Question

下面是多线程函数的代码和输出，其中有一个计数器和加，减，乘和除的函数。我正在使用Eclipse。

每个数学函数的4个线程：

public class Maths {

    public static void main(String args[]){

        CalculationThread T1 = new CalculationThread("Addition");
        T1.start();

        CalculationThread T2 = new CalculationThread("Subtraction");
        T2.start();

        CalculationThread T3 = new CalculationThread("Multiplication");
        T3.start();

        CalculationThread T4 = new CalculationThread("Division");
        T4.start();
    }
}

class CalculationThread extends Thread{
    private Thread t;
    private String maths;
    private int count = 0;
    private int resultplus, resultminus, resulttimes, resultdivide = 0;

    CalculationThread(String answer){
        maths = answer;
    }

    public void start(){
        System.out.println("Starting calculation of " + maths + "\n");
        if(t == null){
            t = new Thread (this, maths);
            t.start();
        }
    }

这是函数发生的地方，它将使用计数器作为2个数字来执行等式。

    public void run(){
        try {
            for (int x=0; x<=3 ; x++){

                if(maths == "Addition"){
                System.out.println("Calculating: " + maths + " of " + count + 
                        " + "+ count + " = " + resultplus + "\n");
                Thread.sleep(3000);
                count++;
                resultplus = count + count;
                }

                else if(maths == "Subtraction"){
                    System.out.println("Calculating: " + maths + " of " + count + 
                            " - "+ count + " = " + resultminus + "\n");
                    Thread.sleep(3000);
                    count++;
                    resultminus = count - count;
                }

                else if(maths == "Multiplication"){
                    System.out.println("Calculating: " + maths + " of " + count + 
                            " * "+ count + " = " + resulttimes + "\n");
                    Thread.sleep(3000);
                    count++;
                    resulttimes = count * count;
                }

                else if(maths == "Division"){
                    System.out.println("Calculating: " + maths + " of " + count + 
                            " / "+ count + " = " + resultdivide + "\n");
                    Thread.sleep(3000);
                    count++;
                    resultdivide = count / count;
                }

            }

        }
            catch (InterruptedException e){
                System.out.println("Math function failed");
            }
                if(maths == "Addition"){
                System.out.println("Addition completed.");
                }
                else if(maths == "Subtraction"){
                    System.out.println("Subtraction completed.");
                }
                else if(maths == "Multiplication"){
                    System.out.println("Multiplication completed.");
                }
                else if(maths == "Division"){
                    System.out.println("Division completed.");
                }
    }
}

输出：

Starting calculation of Addition

Starting calculation of Subtraction

Calculating: Addition of 0 + 0 = 0

Starting calculation of Multiplication

Calculating: Subtraction of 0 - 0 = 0

Starting calculation of Division

Calculating: Multiplication of 0 * 0 = 0

Calculating: Division of 0 / 0 = 0

Calculating: Subtraction of 1 - 1 = 0

Calculating: Addition of 1 + 1 = 2

Calculating: Multiplication of 1 * 1 = 1

Calculating: Division of 1 / 1 = 1

Calculating: Addition of 2 + 2 = 4

Calculating: Subtraction of 2 - 2 = 0

Calculating: Division of 2 / 2 = 1

Calculating: Multiplication of 2 * 2 = 4

Calculating: Subtraction of 3 - 3 = 0

Calculating: Addition of 3 + 3 = 6

Calculating: Division of 3 / 3 = 1

Calculating: Multiplication of 3 * 3 = 9

Subtraction completed.
Addition completed.
Division completed.
Multiplication completed.

上面的代码可以同时完成所有4个函数，但每当我尝试为用户输入而不是自动计数器包含JOptionPane时，4个线程中的每一个都会一次请求。因此，如果函数等待我输入2个数字，则不计入多线程。如何以及以何种方式包含用户输入，只需要用户在开头输入，以便所有函数都可以使用2个变量。

Answer 1

我不确定我是否理解正确。

如果您只想阻止计算线程并等待初始用户输入，您可以使用Semaphore。

等待用户输入的UI线程显示对话框，并通过设置许可/线程数来释放等待计算线程。

这是一个如何看起来的例子（它也使用更面向对象的方法）。为简单起见，我跳过了乘法和除法任务

import java.util.concurrent.Semaphore;

import javax.swing.JOptionPane;

public class MathSample {

    // because updated / read from different threads mark as volatile 
    private volatile int a, b;

    // semaphore with no initial permits i.e.
    // the calculations will wait until permits are available.
    private Semaphore available = new Semaphore(0);

    private abstract class Task implements Runnable {

        public abstract void doCalculation();

        public abstract String getName();

