应用篇
效率
1.使用多线程充分利用 CPU
1)环境搭建
基准测试工具选择，使用了比较靠谱的 JMH，它会执行程序预热，执行多次测试并平均
cpu 核数限制，有两种思路
使用虚拟机，分配合适的核
使用 msconfig，分配合适的核，需要重启比较麻烦
并行计算方式的选择
最初想直接使用 parallel stream，后来发现它有自己的问题
改为了自己手动控制 thread，实现简单的并行计算
测试代码如下
mvn archetype:generate -DinteractiveMode=false -DarchetypeGroupId=org.openjdk.jmh -
DarchetypeArtifactId=jmh-java-benchmark-archetype -DgroupId=org.sample -DartifactId=test -
Dversion= 1.0
代码
package org.sample;
import java.util.Arrays;
import java.util.concurrent.FutureTask;
import org.openjdk.jmh.annotations.Benchmark;
import org.openjdk.jmh.annotations.BenchmarkMode;
import org.openjdk.jmh.annotations.Fork;
import org.openjdk.jmh.annotations.Measurement;
import org.openjdk.jmh.annotations.Mode;
import org.openjdk.jmh.annotations.Warmup;
@Fork( 1 )
@BenchmarkMode(Mode.AverageTime)
@Warmup(iterations= 3 )
@Measurement(iterations= 5 )
public class MyBenchmark {
static int[] ARRAY = new int[1000_000_00];//对一亿个数字进行求和
static {
Arrays.fill(ARRAY, 1 );
}
@Benchmark
public int c() throws Exception {
int[] array = ARRAY;
FutureTask<Integer> t1 = new FutureTask<>(()->{
int sum = 0;
for(int i = 0; i < 250_000_00;i++) {
sum += array[0+i];
}
return sum;
});
FutureTask<Integer> t2 = new FutureTask<>(()->{
int sum = 0;
for(int i = 0; i < 250_000_00;i++) {
sum += array[250_000_00+i];
}
return sum;
});
FutureTask<Integer> t3 = new FutureTask<>(()->{
int sum = 0;
for(int i = 0; i < 250_000_00;i++) {
sum += array[500_000_00+i];
}
return sum;
});
FutureTask<Integer> t4 = new FutureTask<>(()->{
int sum = 0;
for(int i = 0; i < 250_000_00;i++) {
sum += array[750_000_00+i];
}
return sum;
});
new Thread(t1).start();
new Thread(t2).start();
new Thread(t3).start();
new Thread(t4).start();
return t1.get() + t2.get() + t3.get()+ t4.get();
}
@Benchmark
public int d() throws Exception {
int[] array = ARRAY;
FutureTask<Integer> t1 = new FutureTask<>(()->{
int sum = 0;
for(int i = 0; i < 1000_000_00;i++) {
sum += array[0+i];
}
return sum;
});
new Thread(t1).start();
return t1.get();
}
}
2) 双核 CPU（ 4 个逻辑CPU）
C:\Users\lenovo\eclipse-workspace\test>java -jar target/benchmarks.jar
# VM invoker: C:\Program Files\Java\jdk-11\bin\java.exe
# VM options: <none>
# Warmup: 3 iterations, 1 s each
# Measurement: 5 iterations, 1 s each
# Threads: 1 thread, will synchronize iterations
# Benchmark mode: Average time, time/op
# Benchmark: org.sample.MyBenchmark.c
# Run progress: 0.00% complete, ETA 00:00:16
# Fork: 1 of 1
# Warmup Iteration 1: 0.022 s/op
# Warmup Iteration 2: 0.019 s/op
# Warmup Iteration 3: 0.020 s/op
Iteration 1: 0.020 s/op
Iteration 2: 0.020 s/op
Iteration 3: 0.020 s/op
Iteration 4: 0.020 s/op
Iteration 5: 0.020 s/op
Result: 0.020 ±(99.9%) 0.001 s/op [Average]
Statistics: (min, avg, max) = (0.020, 0.020, 0.020), stdev = 0.000
Confidence interval (99.9%): [0.019, 0.021]
# VM invoker: C:\Program Files\Java\jdk-11\bin\java.exe
# VM options: <none>
# Warmup: 3 iterations, 1 s each
# Measurement: 5 iterations, 1 s each
# Threads: 1 thread, will synchronize iterations
# Benchmark mode: Average time, time/op
# Benchmark: org.sample.MyBenchmark.d
# Run progress: 50.00% complete, ETA 00:00:10
# Fork: 1 of 1
# Warmup Iteration 1: 0.042 s/op
# Warmup Iteration 2: 0.042 s/op
# Warmup Iteration 3: 0.041 s/op
Iteration 1: 0.043 s/op
Iteration 2: 0.042 s/op
Iteration 3: 0.042 s/op
Iteration 4: 0.044 s/op
Iteration 5: 0.042 s/op
Result: 0.043 ±(99.9%) 0.003 s/op [Average]
Statistics: (min, avg, max) = (0.042, 0.043, 0.044), stdev = 0.001
Confidence interval (99.9%): [0.040, 0.045]
# Run complete. Total time: 00:00:20
Benchmark Mode Samples Score Score error Units
o.s.MyBenchmark.c avgt 5 0.020 0.001 s/op
o.s.MyBenchmark.d avgt 5 0.043 0.003 s/op
可以看到多核下，效率提升还是很明显的，快了一倍左右
3) 单核 CPU
Benchmark Mode Samples Score Score error Units
o.s.MyBenchmark.c avgt 5 0 .020 0 .001 s/op
o.s.MyBenchmark.d avgt 5 0 .043 0 .003 s/op
C:\Users\lenovo\eclipse-workspace\test>java -jar target/benchmarks.jar
# VM invoker: C:\Program Files\Java\jdk-11\bin\java.exe
# VM options: <none>
# Warmup: 3 iterations, 1 s each
# Measurement: 5 iterations, 1 s each
# Threads: 1 thread, will synchronize iterations
# Benchmark mode: Average time, time/op
# Benchmark: org.sample.MyBenchmark.c
# Run progress: 0.00% complete, ETA 00:00:
# Fork: 1 of 1
# Warmup Iteration 1: 0.064 s/op
# Warmup Iteration 2: 0.052 s/op
# Warmup Iteration 3: 1.127 s/op
Iteration 1 : 0 .053 s/op
Iteration 2 : 0 .052 s/op
Iteration 3 : 0 .053 s/op
Iteration 4 : 0 .057 s/op
Iteration 5 : 0 .088 s/op
Result: 0 .061 ±(99.9%) 0 .060 s/op [Average]
Statistics: (min, avg, max) = (0.052, 0 .061, 0 .088), stdev = 0.
Confidence interval (99.9%): [0.001, 0 .121]
# VM invoker: C:\Program Files\Java\jdk-11\bin\java.exe
# VM options: <none>
# Warmup: 3 iterations, 1 s each
# Measurement: 5 iterations, 1 s each
# Threads: 1 thread, will synchronize iterations
# Benchmark mode: Average time, time/op
# Benchmark: org.sample.MyBenchmark.d
# Run progress: 50.00% complete, ETA 00:00:
# Fork: 1 of 1
# Warmup Iteration 1: 0.054 s/op
# Warmup Iteration 2: 0.053 s/op
# Warmup Iteration 3: 0.051 s/op
Iteration 1 : 0 .096 s/op
Iteration 2 : 0 .054 s/op
Iteration 3 : 0 .065 s/op
Iteration 4 : 0 .050 s/op
Iteration 5: 0.055 s/op
Result: 0.064 ±(99.9%) 0.071 s/op [Average]
Statistics: (min, avg, max) = (0.050, 0.064, 0.096), stdev = 0.018
Confidence interval (99.9%): [-0.007, 0.135]
# Run complete. Total time: 00:00:22
Benchmark Mode Samples Score Score error Units
o.s.MyBenchmark.c avgt 5 0.061 0.060 s/op
o.s.MyBenchmark.d avgt 5 0.064 0.071 s/op
性能几乎是一样的
限制
1. 限制对 CPU 的使用
sleep 实现
在没有利用 cpu 来计算时，不要让 while(true) 空转浪费 cpu，这时可以使用 yield 或 sleep 来让出 cpu 的使用权 给其他程序
while(true) {
try {
Thread.sleep(50);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
- 可以用 wait 或 条件变量达到类似的效果
- 不同的是，后两种都需要加锁，并且需要相应的唤醒操作，一般适用于要进行同步的场景
- sleep 适用于无需锁同步的场景
wait 实现
synchronized(锁对象) {
while(条件不满足) {
try {
锁对象.wait();
} catch(InterruptedException e) {
e.printStackTrace();
}
}
// do sth...
}
条件变量实现
lock.lock();
try {
while(条件不满足) {
try {
条件变量.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
// do sth...
} finally {
lock.unlock();
}
2. 限制对共享资源的使用
semaphore 实现
- 使用 Semaphore 限流，在访问高峰期时，让请求线程阻塞，高峰期过去再释放许可，当然它只适合限制单机 线程数量，并且仅是限制线程数，而不是限制资源数（例如连接数，请对比 Tomcat LimitLatch 的实现）
用 Semaphore 实现简单连接池，对比『享元模式』下的实现（用wait notify），性能和可读性显然更好， 注意下面的实现中线程数和数据库连接数是相等的
@Slf4j(topic = "c.Pool") class Pool { // 1. 连接池大小 private final int poolSize; // 2. 连接对象数组 private Connection[] connections; // 3. 连接状态数组 0 表示空闲， 1 表示繁忙 private AtomicIntegerArray states; private Semaphore semaphore; // 4. 构造方法初始化 public Pool(int poolSize) { this.poolSize = poolSize; // 让许可数与资源数一致 this.semaphore = new Semaphore(poolSize); this.connections = new Connection[poolSize]; this.states = new AtomicIntegerArray(new int[poolSize]); for (int i = 0; i < poolSize; i++) { connections[i] = new MockConnection("连接" + (i+1)); } } // 5. 借连接 public Connection borrow() {// t1, t2, t3 // 获取许可 try { semaphore.acquire(); // 没有许可的线程，在此等待 } catch (InterruptedException e) { e.printStackTrace(); } for (int i = 0; i < poolSize; i++) { // 获取空闲连接 if(states.get(i) == 0) { if (states.compareAndSet(i, 0, 1)) { log.debug("borrow {}", connections[i]); return connections[i]; } } } // 不会执行到这里 return null; } // 6. 归还连接 public void free(Connection conn) { for (int i = 0; i < poolSize; i++) { if (connections[i] == conn) { states.set(i, 0); log.debug("free {}", conn); semaphore.release(); break; } } } }
3. 单位时间内限流
guava 实现
@RestController
public class TestController {
private RateLimiter limiter = RateLimiter.create(50);
@GetMapping("/test")
public String test() {
// limiter.acquire();
return "ok";
}
}
没有限流之前
ab -c 10 -t 10 http://localhost:8080/test
结果
This is ApacheBench, Version 2.3 <$Revision: 1843412 $>
Copyright 1996 Adam Twiss, Zeus Technology Ltd, http://www.zeustech.net/
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking localhost (be patient)
Completed 5000 requests
Completed 10000 requests
Completed 15000 requests
Completed 20000 requests
Finished 24706 requests
Server Software:
Server Hostname: localhost
Server Port: 8080
Document Path: /test
Document Length: 2 bytes
Concurrency Level: 10
Time taken for tests: 10.005 seconds
Complete requests: 24706
Failed requests: 0
Total transferred: 3311006 bytes
HTML transferred: 49418 bytes
Requests per second: 2469.42 [#/sec] (mean)
Time per request: 4.050 [ms] (mean)
Time per request: 0.405 [ms] (mean, across all concurrent requests)
Transfer rate: 323.19 [Kbytes/sec] received
Connection Times (ms)
min mean[+/-sd] median max
Connect: 0 0 1.4 0 16
Processing: 0 4 7.6 0 323
Waiting: 0 3 6.9 0 323
Total: 0 4 7.6 0 323
Percentage of the requests served within a certain time (ms)
50% 0
66% 2
75% 8
80% 8
90% 10
95% 16
98% 16
99% 16
100% 323 (longest request)
限流之后
This is ApacheBench, Version 2.3 <$Revision: 1843412 $>
Copyright 1996 Adam Twiss, Zeus Technology Ltd, http://www.zeustech.net/
Licensed to The Apache Software Foundation, http://www.apache.org/
Benchmarking localhost (be patient)
Finished 545 requests
Server Software:
Server Hostname: localhost
Server Port: 8080
Document Path: /test
Document Length: 2 bytes
Concurrency Level: 10
Time taken for tests: 10.007 seconds
Complete requests: 545
Failed requests: 0
Total transferred: 73030 bytes
HTML transferred: 1090 bytes
Requests per second: 54.46 [#/sec] (mean)
Time per request: 183.621 [ms] (mean)
Time per request: 18.362 [ms] (mean, across all concurrent requests)
Transfer rate: 7.13 [Kbytes/sec] received
Connection Times (ms)
min mean[+/-sd] median max
Connect: 0 0 1.1 0 16
Processing: 0 179 57.0 199 211
Waiting: 0 178 57.6 198 211
Total: 0 179 56.9 199 211
Percentage of the requests served within a certain time (ms)
50% 199
66% 200
75% 200
80% 200
90% 201
95% 201
98% 202
99% 203
100% 211 (longest request)
互斥
1. 悲观互斥
互斥实际是悲观锁的思想
例如，有下面取款的需求
interface Account {
// 获取余额
Integer getBalance();
// 取款
void withdraw(Integer amount);
/**
* 方法内会启动 1000 个线程，每个线程做 -10 元 的操作
* 如果初始余额为 10000 那么正确的结果应当是 0
*/
static void demo(Account account) {
List<Thread> ts = new ArrayList<>();
for (int i = 0; i < 1000; i++) {
ts.add(new Thread(() -> {
account.withdraw(10);
}));
}
long start = System.nanoTime();
ts.forEach(Thread::start);
ts.forEach(t -> {
try {
t.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
});
long end = System.nanoTime();
System.out.println(account.getBalance()
+ " cost: " + (end-start)/1000_000 + " ms");
}
}
用互斥来保护
class AccountSync implements Account {
private Integer balance;
public AccountUnsafe(Integer balance) {
this.balance = balance;
}
@Override
public Integer getBalance() {
synchronized (this) {
return this.balance;
}
}
@Override
public void withdraw(Integer amount) {
synchronized (this) {
this.balance -= amount;
}
}
}
2. 乐观重试
另外一种是乐观锁思想，它其实不是互斥
class AccountCas implements Account {
private AtomicInteger balance;
public AccountCas(int balance) {
this.balance = new AtomicInteger(balance);
}
@Override
public Integer getBalance() {
return balance.get();
}
@Override
public void withdraw(Integer amount) {
while(true) {
// 获取余额的最新值
int prev = balance.get();
// 要修改的余额
int next = prev - amount;
// 真正修改
if(balance.compareAndSet(prev, next)) {
break;
}
}
}
}
同步和异步
1. 需要等待结果
这时既可以使用同步处理，也可以使用异步来处理
1.join 实现（同步）
static int result = 0;
private static void test1() throws InterruptedException {
log.debug("开始");
Thread t1 = new Thread(() -> {
log.debug("开始");
sleep(1);
log.debug("结束");
result = 10;
}, "t1");
t1.start();
t1.join();
log.debug("结果为:{}", result);
}
评价
需要外部共享变量，不符合面向对象封装的思想
必须等待线程结束，不能配合线程池使用
2.Future 实现（同步）
private static void test2() throws InterruptedException, ExecutionException {
log.debug("开始");
FutureTask<Integer> result = new FutureTask<>(() -> {
log.debug("开始");
sleep(1);
log.debug("结束");
return 10;
});
new Thread(result, "t1").start();
log.debug("结果为:{}", result.get());
}
评价
- 规避了使用 join 之前的缺点
- 可以方便配合线程池使用
private static void test3() throws InterruptedException, ExecutionException {
ExecutorService service = Executors.newFixedThreadPool(1);
log.debug("开始");
Future<Integer> result = service.submit(() -> {
log.debug("开始");
sleep(1);
log.debug("结束");
return 10;
});
log.debug("结果为:{}, result 的类型:{}", result.get(), result.getClass());
service.shutdown();
}
评价
- 仍然是 main 线程接收结果
- get 方法是让调用线程同步等
3. 自定义实现（同步）
见模式篇：保护性暂停模式
CompletableFuture 实现（异步）
private static void test4() {
// 进行计算的线程池
ExecutorService computeService = Executors.newFixedThreadPool(1);
// 接收结果的线程池
ExecutorService resultService = Executors.newFixedThreadPool(1);
log.debug("开始");
CompletableFuture.supplyAsync(() -> {
log.debug("开始");
sleep(1);
log.debug("结束");
return 10;
}, computeService).thenAcceptAsync((result) -> {
log.debug("结果为:{}", result);
}, resultService);
}
输出
10:36:28.114 c.TestSync [main] - 开始
10:36:28.164 c.TestSync [pool-1-thread-1] - 开始
10:36:29.165 c.TestSync [pool-1-thread-1] - 结束
10:36:29.165 c.TestSync [pool-2-thread-1] - 结果为:10
评价
可以让调用线程异步处理结果，实际是其他线程去同步等待
可以方便地分离不同职责的线程池
以任务为中心，而不是以线程为中心
BlockingQueue 实现（异步）
private static void test6() {
ExecutorService consumer = Executors.newFixedThreadPool(1);
ExecutorService producer = Executors.newFixedThreadPool(1);
BlockingQueue<Integer> queue = new SynchronousQueue<>();
log.debug("开始");
producer.submit(() -> {
log.debug("开始");
sleep(1);
log.debug("结束");
try {
queue.put(10);
} catch (InterruptedException e) {
e.printStackTrace();
}
});
consumer.submit(() -> {
try {
Integer result = queue.take();
log.debug("结果为:{}", result);
} catch (InterruptedException e) {
e.printStackTrace();
}
});
}
2. 不需等待结果
这时最好是使用异步来处理
1. 普通线程实现
@Slf4j(topic = "c.FileReader")
public class FileReader {
public static void read(String filename) {
int idx = filename.lastIndexOf(File.separator);
String shortName = filename.substring(idx + 1);
try (FileInputStream in = new FileInputStream(filename)) {
long start = System.currentTimeMillis();
log.debug("read [{}] start ...", shortName);
byte[] buf = new byte[1024];
int n = -1;
do {
n = in.read(buf);
} while (n != -1);
long end = System.currentTimeMillis();
log.debug("read [{}] end ... cost: {} ms", shortName, end - start);
} catch (IOException e) {
e.printStackTrace();
}
}
}
没有用线程时，方法的调用是同步的：
@Slf4j(topic = "c.Sync")
public class Sync {
public static void main(String[] args) {
String fullPath = "E:\\1.mp4";
FileReader.read(fullPath);
log.debug("do other things ...");
}
}
输出
18:39:15 [main] c.FileReader - read [1.mp4] start ...
18:39:19 [main] c.FileReader - read [1.mp4] end ... cost: 4090 ms
18:39:19 [main] c.Sync - do other things ...
使用了线程后，方法的调用时异步的：
输出
18:39:15 [main] c.FileReader - read [1.mp4] start ...
18:39:19 [main] c.FileReader - read [1.mp4] end ... cost: 4090 ms
18:39:19 [main] c.Sync - do other things ...
使用了线程后，方法的调用时异步的：
private static void test1() {
new Thread(() -> FileReader.read(Constants.MP4_FULL_PATH)).start();
log.debug("do other things ...");
}
输出
18:41:53 [main] c.Async - do other things ...
18:41:53 [Thread-0] c.FileReader - read [1.mp4] start ...
18:41:57 [Thread-0] c.FileReader - read [1.mp4] end ... cost: 4197 ms
2. 线程池实现
private static void test2() {
ExecutorService service = Executors.newFixedThreadPool(1);
service.execute(() -> FileReader.read(Constants.MP4_FULL_PATH));
log.debug("do other things ...");
service.shutdown();
}
输出
11:03:31.245 c.TestAsyc [main] - do other things ...
11:03:31.245 c.FileReader [pool-1-thread-1] - read [1.mp4] start ...
11:03:33.479 c.FileReader [pool-1-thread-1] - read [1.mp4] end ... cost: 2235 ms
缓存
1. 缓存更新策略
更新时，是先清缓存还是先更新数据库
先清缓存
sequenceDiagram
participant B
participant A
participant 缓存
participant 数据库
B->>缓存:1)清空缓存
A->>数据库:2)查询数据库(x=1)
A->>缓存:3)将查询结果放入缓存(x=1)
B->>数据库:4)将新数据存入库(x=2)
A->>缓存:5)后续查询将一直是旧值(x=1)!
先更新数据库
sequenceDiagram
participant B
participant A
participant 缓存
participant 数据库
B->>数据库:1)将新数据存入库(x=2)
A->>缓存:2)查询缓存(x=1)！！
B->>缓存:3)清空缓存(x=1)
A->>数据库:4)查询数据库(x=2)
A->>缓存:5)后续查询将一直是旧值(x=2)
private static void test3() throws IOException {
CompletableFuture.runAsync(() -> FileReader.read(Constants.MP4_FULL_PATH));
log.debug("do other things ...");
System.in.read();
}
补充一种情况，假设查询线程 A 查询数据时恰好缓存数据由于时间到期失效，或是第一次查询
sequenceDiagram
participant B
participant A
participant 缓存
participant 数据库
A->>数据库:1)缓存没有，查询数据库（x=1）
B->>数据库:2)将新数据村入库(x=2)
B->>缓存:3)清空缓存
A->>缓存:4)将查询结果放入缓存(x=1)
A->>缓存:5)后续查询将一直是旧值(x=1)!!
这种情况的出现几率非常小，见 facebook 论文
2. 读写锁实现一致性缓存
使用读写锁实现一个简单的按需加载缓存
class GenericCachedDao<T> {
// HashMap 作为缓存非线程安全, 需要保护
HashMap<SqlPair, T> map = new HashMap<>();
ReentrantReadWriteLock lock = new ReentrantReadWriteLock();
GenericDao genericDao = new GenericDao();
public int update(String sql, Object... params) {
SqlPair key = new SqlPair(sql, params);
// 加写锁, 防止其它线程对缓存读取和更改
lock.writeLock().lock();
try {
int rows = genericDao.update(sql, params);
map.clear();
return rows;
} finally {
lock.writeLock().unlock();
}
}
public T queryOne(Class<T> beanClass, String sql, Object... params) {
SqlPair key = new SqlPair(sql, params);
// 加读锁, 防止其它线程对缓存更改
lock.readLock().lock();
try {
T value = map.get(key);
if (value != null) {
return value;
}
} finally {
lock.readLock().unlock();
}
// 加写锁, 防止其它线程对缓存读取和更改
lock.writeLock().lock();
try {
// get 方法上面部分是可能多个线程进来的, 可能已经向缓存填充了数据
// 为防止重复查询数据库, 再次验证
T value = map.get(key);
if (value == null) {
// 如果没有, 查询数据库
value = genericDao.queryOne(beanClass, sql, params);
map.put(key, value);
}
return value;
} finally {
lock.writeLock().unlock();
}
}
// 作为 key 保证其是不可变的
class SqlPair {
private String sql;
private Object[] params;
public SqlPair(String sql, Object[] params) {
this.sql = sql;
this.params = params;
}
@Override
public boolean equals(Object o) {
if (this == o) {return true;
}
if (o == null || getClass() != o.getClass()) {
return false;
}
SqlPair sqlPair = (SqlPair) o;
return sql.equals(sqlPair.sql) &&
Arrays.equals(params, sqlPair.params);
}
@Override
public int hashCode() {
int result = Objects.hash(sql);
result = 31 * result + Arrays.hashCode(params);
return result;
}
}
}
注意
- 以上实现体现的是读写锁的应用，保证缓存和数据库的一致性，但有下面的问题没有考虑适合读多写少，如果写操作比较频繁，以上实现性能低
