Question

我必须创建一个与用户界面交互的程序，该用户界面要求用户输入数字和基数以转换为二进制，十六进制或八进制。我制作了一个有效的程序但是当用户输入＆＃34; 0 0＆＃34;时，假设终止并结束程序。另一方面，我的程序不会这样做，但继续进行while循环。

以下是代码：

import java.util.*;  //for Scanner

public class BaseConversionApp {
public static void main(String[] args) {
    Scanner console = new Scanner(System.in);
    String combo = numbers(console);

    while(combo != "0 0") {
        if(combo.length() > 0) {     //breaks the string into integers to do math to it
            Scanner s = new Scanner(combo);
            int count = s.nextInt();
            int countSecond = s.nextInt();
            s.close();
            conversion(count,countSecond);
            System.out.println();
            //now if it goes 0 0 or

            String again = numbers(console);
            //  conversion(count,countSecond);
        }
    }


//prompts the user for two numbers and checks if the bases are 16, 8, 2.
public static String numbers(Scanner console) {
        String combination = "";
        String nothing = "0 0";
        System.out.print("Enter an integer and the new base: ");
        int integer = console.nextInt();
        int base = console.nextInt();
        if(base == 16 || base == 2 || base == 8) {
            combination = integer + " " + base;
            return combination;
        } else if (base == 0 && integer == 0){
            System.out.println("Good bye!");
            return nothing;
        } else {
            System.out.println("Sorry, that is an invalid base. Please enter 2, 8, or 16 
only.");
        }
        return "";
    }

    //converts the integers into binary, hexa, or octo.
    public static void conversion (int integer, int base) {

        //takes cares of the special case if the user wants to know hexidecimal
        if(base <= 16) {
            String calculations = Integer.toString(integer, base);


            if(integer > 0 && base > 0) {
                System.out.println(integer + " in binary -> " + Integer.toString(integer, 
base));
            }
        }
    }
}

Answer 1

您无法比较这样的字符串，您必须使用String对象的equals()方法。所以，你应该：

while(!"0 0".equals(combo)) {
    ...
}

请注意，我已将常量"0 0"置于首位 - 这可以保护您免受combo null !combo.equals("0 0")的攻击。否则，您有null，如果组合为NullPointerException，当您尝试在{{1}上拨打equals()时，您将获得null }值。

Answer 2

请尝试使用此代码。你的看起来很复杂。顺便说一下，你的代码适合我。

import java.util.Scanner;

public class NewClass {

    static Scanner inp = new Scanner(System.in);
    static String line1 = "";
    static String line2 = "";
    static String exit = "exit";

    public static void main(String[] args) {
        System.out.println("Enter int first and then base...");
        System.out
                .println("Enter the word exit to exit!");

        while (true) {

            line1 = inp.next();
            if (line1.equalsIgnoreCase(exit)) {
                break;
            }
            line2 = inp.next();

            try {
                conversion(Integer.parseInt(line1), Integer.parseInt(line2));
            } catch (NumberFormatException e) {
                e.printStackTrace();
            }

        }

        System.out.println("Bye...");

    }

    public static void conversion(int integer, int base) {

        // takes cares of the special case if the user wants to know hexadecimal
        if (base <= 16) {
            String calculations = Integer.toString(integer, base);

            if (integer > 0 && base > 0) {
                System.out.println(integer + " in binary -> "
                        + Integer.toString(integer, base));
            }
        }
    }

}

我希望它停止时，我在Java中的方法一直在运行

2 个答案: