java 字节序列_java – 这个线程安全的字节序列生成器有什么问题？
我需要一個字節生成器,它將生成從Byte.MIN_VALUE到Byte.MAX_VALUE的值.當它達到MAX_VALUE時,它應該從MIN_VALUE重新開始.
我使用AtomicInteger編寫了代碼(見下文);但是,如果同時訪問并且如果使用Thread.sleep()人為地減慢了代碼,那么代碼似乎沒有正常運行(如果沒有睡眠,它運行正常;但是,我懷疑它對于出現并發問題來說太快了).
代碼(添加了一些調試代碼)：
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;
}
}
并且,測試：
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核機器上的結果是值#-128得到146次點擊(所有這些都應該得到100次點擊,因為我們有100個線程).
如果有人有任何想法,這個代碼有什么問題,我都是耳朵/眼睛.
更新：對于那些趕時間并且不想向下滾動的人來說,在Java中解決這個問題的正確(以及最短和最優雅)方式將是這樣的：
public byte nextValue() {
return (byte) counter.incrementAndGet();
}
謝謝,亨氏！
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