- 没有考虑缓存容量
- 没有考虑缓存过期
- 只适合单机
- 并发性还是低，目前只会用一把锁
- 更新方法太过简单粗暴，清空了所有 key（考虑按类型分区或重新设计 key）
- 乐观锁实现：用 CAS 去更新
分治
1. 案例 - 单词计数
private static <V> void demo(Supplier<Map<String, V>> supplier, BiConsumer<Map<String, V>,
List<String>> consumer) {
Map<String, V> counterMap = supplier.get();
List<Thread> ts = new ArrayList<>();
for (int i = 1; i <= 26; i++) {
int idx = i;
Thread thread = new Thread(() -> {
List<String> words = readFromFile(idx);
consumer.accept(counterMap, words);
});
ts.add(thread);
}
ts.forEach(t -> t.start());
ts.forEach(t -> {
try {t.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
});
System.out.println(counterMap);
}
public static List<String> readFromFile(int i) {
ArrayList<String> words = new ArrayList<>();
try (BufferedReader in = new BufferedReader(new InputStreamReader(new FileInputStream("tmp/"
+ i + ".txt")))) {
while (true) {
String word = in.readLine();
if (word == null) {
break;
}
words.add(word);
}
return words;
} catch (IOException e) {
throw new RuntimeException(e);
}
}
解法 1 ：
demo(
() -> new ConcurrentHashMap<String, LongAdder>(),
(map, words) -> {
for (String word : words) {
map.computeIfAbsent(word, (key) -> new LongAdder()).increment();
}
}
);
解法 2 ：
Map<String, Integer> collect = IntStream.range(1, 27).parallel()
.mapToObj(idx -> readFromFile(idx))
.flatMap(list -> list.stream())
.collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(w -> 1)));
2. 案例 - 求和
class AddTask3 extends RecursiveTask<Integer> {
int begin;
int end;
public AddTask3(int begin, int end) {
this.begin = begin;
this.end = end;
}
@Override
public String toString() {
return "{" + begin + "," + end + '}';
}
@Override
protected Integer compute() {
// 5, 5
if (begin == end) {
log.debug("join() {}", begin);
return begin;
}
// 4, 5
if (end - begin == 1) {
log.debug("join() {} + {} = {}", begin, end, end + begin);
return end + begin;
}
// 1 5
int mid = (end + begin) / 2; // 3
AddTask3 t1 = new AddTask3(begin, mid); // 1,3
t1.fork();
AddTask3 t2 = new AddTask3(mid + 1, end); // 4,5
t2.fork();
log.debug("fork() {} + {} = ?", t1, t2);
int result = t1.join() + t2.join();
log.debug("join() {} + {} = {}", t1, t2, result);
return result;
}
}
然后提交给 ForkJoinPool 来执行
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool(4);
System.out.println(pool.invoke(new AddTask3(1, 10)));
}
结果
[ForkJoinPool-1-worker-0] - join() 1 + 2 = 3
[ForkJoinPool-1-worker-3] - join() 4 + 5 = 9
[ForkJoinPool-1-worker-0] - join() 3
[ForkJoinPool-1-worker-1] - fork() {1,3} + {4,5} = ?
[ForkJoinPool-1-worker-2] - fork() {1,2} + {3,3} = ?
