java 代码重用需要注意的事项_程序员笔记|编写高性能的Java代码需要注意的4个问题...

一、并发

无法创建新的本机线程......

问题1：Java的中创建一个线程消耗多少内存？

每个线程有独自的栈内存，共享堆内存

问题2：一台机器可以创建多少线程？

CPU，内存，操作系统，JVM，应用服务器

我们编写一段示例代码，来验证下线程池与非线程池的区别：

//线程池和非线程池的区别public class ThreadPool {  public static int times = 100;//100,1000,10000  public static ArrayBlockingQueue arrayWorkQueue = new ArrayBlockingQueue(1000); public static ExecutorService threadPool = new ThreadPoolExecutor(5, //corePoolSize线程池中核心线程数 10, 60, TimeUnit.SECONDS, arrayWorkQueue, new ThreadPoolExecutor.DiscardOldestPolicy() );  public static void useThreadPool() { Long start = System.currentTimeMillis(); for (int i = 0; i < times; i++) { threadPool.execute(new Runnable() { public void run() { System.out.println("说点什么吧..."); } }); } threadPool.shutdown(); while (true) { if (threadPool.isTerminated()) { Long end = System.currentTimeMillis(); System.out.println(end - start); break; } } }  public static void createNewThread() { Long start = System.currentTimeMillis(); for (int i = 0; i < times; i++) {  new Thread() { public void run() { System.out.println("说点什么吧..."); } }.start(); } Long end = System.currentTimeMillis(); System.out.println(end - start); }  public static void main(String args[]) { createNewThread(); //useThreadPool(); } }

启动不同数量的线程，然后比较线程池和非线程池的执行结果：

非线程池线程池100次16毫秒5ms的1000次90毫秒28ms10000次1329ms164ms

结论：不要新的Thread()，采用线程池

非线程池的缺点：

每次创建性能消耗大
无序，缺乏管理。容易无限制创建线程，引起OOM和死机

1.1 使用线程池要注意的问题

避免死锁，请尽量使用CAS

我们编写一个乐观锁的实现示例：

public class CASLock {  public static int money = 2000;  public static boolean add2(int oldm, int newm) { try { Thread.sleep(2000); } catch (InterruptedException e) { e.printStackTrace(); } if (money == oldm) { money = money + newm; return true; } return false; }  public synchronized static void add1(int newm) { try { Thread.sleep(3000); } catch (InterruptedException e) { e.printStackTrace(); } money = money + newm; }  public static void add(int newm) { try { Thread.sleep(3000); } catch (InterruptedException e) { e.printStackTrace(); } money = money + newm; }  public static void main(String args[]) { Thread one = new Thread() { public void run() { //add(5000) while (true) { if (add2(money, 5000)) { break; } } } }; Thread two = new Thread() { public void run() { //add(7000) while (true) { if (add2(money, 7000)) { break; } } } }; one.start(); two.start(); try { one.join(); two.join(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println(money); } }

使用ThreadLocal中要注意

ThreadLocalMap使用ThreadLocal的弱引用作为key，如果一个ThreadLocal没有外部强引用来引用它，那么系统GC的时候，这个ThreadLocal势必会被回收，这样一来，ThreadLocalMap中就会出现key为null的条目，就没有办法访问这些键为null的条目的值，如果当前线程再迟迟不结束的话，这些键为null的条目的值就会一直存在一条强引用链：线程参考 - >线程 - > ThreaLocalMap - >条目 - >值永远无法回收，造成内存泄漏。

我们编写一个ThreadLocalMap正确使用的示例：

//ThreadLocal应用实例public class ThreadLocalApp {  public static final ThreadLocal threadLocal = new ThreadLocal();  public static void muti2() { int i[] = (int[]) threadLocal.get(); i[1] = i[0] * 2; threadLocal.set(i); }  public static void muti3() { int i[] = (int[]) threadLocal.get(); i[2] = i[1] * 3; threadLocal.set(i); }  public static void muti5() { int i[] = (int[]) threadLocal.get(); i[3] = i[2] * 5; threadLocal.set(i); }  public static void main(String args[]) { for (int i = 0; i < 5; i++) { new Thread() { public void run() { int start = new Random().nextInt(10); int end[] = {0, 0, 0, 0}; end[0] = start; threadLocal.set(end); ThreadLocalApp.muti2(); ThreadLocalApp.muti3(); ThreadLocalApp.muti5(); //int end = (int) threadLocal.get(); System.out.println(end[0] + " " + end[1] + " " + end[2] + " " + end[3]); threadLocal.remove(); } }.start(); } } }

1.2 线程交互 - 线程不安全造成的问题

经典的HashMap的死循环造成CPU100％问题

我们模拟一个HashMap的死循环的示例：

//HashMap死循环示例public class HashMapDeadLoop {  private HashMap hash = new HashMap();  public HashMapDeadLoop() { Thread t1 = new Thread() { public void run() { for (int i = 0; i < 100000; i++) { hash.put(new Integer(i), i); } System.out.println("t1 over"); } };  Thread t2 = new Thread() { public void run() { for (int i = 0; i < 100000; i++) { hash.put(new Integer(i), i); } System.out.println("t2 over"); } }; t1.start(); t2.start(); }  public static void main(String[] args) { for (int i = 0; i < 1000; i++) { new HashMapDeadLoop(); } System.out.println("end"); } }https://coolshell.cn/articles/9606.html

