此线程安全字节序列生成器有什么问题? [英] what is wrong with this thread-safe byte sequence generator?
问题描述
我需要一个字节生成器，它将生成从Byte.MIN_VALUE到Byte.MAX_VALUE的值.达到MAX_VALUE时，应从MIN_VALUE重新开始.
I need a byte generator that would generate values from Byte.MIN_VALUE to Byte.MAX_VALUE. When it reaches MAX_VALUE, it should start over again from MIN_VALUE.
我已经使用AtomicInteger编写了代码(见下文)；但是，如果并发访问并使用Thread.sleep()人为地使其变慢，则代码似乎无法正常工作(如果没有睡眠，则运行良好；但是，我怀疑它太快了，无法显示并发问题).
I have written the code using AtomicInteger (see below); however, the code does not seem to behave properly if accessed concurrently and if made artificially slow with Thread.sleep() (if no sleeping, it runs fine; however, I suspect it is just too fast for concurrency problems to show up).
代码(带有一些添加的调试代码):
The code (with some added debug code):
public class ByteGenerator {
private static final int INITIAL_VALUE = Byte.MIN_VALUE-1;
private AtomicInteger counter = new AtomicInteger(INITIAL_VALUE);
private AtomicInteger resetCounter = new AtomicInteger(0);
private boolean isSlow = false;
private long startTime;
public byte nextValue() {
int next = counter.incrementAndGet();
//if (isSlow) slowDown(5);
if (next > Byte.MAX_VALUE) {
synchronized(counter) {
int i = counter.get();
//if value is still larger than max byte value, we reset it
if (i > Byte.MAX_VALUE) {
counter.set(INITIAL_VALUE);
resetCounter.incrementAndGet();
if (isSlow) slowDownAndLog(10, "resetting");
} else {
if (isSlow) slowDownAndLog(1, "missed");
}
next = counter.incrementAndGet();
}
}
return (byte) next;
}
private void slowDown(long millis) {
try {
Thread.sleep(millis);
} catch (InterruptedException e) {
}
}
private void slowDownAndLog(long millis, String msg) {
slowDown(millis);
System.out.println(resetCounter + " "
+ (System.currentTimeMillis()-startTime) + " "
+ Thread.currentThread().getName() + ": " + msg);
}
public void setSlow(boolean isSlow) {
this.isSlow = isSlow;
}
public void setStartTime(long startTime) {
this.startTime = startTime;
}
}
然后，测试:
public class ByteGeneratorTest {
@Test
public void testGenerate() throws Exception {
ByteGenerator g = new ByteGenerator();
for (int n = 0; n < 10; n++) {
for (int i = Byte.MIN_VALUE; i <= Byte.MAX_VALUE; i++) {
assertEquals(i, g.nextValue());
}
}
}
@Test
public void testGenerateMultiThreaded() throws Exception {
final ByteGenerator g = new ByteGenerator();
g.setSlow(true);
final AtomicInteger[] counters = new AtomicInteger[Byte.MAX_VALUE-Byte.MIN_VALUE+1];
for (int i = 0; i < counters.length; i++) {
counters[i] = new AtomicInteger(0);
}
Thread[] threads = new Thread[100];
final CountDownLatch latch = new CountDownLatch(threads.length);
for (int i = 0; i < threads.length; i++) {
threads[i] = new Thread(new Runnable() {
public void run() {
try {
for (int i = Byte.MIN_VALUE; i <= Byte.MAX_VALUE; i++) {
byte value = g.nextValue();
counters[value-Byte.MIN_VALUE].incrementAndGet();
}
} finally {
latch.countDown();
}
}
}, "generator-client-" + i);
threads[i].setDaemon(true);
}
g.setStartTime(System.currentTimeMillis());
for (int i = 0; i < threads.length; i++) {
threads[i].start();
}
latch.await();
for (int i = 0; i < counters.length; i++) {
System.out.println("value #" + (i+Byte.MIN_VALUE) + ": " + counters[i].get());
}
//print out the number of hits for each value
for (int i = 0; i < counters.length; i++) {
assertEquals("value #" + (i+Byte.MIN_VALUE), threads.length, counters[i].get());
}
}
}
在我的2核计算机上，结果是#-128值获得146次命中(由于我们有100个线程，因此它们都应平均获得100次命中).
The result on my 2-core machine is that value #-128 gets 146 hits (all of them should get 100 hits equally as we have 100 threads).
如果有人有任何想法，那么这段代码出了什么问题，我真是无所适从.
If anyone has any ideas, what's wrong with this code, I'm all ears/eyes.
更新:对于那些急于不想向下滚动的人，用Java解决此问题的正确(且最短，最优雅)的方法将是这样的:
UPDATE: for those who are in a hurry and do not want to scroll down, the correct (and shortest and most elegant) way to solve this in Java would be like this:
public byte nextValue() {
return (byte) counter.incrementAndGet();
}
谢谢，亨氏！
推荐答案
您根据counter.get()的旧值来决定是否要递增And.Get().在对计数器执行crementAndAndGet()操作之前，计数器的值可以再次达到MAX_VALUE.
You make the decision to incrementAndGet() based on a old value of counter.get(). The value of the counter can reach MAX_VALUE again before you do the incrementAndGet() operation on the counter.
if (next > Byte.MAX_VALUE) {
synchronized(counter) {
int i = counter.get(); //here You make sure the the counter is not over the MAX_VALUE
if (i > Byte.MAX_VALUE) {
counter.set(INITIAL_VALUE);
resetCounter.incrementAndGet();
if (isSlow) slowDownAndLog(10, "resetting");
} else {
if (isSlow) slowDownAndLog(1, "missed"); //the counter can reach MAX_VALUE again if you wait here long enough
}
next = counter.incrementAndGet(); //here you increment on return the counter that can reach >MAX_VALUE in the meantime
}
}
要使其生效，必须确保对过时的信息不做任何决定.重置计数器或返回旧值.
To make it work one has to make sure the no decisions are made on stale info. Either reset the counter or return the old value.
public byte nextValue() {
int next = counter.incrementAndGet();
if (next > Byte.MAX_VALUE) {
synchronized(counter) {
next = counter.incrementAndGet();
//if value is still larger than max byte value, we reset it
if (next > Byte.MAX_VALUE) {
counter.set(INITIAL_VALUE + 1);
next = INITIAL_VALUE + 1;
resetCounter.incrementAndGet();
if (isSlow) slowDownAndLog(10, "resetting");
} else {
if (isSlow) slowDownAndLog(1, "missed");
}
}
}
return (byte) next;
}
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