[ForkJoinPool-1-worker-2] - join() {1,2} + {3,3} = 6
[ForkJoinPool-1-worker-1] - join() {1,3} + {4,5} = 15
15
统筹
案例 - 烧水泡茶
解法 1 ：join
Thread t1 = new Thread(() -> {
log.debug("洗水壶");
sleep(1);
log.debug("烧开水");
sleep(15);
}, "老王");
Thread t2 = new Thread(() -> {
log.debug("洗茶壶");
sleep(1);
log.debug("洗茶杯");
sleep(2);
log.debug("拿茶叶");
sleep(1);
try {
t1.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
log.debug("泡茶");
}, "小王");
t1.start();
t2.start();
输出
19:19:37.547 [小王] c.TestMakeTea - 洗茶壶
19:19:37.547 [老王] c.TestMakeTea - 洗水壶
19:19:38.552 [小王] c.TestMakeTea - 洗茶杯
19:19:38.552 [老王] c.TestMakeTea - 烧开水
19:19:40.553 [小王] c.TestMakeTea - 拿茶叶
19:19:53.553 [小王] c.TestMakeTea - 泡茶
解法 1 的缺陷：
上面模拟的是小王等老王的水烧开了，小王泡茶，如果反过来要实现老王等小王的茶叶拿来了，老王泡茶呢？代码最好能适应两种情况
上面的两个线程其实是各执行各的，如果要模拟老王把水壶交给小王泡茶，或模拟小王把茶叶交给老王泡茶呢
解法 2 ：wait/notify
class S2 {
static String kettle = "冷水";
static String tea = null;
static final Object lock = new Object();
static boolean maked = false;
public static void makeTea() {
new Thread(() -> {
log.debug("洗水壶");
sleep(1);
log.debug("烧开水");
sleep(5);
synchronized (lock) {
kettle = "开水";
lock.notifyAll();
while (tea == null) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (!maked) {
log.debug("拿({})泡({})", kettle, tea);
maked = true;
}
}
}, "老王").start();
new Thread(() -> {
log.debug("洗茶壶");
sleep(1);
log.debug("洗茶杯");
sleep(2);
log.debug("拿茶叶");
sleep(1);
synchronized (lock) {
tea = "花茶";
lock.notifyAll();
while (kettle.equals("冷水")) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (!maked) {
log.debug("拿({})泡({})", kettle, tea);
maked = true;
}
}
}, "小王").start();
}
}
输出
20:04:48.179 c.S2 [小王] - 洗茶壶
20:04:48.179 c.S2 [老王] - 洗水壶
20:04:49.185 c.S2 [老王] - 烧开水
20:04:49.185 c.S2 [小王] - 洗茶杯
20:04:51.185 c.S2 [小王] - 拿茶叶
20:04:54.185 c.S2 [老王] - 拿(开水)泡(花茶)
解法 2 解决了解法 1 的问题，不过老王和小王需要相互等待，不如他们只负责各自的任务，泡茶交给第三人来做
解法 3 ：第三者协调
class S3 {
static String kettle = "冷水";
static String tea = null;
static final Object lock = new Object();
public static void makeTea() {
new Thread(() -> {
log.debug("洗水壶");
sleep(1);
log.debug("烧开水");
sleep(5);
synchronized (lock) {
kettle = "开水";
lock.notifyAll();
}
}, "老王").start();
new Thread(() -> {
log.debug("洗茶壶");
sleep(1);
log.debug("洗茶杯");
sleep(2);
log.debug("拿茶叶");
sleep(1);
synchronized (lock) {
tea = "花茶";
lock.notifyAll();
}
}, "小王").start();
new Thread(() -> {
synchronized (lock) {
while (kettle.equals("冷水") || tea == null) {
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
log.debug("拿({})泡({})", kettle, tea);
}
}, "王夫人").start();
}
}
输出
20:13:18.202 c.S3 [小王] - 洗茶壶
20:13:18.202 c.S3 [老王] - 洗水壶
20:13:19.206 c.S3 [小王] - 洗茶杯
20:13:19.206 c.S3 [老王] - 烧开水
20:13:21.206 c.S3 [小王] - 拿茶叶
20:13:24.207 c.S3 [王夫人] - 拿(开水)泡(花茶)
定时
1. 定期执行
如何让每周四 18:00:00 定时执行任务？
// 获得当前时间
LocalDateTime now = LocalDateTime.now();
// 获取本周四 18:00:00.000
//LocalDateTime jdk1.8新增的线程安全类
LocalDateTime thursday =
now.with(DayOfWeek.THURSDAY).withHour(18).withMinute(0).withSecond(0).withNano(0);
// 如果当前时间已经超过 本周四 18:00:00.000， 那么找下周四 18:00:00.000
if(now.compareTo(thursday) >= 0) {
thursday = thursday.plusWeeks(1);
}
// 计算时间差，即延时执行时间
long initialDelay = Duration.between(now, thursday).toMillis();
// 计算间隔时间，即 1 周的毫秒值
long oneWeek = 7 * 24 * 3600 * 1000;
ScheduledExecutorService executor = Executors.newScheduledThreadPool(2);
System.out.println("开始时间：" + new Date());
executor.scheduleAtFixedRate(() -> {
System.out.println("执行时间：" + new Date());
}, initialDelay, oneWeek, TimeUnit.MILLISECONDS);