        @Override
        public void run() {
            try {
                // wait until a permit becomes available
                available.acquire();
                // not sure what should happen here
                // wait again for user input? 
                for (int x = 0; x < 50; ++x) {
                    a = a + x;
                    doCalculation();
                }

            } catch (InterruptedException e) {
                Thread.currentThread().interrupt();
            }

            System.out.println(String.format("Task %s completed.", getName()));
        }

    }


    private  class AdditionTask extends Task {

        public void doCalculation() {
            System.out.println(String.format("Calculating: Addition of + %d + %d = %d", a, b, a+b));
        }
        public String getName() {
            return "Addition";
        }
    }

    private  class SubstractionTask extends Task {
        public void doCalculation() {
            System.out.println(String.format("Calculating: Substraction of + %d - %d = %d", a, b, a-b));
        }

        public String getName() {
            return "Substraction";
        }
    }



    private void run() {
        new Thread(new AdditionTask()).start();
        new Thread(new SubstractionTask()).start();


        a = Integer.parseInt(JOptionPane.showInputDialog("First value"));
        b = Integer.parseInt(JOptionPane.showInputDialog("Second value"));
        available.release(2); // let the 2 calculation threads run

    }


    public static void main(String ...args) {
        new MathSample().run();

    }
}

正如您所看到的，您不必覆盖线程的start方法来运行不同的线程。

您start的{{1}}方法至少是奇怪的，因为您覆盖了CalculationThread类的start方法，并在其中创建了另一个Thread实例，您传递了Thread作为CalculationThread。

更容易/更好：

Runnable

Answer 2

您可以参考以下代码。根据我对你问题的理解，也许它会满足你的需求。我还根据Hovercraft Full Of Eels建议将.equals()应用于代码。

import javax.swing.JOptionPane;

public class Maths {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String num1 = JOptionPane.showInputDialog("Num1: ");
        String num2 = JOptionPane.showInputDialog("Num2: ");
        int num11 = Integer.parseInt(num1);
        int num22 = Integer.parseInt(num2);

        calculationThread T1 = new calculationThread("Addition");
        T1.getNumber(num11, num22);
        T1.start();

        calculationThread T2 = new calculationThread("Subtraction");
        T2.getNumber(num11, num22);
        T2.start();

        calculationThread T3 = new calculationThread("Multiplication");
        T3.getNumber(num11, num22);
        T3.start();

        calculationThread T4 = new calculationThread("Division");
        T4.getNumber(num11, num22);
        T4.start();
    }
}

class calculationThread extends Thread{
    private Thread t;
    private String maths;
    private int a;
    private int b;
    private int resultplus = 0;
    private int resultminus = 0;
    private int resulttimes = 0;
    private int resultdivide = 0;

    public void getNumber(int num1, int num2){
        a = num1;
        b = num2;
    }

    calculationThread(String answer){
        maths = answer;
    }

    public void start(){
        System.out.println("Starting calculation of " + maths + "\n");
        if(t == null){
            t = new Thread(this, maths);
            t.start();
        }
    }

    public void run(){
        try {
            for (int x=0; x<=3 ; x++){
                if(maths.equals("Addition")){
                System.out.println("Calculating: " + maths + " of " + a + 
                        " + "+ b + " = " + resultplus + "\n");
                Thread.sleep(3000);
                resultplus = a + b;
                }
                else if(maths.equals("Subtraction")){
                    System.out.println("Calculating: " + maths + " of " + a + 
                            " - "+ b + " = " + resultminus + "\n");
                    Thread.sleep(3000);
                    resultminus = a - b;
                }
                else if(maths.equals("Multiplication")){
                    System.out.println("Calculating: " + maths + " of " + a + 
                            " * "+ b + " = " + resulttimes + "\n");
                    Thread.sleep(3000);
                    resulttimes = a * b;
                }   
                else if(maths.equals("Division")){
                    System.out.println("Calculating: " + maths + " of " + a + 
                            " / "+ b + " = " + resultdivide + "\n");
                    Thread.sleep(3000);
                    resultdivide = a / b;
                }   
            }   
        }
        catch (InterruptedException e){
            System.out.println("Math function failed");
        }
        finally{
            if(maths.equals("Addition")){
                System.out.println("Addition completed.");
            }
            else if(maths.equals("Subtraction")){
                System.out.println("Subtraction completed.");
            }
            else if(maths.equals("Multiplication")){
                System.out.println("Multiplication completed.");
            }
            else if(maths.equals("Division")){
                System.out.println("Division completed.");
            }           
        }
    }
}

多线程 - 为所有变量请求输入一次（Java）

2 个答案: