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普通int值在多线程下的递增操作
Java针对多线程下的数值安全计数器设计了一些类,这些类叫做原子类,其中一部分如下:
java.util.concurrent.atomic.AtomicBoolean;
java.util.concurrent.atomic.AtomicInteger;
java.util.concurrent.atomic.AtomicLong;
java.util.concurrent.atomic.AtomicReference;
下面是一个对比 AtomicInteger 与 普通 int 值在多线程下的递增测试,使用的是 junit4;
完整代码:
package test.java;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.atomic.AtomicInteger;
import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;
/**
* 测试AtomicInteger与普通int值在多线程下的递增操作
*/
public class TestAtomic {
// 原子Integer递增对象
public static AtomicInteger counter_integer;// = new AtomicInteger(0);
// 一个int类型的变量
public static int count_int = 0;
@Before
public void setUp() {
// 所有测试开始之前执行初始设置工作
counter_integer = new AtomicInteger(0);
}
@Test
public void testAtomic() throws InterruptedException {
// 创建的线程数量
int threadCount = 100;
// 其他附属线程内部循环多少次
int loopCount = 10000600;
// 控制附属线程的辅助对象;(其他await的线程先等着主线程喊开始)
CountDownLatch latch_1 = new CountDownLatch(1);act答案
// 控制主线程的辅助对象;(主线程等着所有附属线程都运行完毕再继续)
CountDownLatch latch_n = new CountDownLatch(threadCount);
// 创建并启动其他附属线程
for (int i = 0; i < threadCount; i++) {
Thread thread = new AtomicIntegerThread(latch_1, latch_n, loopCount);
thread.start();
}
long startNano = System.nanoTime();
// 让其他等待的线程统一开始
latch_1.countDown();
// 等待其他线程执行完
latch_n.await();
//
long endNano = System.nanoTime();
int sum = counter_integer.get();
//
Assert.assertEquals("sum 不等于 threadCount * loopCount,测试失败",
sum, threadCount * loopCount);
System.out.println("--------testAtomic(); 预期两者相等------------");
System.out.println("耗时: " + ((endNano - startNano) / (1000 * 1000)) + "ms");
System.out.println("threadCount = " + (threadCount) + ";");
System.out.println("loopCount = " + (loopCount) + ";");
System.out.println("sum = " + (sum) + ";");
}
@Test
public void testIntAdd() throws InterruptedException {
// 创建的线程数量
int threadCount = 100;
// 其他附属线程内部循环多少次
int loopCount = 10000600;
// 控制附属线程的辅助对象;(其他await的线程先等着主线程喊开始)
CountDownLatch latch_1 = new CountDownLatch(1);
// 控制主线程的辅助对象;(主线程等着所有附属线程都运行完毕再继续)
CountDownLatch latch_n = new CountDownLatch(threadCount);
// 创建并启动其他附属线程SAT答案
for (int i = 0; i < threadCount; i++) {
Thread thread = new IntegerThread(latch_1, latch_n, loopCount);
thread.start();
}
long startNano = System.nanoTime();
// 让其他等待的线程统一开始
latch_1.countDown();
// 等待其他线程执行完
latch_n.await();
//
long endNano = System.nanoTime();
int sum = count_int;
//
Assert.assertNotEquals(
"sum 等于 threadCount * loopCount,testIntAdd()测试失败",
sum, threadCount * loopCount);
System.out.println("-------testIntAdd(); 预期两者不相等---------");
System.out.println("耗时: " + ((endNano - startNano) / (1000*1000))+ "ms");
System.out.println("threadCount = " + (threadCount) + ";");
System.out.println("loopCount = " + (loopCount) + ";");
System.out.println("sum = " + (sum) + ";");
}
// 线程
class AtomicIntegerThread extends Thread {
private CountDownLatch latch = null;
private CountDownLatch latchdown = null;
private int loopCount;
public AtomicIntegerThread(CountDownLatch latch,
CountDownLatch latchdown, int loopCount) {
this.latch = latch;
this.latchdown = latchdown;
this.loopCount = loopCount;
}
@Override
public void run() {
// 等待信号同步
try {
this.latch.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
//
for (int i = 0; i < loopCount; i++) {
counter_integer.getAndIncrement();
}
// 通知递减1次
latchdown.countDown();
}
}
// 线程
class IntegerThread extends Thread {
private CountDownLatch latch = null;
private CountDownLatch latchdown = null;
private int loopCount;
public IntegerThread(CountDownLatch latch,
CountDownLatch latchdown, int loopCount) {
this.latch = latch;
this.latchdown = latchdown;
this.loopCount = loopCount;
}
@Override
public void run() {
// 等待信号同步
try {
this.latch.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
//
for (int i = 0; i < loopCount; i++) {
count_int++;
}
// 通知递减1次
latchdown.countDown();
}
}
}
普通PC机上的执行结果类似如下:
--------------testAtomic(); 预期两者相等-------------------
耗时: 85366ms
threadCount = 100;
loopCount = 10000600;
sum = 1000060000;
--------------testIntAdd(); 预期两者不相等-------------------
耗时: 1406ms
threadCount = 100;
loopCount = 10000600;
sum = 119428988;
从中可以看出, AtomicInteger操作 与 int操作的效率大致相差在50-80倍上下,当然,int很不消耗时间,这个对比只是提供一个参照。
如果确定是单线程执行,那应该使用 int; 而int在多线程下的操作执行的效率还是蛮高的, 10亿次只花了1.5秒钟;
(假设CPU是 2GHZ,双核4线程,理论最大8GHZ,则每秒理论上有80亿个时钟周期,10亿次Java的int增加消耗了1.5秒,即 120亿次运算, 算下来每次循环消耗CPU周期 12个;
个人觉得效率不错, C 语言也应该需要4个以上的时钟周期(判断,执行内部代码,自增判断,跳转)
前提是: JVM和CPU没有进行激进优化。托福答案 www.daan678.com
而 AtomicInteger 效率其实也不低,10亿次消耗了80秒, 那100万次大约也就是千分之一,80毫秒的样子。
普通int值在多线程下的递增操作