我用线程失去了理智

Question

我想要这个类的对象:

public class Chromosome implements Runnable, Comparable<Chromosome> {
    private String[] chromosome;
    public double fitness;
    private Random chromoGen;

    public Chromosome(double[] candidate) {
        super();
        //encode candidate PER parameter; using Matrix as storage
        chromosome = encode(candidate);
        chromoGen = new Random();
    }

    //De-fault
    public Chromosome() {
        super();
        chromoGen = new Random();

        //de-fault genotype
        chromosome = new String[6];
    }

    /**
     * IMPLEMENTED
     */
    public void run() {
        //I hope leaving it empty works...
    }

    public int compareTo(Chromosome c) {
        return (int) (fitness - c.fitness);
    }

    /**
     * Fitness stored in chromosome!
     */
    public void setFitness(ArrayList<double[]> target) {
        fitness = FF.fitness(this, target);
    }

    public double getFitness() {
        return fitness;
    }

    /**
     * ENCODERS/DECODERS
     */
    public String[] encode(double[] solution) {
        //subtract 2^n from solution until you reach 2^-n

        /**
         * LENGTH: 51 BITS!!
         *
         * 1st BIT IS NEGATIVE/POSITIVE
         *
         * THE PRECISION IS [2^30 <-> 2^-20]!!!
         *
         * RANGE: -2.14748...*10^9 <-> 2.14748...*10^9
         */
        String[] encoded = new String[6];

        //PER PARAMETER
        for (int j = 0; (j < 6); j++) {
            encoded[j] = encode(solution[j]);
        }

        return encoded;
    }

    public String encode(double sol) {
        /**
         * THE PRECISION IS [2^30 <-> 2^-20]!!!
         */
        double temp = sol;
        String row = "";
        //NEGATIVITY CHECK
        if (temp < 0) {
            //negative
            row = "1";
        } else {
            //positive
            row = "0";
        }
        //main seq.
        for (int n = 30; (n > (-21)); n--) {
            if ((temp - Math.pow(2, n)) >= 0) {
                temp = temp - Math.pow(2, n);
                row = row + "1";
            } else {
                row = row + "0";
            }
        }
        return row;
    }

    public double decoded(int position) {
        //returns UN-ENCODED solution
        double decoded = 0.00;
        char[] encoded = (chromosome[position]).toCharArray();
        /**
         * [n?][<--int:30-->][.][<--ratio:20-->]
         */
        int n = 30;
        for (int i = 1; (i < encoded.length); i++) {
            if (encoded[i] == '1') {
                decoded += Math.pow(2, n);
            }
            //next binary-place
            n--;
        }
        //NEGATIVE??
        if (encoded[0] == '1') {
            decoded = ((-1) * decoded);
        }
        //Output
        return decoded;
    }

    /**
     * GETTERS
     * ---------------\/--REDUNDANT!!
     */
    public double getParameter(int parameter) {
        //decoded solution
        return decoded(parameter);
    }

    /**
     * Used in E-algrm.
     */
    public String getRow(int row) {
        //encoded solution
        return chromosome[row];
    }

    /**
     * SETTERS
     */
    public void setRow(String encoded, int row) {
        chromosome[row] = encoded;
    }

    public void setRow(double decoded, int row) {
        chromosome[row] = encode(decoded);
    }

    /**
     * MUTATIONS
     */
    public void mutate(double mutationRate) {
        //range of: 51
        double ran = 0;
        int r;
        char[] temp;
        for (int m = 0; (m < 6); m++) {
            temp = (chromosome[m]).toCharArray();
            ran = chromoGen.nextDouble();
            if (ran <= mutationRate) {
                r = chromoGen.nextInt(51);
                if (temp[r] == '1') {
                    temp[r] = '0';
                } else {
                    temp[r] = '1';
                }
            }
            //output
            chromosome[m] = new String(temp);
        }
    }
}

...处于SEPARATE线程中; 但是我不需要这种方法run(); 但是当我尝试这样做时:

child1 = new Chromosome();
(new Thread(child1)).start();

不过,我在运行时看到的唯一线程是main().那么,我怎样才能使它成为单独的线程？

Answer 1

我看到你对线程如何工作的理解存在问题.

创建线程时,它会查找该方法run().有几种创建线程的方法.我通过传递一个Runnable对象来做到这一点.

Thread t=new Thread (new Runnable);

你知道一个线程有多长？

• 只要方法run()存在并运行,线程就会存在.线程只执行run()方法内的代码.它不是为了执行任何外部操作而设计的run().当控件移出时线程死亡run().

你的情况:

你把run()方法留空了.因此,线程不执行任何操作,并在创建时死亡.

你能做什么？

将程序的其余部分包含在内,run()以便run()保留在堆上,因此,新创建的线程运行程序.

您不需要进行插入run(),只需将第一个方法调用转移到run()堆上即可.

我们举一个例子:

public class threading implements Runnable
{
     public static void main(String args[])
     {
       Thread t = new Thread (new Runnable);
        t.setName("thread1");
        t.start();
         print1();
         print2();

     }    
      public static void print2()
      {
      System.out.println(Thread.getName());
       }
       public static void print1()
      {
      System.out.println(Thread.getName());
       }
       public void run()
       {
      System.out.println(Thread.getName());
       }
}

.OUPUTS:

• 线程1
• 主要
• 主要

是时候让你的新线程保持活着直到结束.

public class threading implements Runnable
{
     public static void main(String args[])
     {
       Thread t = new Thread (new Runnable);
        t.setName("thread1");
        t.start();


     }    
      public static void print2()
      {
      System.out.println(Thread.getName());
       }
       public static void print1()
      {
      System.out.println(Thread.getName());
       print2();
       }
       public void run()
       {
      System.out.println(Thread.getName());
        print1();
       }
}

输出:

• 线程1
• 线程1
• 线程1

我们run()通过调用方法调用将方法保留在堆上run().该方法是进一步维持流动的方法.