为什么单线程比Java中的多线程更快？

Question

据我所知,我已经编写了下面简单的单线程和多线程程序来检查执行速度.但我的单线程程序执行速度比多线程快,请看下面的程序并提及是否有任何错误.

单线程:

import java.util.Calendar;

public class NormalJava {
    public static void main(String[] args) {
        System.out.println("Single Thread");
        int a = 1000;
        int b = 200;
        NormalJava nj = new NormalJava();
        nj.Add(a, b);
        nj.Sub(a, b);
        nj.Mul(a, b);
        nj.Div(a, b);
        Calendar lCDateTime = Calendar.getInstance();
        System.out.println("Calender - Time in milliseconds :"
                + lCDateTime.getTimeInMillis());

    }

    private void Add(int a, int b) {
        System.out.println("Add :::" + (a + b));
    }

    private void Sub(int a, int b) {
        System.out.println("Sub :::" + (a - b));
    }

    private void Mul(int a, int b) {
        System.out.println("Mul :::" + (a * b));
    }

    private void Div(int a, int b) {
        System.out.println("Mul :::" + (a / b));
    }
}

输出:
Single Thread
Add ::: 1200
Sub ::: 800
Mul ::: 200000
Mul ::: 5
Calender - 以毫秒为单位的时间:138 415 866 7863

多线程程序:

package runnableandcallable;

import java.util.ArrayList;
import java.util.Calendar;
import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.TimeoutException;

public class MainThread {

    private static ExecutorService service = Executors.newFixedThreadPool(10); // connection
                                                                               // pool
    @SuppressWarnings("unchecked")
    public static void main(String[] args) throws InterruptedException {
        System.out.println("Multithreading");
        MainThread mt = new MainThread();
        mt.testThread(1000, 200);
        Calendar lCDateTime = Calendar.getInstance();
        System.out.println("Calender - Time in milliseconds :"
                + lCDateTime.getTimeInMillis());
    }

    public void testThread(final int a, final int b) {
        // create a callable for each method
        Callable<Void> callableAdd = new Callable<Void>() {
            @Override
            public Void call() throws Exception {
                Add(a, b);
                return null;
            }
        };

        Callable<Void> callableSub = new Callable<Void>() {
            @Override
            public Void call() throws Exception {
                Sub(a, b);
                return null;
            }
        };

        Callable<Void> callableMul = new Callable<Void>() {
            @Override
            public Void call() throws Exception {
                Mul(a, b);
                return null;
            }
        };

        Callable<Void> callableDiv = new Callable<Void>() {
            @Override
            public Void call() throws Exception {
                Div(a, b);
                return null;
            }
        };

        // add to a list
        List<Callable<Void>> taskList = new ArrayList<Callable<Void>>();
        taskList.add(callableAdd);
        taskList.add(callableSub);
        taskList.add(callableMul);
        taskList.add(callableDiv);

        // create a pool executor with 3 threads
        ExecutorService executor = Executors.newFixedThreadPool(3);

        try {
            // start the threads
            List<Future<Void>> futureList = executor.invokeAll(taskList);

            for (Future<Void> voidFuture : futureList) {
                try {
                    // check the status of each future. get will block until the
                    // task
                    // completes or the time expires
                    voidFuture.get(100, TimeUnit.MILLISECONDS);
                } catch (ExecutionException e) {
                    System.err
                            .println("Error executing task " + e.getMessage());
                } catch (TimeoutException e) {
                    System.err.println("Timed out executing task"
                            + e.getMessage());
                }

            }

        } catch (InterruptedException ie) {
            // do something if you care about interruption;
        }

    }

    private void Add(int a, int b) {
        System.out.println("Add :::" + (a + b));
    }

    private void Sub(int a, int b) {
        System.out.println("Sub :::" + (a - b));
    }

    private void Mul(int a, int b) {
        System.out.println("Multiply :::" + (a * b));
    }

    private void Div(int a, int b) {
        System.out.println("Division :::" + (a / b));
    }

}

多线程输出:多线程
子::: 800
Division ::: 5
Add ::: 1200
Multiply ::: 200000
Calender - 以毫秒为单位的时间:138 415 868 0821

这里单线程执行138 415 866 7863毫秒,多线程执行138 415 868 0821毫秒.那么多线程的真正目的是什么？

Answer 1

您正在进行的处理非常简单,因此创建线程的开销更加昂贵.

如果你有可以并行完成的昂贵操作,那么多线程是有意义的.

Answer 2

第一:因为创建线程的开销超过了它们执行的有用工作.如果你在线程中运行更多的工作,它将使它比一个线程更快.必须在一个线程中运行必须的代码.

第二:对于创建微基准,你应该使用JMH

Answer 3

1,384,158,667,863毫秒约为44年.所以你告诉我们你等了44年这次行动的结果？或者你测量执行速度的方式有问题吗？

要测量两次之间的差异,您需要至少两次,而您只能在程序结束时获得当前日期,这甚至不准确.

简单的时间测量类:

public class StopWatch {
  private long startTime = -1;

  public void start() {
    this.startTime = System.nanoTime();
  }

  public long timeNanos() {
    return System.nanoTime() - this.startTime;
  }

  public double timeMillis() {
    return this.timeNanos() / 1000000.0;
  }
}

使用此秒表来测量执行的时间(就像你使用秒表一样),然后做3次,并意识到每次你得到完全不同的结果.这是因为测量精确的执行时间根本不是微不足道的.操作系统不断地用其他任务中断程序的执行,看似简单的命令可以有一整套需要运行的后台命令.

您所能做的就是通过运行该任务一百万次来估算所需的时间,然后取平均值.