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java - DelayQueue 意外行为。 DrainTo 仅从队列中删除 1 个过期项

转载 作者:塔克拉玛干 更新时间:2023-11-02 19:50:03 24 4
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我想遍历我的 DelayQueue 中未过期的元素。类Transaction实现了Delayed,有一个字段timestamp,代表一笔交易发起时的UTC时间戳(不是当前时间戳)

public class Transaction implements Delayed {
private final Double amount;
private final Long timestamp; //timestamp of a time when the item was created and send here


public Transaction(double amount, long timestamp) {
this.amount = amount;
this.timestamp = timestamp;
}

@Override
public long getDelay(TimeUnit unit) {
long delay = unit.convert(ONEMINUTE - (System.currentTimeMillis() - timestamp), TimeUnit.MILLISECONDS);
return delay;
}

@Override
public int compareTo(Delayed delayed) {
if (delayed == this) {
return 0;
}

if (delayed instanceof Transaction) {
return 0;
}

long diff = (getDelay(TimeUnit.MILLISECONDS) - delayed.getDelay(TimeUnit.MILLISECONDS));
return ((diff == 0) ? 0 : ((diff < 0) ? -1 : 1));
}

@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + timestamp.hashCode();
return result;
}

@Override
public boolean equals( Object obj ) {
if( this == obj ) {
return true;
}

if( obj == null ) {
return false;
}

if( !( obj instanceof Transaction ) ) {
return false;
}

final Transaction other = ( Transaction )obj;
return timestamp.equals(other.timestamp);
}
}

下面的 TransactionManager 类将传入的事务添加到队列中,如果新的传入事务小于 1 分钟。在 getStatistics 上,旧的交易应该从队列中移除,队列应该只包含小于 1 分钟的交易

public class TransactionManager {

private DelayQueue<Transaction> transactions;


public TransactionManager() {
transactions = new DelayQueue<>();
System.setProperty("user.timezone", "UTC");
}

public Object createTransaction(String json) {
JSONObject jsonObject = null;
try {
jsonObject = JsonValidator.validateTransactionJson(json);
} catch (Exception ex) {
return new ResponseEntity(HttpStatus.UNPROCESSABLE_ENTITY);
}
long delay = System.currentTimeMillis() - ((Long) jsonObject.get(TIMESTAMP));
if (delay > ONEMINUTE) {
return new ResponseEntity(HttpStatus.NO_CONTENT);
}
transactions.add(new Transaction((Double) jsonObject.get(AMOUNT), (Long) jsonObject.get(TIMESTAMP)));
return new ResponseEntity(HttpStatus.OK);
}

public long getStatistics() {
List<Transaction> tempForCleaning = new ArrayList<>();
transactions.drainTo(tempForCleaning);
tempForCleaning.clear();
StatisticJSON statistics = new StatisticJSON();
transactions.stream().forEach(transaction -> {
statistics.setCount(statistics.getCount() + 1);
});
return statistics.getCount();
}
}

在此测试中,我创建了 40 秒之前的 5 个事务和 10 秒之前的 3 个事务。所以在等待 45 秒后,前 5 个应该被排出,队列应该只包含 3 个事务,然而,方法 drainTo 只删除 1 个旧事务。

@Test
public void test() {
DateTime dateTime = new DateTime(DateTimeZone.UTC);
long fortyMilliSecondsAgo = dateTime.minusSeconds(40).getMillis();
long twentyMilliSecondsAgo = dateTime.minusSeconds(10).getMillis();
for (int i = 0; i < 5; i++) {
createTransaction(fortyMilliSecondsAgo);
}
for (int i = 0; i < 3; i++) {
createTransaction(twentyMilliSecondsAgo);
}
Assert.assertTrue(transactionManager.getStatistics() == 8);
try {
TimeUnit.SECONDS.sleep(45);
System.out.println("\n\n\n");
Assert.assertTrue(transactionManager.getStatistics() == 3);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

private void createTransaction(long timestamp) {
transactionManager.createTransaction("{\"amount\":100.0,\"timestamp\":" + timestamp + "}");
}

我遗漏了一些东西,为什么 drainTo 只删除一个过期的项目,即使还剩下 4 个,但无法捕捉到哪里......

最佳答案

if (delayed instanceof Transaction) { return 0; }

看起来不对 - 如果您希望 compareTogetDelay() 一致,您可能应该删除该位。所以你的方法应该看起来像这样(使用静态导入):

public int compareTo(Delayed delayed) {
return delayed == this
? 0
: Long.compare(getDelay(MILLISECONDS), delayed.getDelay(MILLISECONDS));
}

关于java - DelayQueue 意外行为。 DrainTo 仅从队列中删除 1 个过期项,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45032330/

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