Question

我有一个程序可以读取一个包含4000 1或0的输入文件，每行一个。

程序编译正常，但运行时我收到以下错误：

ˇ˛0Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:840)
at java.util.Scanner.next(Scanner.java:1461)
at java.util.Scanner.nextInt(Scanner.java:2091)
at java.util.Scanner.nextInt(Scanner.java:2050)
at Project1.main(Project1.java:27)

出于某种原因，当我查看文本文件时，第一行是0，当程序读取它获取的第一行ˇ˛0时，还没有其他内容。我也试过使用缓冲的阅读器，没有运气。任何人都可以提供一些意见。谢谢！

这是我的代码：

import java.io.*;
import java.util.Random;
import java.util.Scanner;

public class Project1 {

    public static void main(String[] args) {
        int [] stream = new int[4000];
        int [] received = new int[4000];
        int corrupt = 0;
        float standardDev = 0;
        float [] averages = new float[4000];
        float averagev = 0;
        float voltage = 0;
        try {
            Scanner st = new Scanner(new File("CS 380 bit feed.txt"));
            System.out.print(st.next());
            FileWriter outFile = new FileWriter("output");
            PrintWriter out = new PrintWriter(outFile);
            int i=0;
            Random rand = new Random();

            stream[i]=st.nextInt();
            if (stream[i] == 0)
                voltage = (float) (2.49 * rand.nextFloat());
            else
                voltage = (float) ((2.5 * rand.nextFloat()) + 2.5);
            if (voltage < 2.5)
                received[i] = 0;
            else
                received[i] = 1;
            averagev = voltage;
            averages[i] = voltage;
            if (stream[i] != received[i])
                corrupt++;
            i++;

            while (i < 4000) {
                stream[i]=st.nextInt();
                if (stream[i] == 0)
                    voltage = (float) 2.49 * rand.nextFloat();
                else
                    voltage = (float) ((2.5 * rand.nextFloat()) + 2.5);
                averagev = ((averagev * i) + voltage)/(i+1);
                if (voltage <= averagev)
                    received[i] = 0;
                else
                    received[i] = 1;
                if (stream[i] != received[i])
                    corrupt++;
                i++;
            }
            averagev = 0;
            int j = 0;
            while (j < 4000) {
                for (int k = 0; k<8; k++) {
                    out.print(received[j]);
                    averagev = averages[i] + averagev;
                    j++;
                }
                out.println("    " + averages[j]);
            }

        }
        catch (FileNotFoundException e) {
              e.printStackTrace();
        } 
        catch (IOException e) {
              e.printStackTrace();
        }
        averagev = averagev / 4000;
        for (int k = 0; k<4000; k++) 
            standardDev = ((averagev - averages[k])*(averagev - averages[k])) + standardDev;
        standardDev = standardDev/4000;

        System.out.println("Errors: " + corrupt);
        System.out.println("Percentage Corrupt: " + corrupt / 4000);
        System.out.println("Average of Average Voltages: " + averagev);
        System.out.println("Standard Deviation: " + standardDev);        
    }
}

Answer 1

在十六进制编辑器中查看文件。它可能已损坏或在开头有一个字节顺序标记，这不一定会显示在文本编辑器中。

无法从java中的输入文件中读取int

1 个答案: