Question

我正在研究一个Java程序，该程序应该表现得像一个“虚拟寻宝游戏”，基本上它会随机进行数字运算，直到找到与您要寻找的动物相匹配的数字，然后显示结果。该信息应该存储在文件中，然后显示在控制台中，但是只要我单击运行，它就会终止。我在想某种逻辑错误，因为我对语法没有任何了解。

import java.io.IOException; 
import java.io.PrintWriter;
import java.io.File;
import java.util.Scanner; 
public class AnimalPopulation {

public static void main(String[] args) throws IOException
{
    // TODO Auto-generated method stub

    int totalTrials = 0;
    int animalsSpotted = 0; 
    double randNum = 0;
    double myAnimal = 4.0;
    double notMyAnimal = 0; 
    String data = "int";

    Scanner in = new Scanner(System.in);
    Scanner inFile = new Scanner ("animals.txt");

    PrintWriter outFile = new PrintWriter(new File("animals.txt"));
    //BufferedReader in = new BufferedReader(new FileReader("animals.txt"));

System.out.println("How many trials will you be running? (It must be greater than or equal to 1,000.)");
totalTrials = in.nextInt();

    while(animalsSpotted < totalTrials) {
        animalsSpotted++; 
        randNum = (1 + (Math.random()* 9));

        if(randNum < myAnimal) {

            outFile.println(notMyAnimal++);
        }else if(randNum > myAnimal) {

            outFile.println(notMyAnimal++);
        }else if(randNum == myAnimal) {

            outFile.println(myAnimal++);


            while(inFile.hasNext()) {

                data = inFile.next();
                System.out.println(data);
            }

            inFile.close();



        }


    }

outFile.close();
}

}

Answer 1

第一件事是您必须将randNum更改为int并使用ThreadLocalRandom，如答案中所述，另一件事是首先将所有数据写入文件，然后读取它不要并行执行，这是一个程序，以您为例：

import java.io.*;
import java.util.Scanner;
import java.util.concurrent.ThreadLocalRandom;

public class AnimalPopulation{
   public static void main(String[] args) throws IOException {
       // TODO Auto-generated method stub

    int totalTrials = 0;
    int animalsSpotted = 0;
    int randNum = 0;
    double myAnimal = 4.0;
    double notMyAnimal = 0;
    String data = "int";

    Scanner in = new Scanner(System.in);

    PrintWriter outFile = new PrintWriter(new File("animals.txt"));
    //BufferedReader in = new BufferedReader(new FileReader("animals.txt"));

    System.out.println("How many trials will you be running? (It must be greater than or equal to 1,000.)");
    totalTrials = in.nextInt();

    while(animalsSpotted < totalTrials) {
        animalsSpotted++;
        randNum = ThreadLocalRandom.current().nextInt(1,11);

        if(randNum < myAnimal) {

            outFile.println(notMyAnimal++);
        }else if(randNum > myAnimal) {

            outFile.println(notMyAnimal++);
        }else if(randNum == myAnimal) {

            outFile.println(myAnimal++);
        }
    }
    outFile.close();
    File file = new File("animals.txt");

    BufferedReader br = new BufferedReader(new FileReader(file));

    String st;
    while ((st = br.readLine()) != null)
        System.out.println(st);
}
}

Answer 2

randNum为double，而(1 + (Math.random()* 9))产生的分数在1到10之间，因此极不可能等于4.0。相反，您可能需要将类型更改为int，并使用ThreadLocalRandom.current().nextInt(1,11)来生成1到10之间的值。

在控制台中显示文本文件中的信息

2 个答案: