我有这段代码,但我不理解对incrementOnly方法调用的部分的第37到43行。
继承了我的理解(这是正确的吗?) t2将在第35行创建一个新线程
t3将在第36行创建一个新线程,然后它将调用方法incrementOnly。
然后在第41行,将执行t2的run方法。在第42行,将为t3执行run方法。
package aa.race;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class ThreadDemoCounter implements Runnable
{
int counter;
int alternate;
String name;
public static int numLoops = 4;
public static int numPrints = 1500;
public ThreadDemoCounter(String n)
{
name = n;
counter = 0;
}
// For bonus -- delete method go. Change main to below code:
public static void main(String[] args) throws Exception
{
ThreadDemoCounter c1 = new ThreadDemoCounter("c1");
//Run the multithreaded demo a few times
for (int foo = 0; foo < numLoops; foo++)
{
c1.counter = 0;
Thread t1 = new Thread(c1);
Thread t2 = new Thread(c1);
Thread t3 = new Thread(c1::incrementOnly);
Thread t4 = new Thread(c1::incrementOnly);
t1.start();
t2.start();
t3.start();
t4.start();
t1.join();
t2.join(); //wait for both
t3.join();
t4.join(); //wait for both
System.out.println("c1 = " + c1.counter);
System.out.println("===== end loop =====");
}
}
public void incrementOnly()
{
for (int i =0 ; i < numPrints; i++)
{
incrementCounter();
}
}
public void run()
{
for (int j = 0; j < numPrints; j++)
{
LockFactory.getLock(name).lock();
System.out.println("counter " + name + " = " + getCounter() + " retrieved by thread: " + Thread.currentThread().getName());
incrementCounter();
LockFactory.getLock(name).unlock();
}
System.out.println();
}
public int getCounter()
{
return counter;
} //start at 0
public void incrementCounter()
{
LockFactory.getLock(name).lock();
counter++;
LockFactory.getLock(name).unlock();
}
}
答案 0 :(得分:4)
所有4个构造函数调用都在调用Thread(Runnable target)
,其中Runnable
是@FunctionalInterface
,方法为void run()
。线程启动后,它将调用run()
的{{1}}方法。
前两个构造函数调用Runnable
正在传递new Thread(c1)
的实例,因此这两个线程将为ThreadDemoCounter
实例调用ThreadDemoCounter.run()
方法。
另外两个构造函数调用正在将method reference传递给c1
的{{1}}方法。这是一个有效的方法,因为它也是一个无参数的void方法。这两个线程将为incrementOnly()
实例调用c1
方法。
总之,你将有4个线程在运行,其中两个执行ThreadDemoCounter.incrementOnly()
方法,其中两个执行c1
方法,所有这些都在run()
的同一个实例上,即incrementOnly()
。
仅供参考:该代码中没有lambda expressions。 method reference expression不是lambda表达式。