HashMap的死循环发生后，我们可以在线程栈中观测到如下信息：

/HashMap死循环产生的线程栈Thread-281" #291 prio=5 os_prio=31 tid=0x00007f9f5f8de000 nid=0x5a37 runnable [0x0000700006349000] java.lang.Thread.State: RUNNABLE at java.util.HashMap$TreeNode.split(HashMap.java:2134) at java.util.HashMap.resize(HashMap.java:713) at java.util.HashMap.putVal(HashMap.java:662) at java.util.HashMap.put(HashMap.java:611) at com.example.demo.HashMapDeadLoop$2.run(HashMapDeadLoop.java:26)

应用停滞的死锁，Spring3.1的死锁问题

我们模拟一个死锁的示例：

//死锁的示例public class DeadLock { public static Integer i1 = 2000; public static Integer i2 = 3000; public static synchronized Integer getI2() { try { Thread.sleep(3000); } catch (InterruptedException e) { e.printStackTrace(); } return i2; } public static void main(String args[]) { Thread one = new Thread() { public void run() { synchronized (i1) { try { Thread.sleep(3000); } catch (InterruptedException e) { e.printStackTrace(); } synchronized (i2) { System.out.println(i1 + i2); } } } }; one.start(); Thread two = new Thread() { public void run() { synchronized (i2) { try { Thread.sleep(3000); } catch (InterruptedException e) { e.printStackTrace(); } synchronized (i1) { System.out.println(i1 + i2); } } } }; two.start(); } }

死锁发生后，我们可以在线程栈中观测到如下信息：

//死锁时产生堆栈"Thread-1": at com.example.demo.DeadLock$2.run(DeadLock.java:47) - waiting to lock (a java.lang.Integer) - locked (a java.lang.Integer)"Thread-0": at com.example.demo.DeadLock$1.run(DeadLock.java:31) - waiting to lock (a java.lang.Integer) - locked (a java.lang.Integer)Found 1 deadlock.

1.3 基于洁悠神的优化示例

一个计数器的优化，我们分别用同步，ReentrantLock的，原子三种不同的方式来实现一个计数器，体会其中的性能差异

//示例代码public class SynchronizedTest {  public static int threadNum = 100; public static int loopTimes = 10000000;  public static void userSyn() { //线程数 Syn syn = new Syn(); Thread[] threads = new Thread[threadNum]; //记录运行时间 long l = System.currentTimeMillis(); for (int i = 0; i < threadNum; i++) { threads[i] = new Thread(new Runnable() { @Override public void run() { for (int j = 0; j < loopTimes; j++) { //syn.increaseLock(); syn.increase(); } } }); threads[i].start(); } //等待所有线程结束 try { for (int i = 0; i < threadNum; i++) threads[i].join(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println("userSyn" + "-" + syn + " : " + (System.currentTimeMillis() - l) + "ms"); }  public static void useRea() { //线程数 Syn syn = new Syn(); Thread[] threads = new Thread[threadNum]; //记录运行时间 long l = System.currentTimeMillis(); for (int i = 0; i < threadNum; i++) { threads[i] = new Thread(new Runnable() { @Override public void run() { for (int j = 0; j < loopTimes; j++) { syn.increaseLock(); //syn.increase(); } } }); threads[i].start(); } //等待所有线程结束 try { for (int i = 0; i < threadNum; i++) threads[i].join(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println("userRea" + "-" + syn + " : " + (System.currentTimeMillis() - l) + "ms"); } public static void useAto() { //线程数 Thread[] threads = new Thread[threadNum]; //记录运行时间 long l = System.currentTimeMillis(); for (int i = 0; i < threadNum; i++) { threads[i] = new Thread(new Runnable() { @Override public void run() { for (int j = 0; j < loopTimes; j++) { Syn.ai.incrementAndGet(); } } }); threads[i].start(); } //等待所有线程结束 try { for (int i = 0; i < threadNum; i++) threads[i].join(); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println("userAto" + "-" + Syn.ai + " : " + (System.currentTimeMillis() - l) + "ms"); }  public static void main(String[] args) { SynchronizedTest.userSyn(); SynchronizedTest.useRea(); SynchronizedTest.useAto(); } }  class Syn { private int count = 0; public final static AtomicInteger ai = new AtomicInteger(0);  private Lock lock = new ReentrantLock();  public synchronized void increase() { count++; }  public void increaseLock() { lock.lock(); count++; lock.unlock(); }  @Override public String toString() { return String.valueOf(count); } }

结论：在并发量高，循环次数多的情况，可重入锁的效率高于同步，但最终原子性能最好。

二、通信

2.1 数据库连接池的高效问题