Question

我正在编写一个Java程序，该程序从用户处获取一个整数并以二进制形式输出该数字。我的代码没有给我任何错误，但是该程序无法正常运行。该程序要求用户输入一个数字，但是当我按Enter键时，它仅转到下一行，用户可以无限输入。我哪里做错了？我尝试调试它，但是似乎找不到问题。提前非常感谢您！

package example;
import java.util.Scanner;    
public class test {

public static void main(String[] args) {
    Scanner inScanner = new Scanner(System.in);
    int input = promptForDecimal(inScanner);
    String output = decimalToBinary(input);
    System.out.println("The decimal value " + input + " is " + output 
                    + " in binary.");
}

public static String decimalToBinary(int value) {
    String binary = "";
    int i = 0;
    while (value > 0) {
        i = value % 2;
        if (i == 1) {
            binary = binary + "1";
        }
        else {
            binary = binary + "0";
        }
    }
    return binary;
}


public static int promptForDecimal(Scanner inScanner) {
    System.out.print("Enter an integer value (negative value to quit): ");
    String val = inScanner.nextLine();
    while (checkForValidDecimal(val) == false)  {
        System.out.println("Error - value must contain only digits");
        System.out.println("Enter an integer value (negative value to quit): ");
        val = inScanner.nextLine();
    }
    return Integer.parseInt(val);
}


public static boolean checkForValidDecimal(String value) {
    int length = value.length();
    int pos = 0;
    boolean a = true;
    while (pos < length) {
        a = Character.isDigit(value.charAt(pos));
        if (a == true) {
            pos++;
        }
        else {
            if (value.charAt(0) == '-') {
                pos++;
            }
            else {
                a = false;
            }
        }

    }
    return a;
}
}

Answer 1

写出二进制文件后，您忘记更新值。

public static String decimalToBinary(int value) {
    String binary = "";
    int i = 0;
    while (value > 0) {
        i = value % 2;
        value = value / 2;
        if (i == 1) {
            binary = binary + "1";
        }
        else {
            binary = binary + "0";
        }
    }
    return binary;
}

Java程序不会停止接受用户输入

1 个答案: