問(wèn)題

線程同步問(wèn)題的解決

Java提供了java.util.concurrent.atomic 包來(lái)提供線程安全的基本類型包裝類,例子如下

package?com.qunar.atomicinteger;

import?java.util.concurrent.atomic.AtomicInteger;

/**
?*?@author?zhenwei.liu?created?on?2013?13-9-2?下午10:18
?*?@version?$Id$
?*/
public?class?SafeTest?{

????private?static?AtomicInteger?count?=?new?AtomicInteger(0);
????private?static?final?int?times?=?Integer.MAX_VALUE;

????public?static?void?main(String[]?args)?{

????????long?curTime?=?System.nanoTime();

????????Thread?decThread?=?new?DecThread();
????????decThread.start();

????????//?使用run()來(lái)運(yùn)行結(jié)果為0,原因是單線程執(zhí)行不會(huì)有線程安全問(wèn)題
????????//?new?DecThread().run();

????????System.out.println("Start?thread:?"?+?Thread.currentThread()?+?"?i++");

????????for?(int?i?=?0;?i?????????????count.incrementAndGet();
????????}

????????//?等待decThread結(jié)束
????????while?(decThread.isAlive());

????????long?duration?=?System.nanoTime()?-?curTime;
????????System.out.println("Result:?"?+?count);
????????System.out.format("Duration:?%.2f\n",?duration?/?1.0e9);
????}

????private?static?class?DecThread?extends?Thread?{

????????@Override
????????public?void?run()?{
????????????System.out.println("Start?thread:?"?+?Thread.currentThread()?+?"?i--");
????????????for?(int?i?=?0;?i?????????????????count.decrementAndGet();
????????????}
????????????System.out.println("End?thread:?"?+?Thread.currentThread()?+?"?i--");
????????}
????}
}

輸出

“

Start thread: Thread[main,5,main] i++Start thread: Thread[Thread-0,5,main] i--End thread: Thread[Thread-0,5,main] i--Result: 0Duration: 105.15

結(jié)論

1. volatile解決了線程間共享變量的可見(jiàn)性問(wèn)題
2. 使用volatile會(huì)增加性能開(kāi)銷
3. volatile并不能解決線程同步問(wèn)題
4. 解決i++或者++i這樣的線程同步問(wèn)題需要使用synchronized或者AtomicXX系列的包裝類,同時(shí)也會(huì)增加性能開(kāi)銷