java - 基本转换错误检查

Question

当我输入正确的值时，我的代码中的所有内容都工作正常。当我输入不正确的值故意导致错误时，错误消息永远不会显示。控制台一片空白，我必须终止。我不确定为什么程序从未真正到达错误消息。相反，它似乎进入了无限循环。

import java.util.Scanner;

public class BaseConversion {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        String theValue;
        String newNum;
        int initialBase;
        int finalBase;

        System.out.println("Please enter a value: ");
        theValue = s.nextLine();
        System.out.println("Please enter original base: ");
        initialBase = s.nextInt();


        System.out.println("Please enter new base: ");
        finalBase = s.nextInt();
        s.close();

        newNum = convertInteger(theValue, initialBase, finalBase);


        System.out.println("new number: " + newNum);
    }

    public static void isValidInteger(String num, int finalBase) {

        if (finalBase < 2 || finalBase > 36) {
            System.out.println("Error: Base must be greater than or equal to 2 & less than or equal to 36");
            System.exit(1);
        }


        char chDigit;

        for(int d = 0; d < num.length(); d++) {
            chDigit = num.toUpperCase().charAt(d);
            if(Character.isDigit(chDigit) && (chDigit - '0') >= finalBase) {
                System.out.println("Error");
                System.exit(1);
            } 
            else if (Character.isLetter(chDigit) && (chDigit - 'A') + 10 >= finalBase) {
                System.out.println("Error");
                System.exit(1);
            }
            else if (!Character.isDigit(chDigit) && !Character.isLetter(chDigit)) {
                System.out.println("Error");
                System.exit(1);
            }

        }
    }

    public static String convertInteger(String theValue, int initialBase, int finalBase) {

        double val = 0;
        double decDigit = 0;
        char chDigit;


        // loop through each digit of the original number
        int L = theValue.length();
        for(int p = 0; p < L; p++) {

            // get the digit character (0-9, A-Z)
            chDigit = Character.toUpperCase(theValue.charAt(L-1-p));

            // get the decimal value of our character
            if(Character.isLetter(chDigit)) {
                decDigit = chDigit - 'A' + 10;
            }
            else if (Character.isDigit(chDigit)) {
                decDigit = chDigit - '0';
            }
            else {
                System.out.println("Error");
                System.exit(1);
            }
            // add value to total
            val += decDigit * Math.pow(initialBase, p);
        }

        // determine number of digits in new base
        int D = 1;

        for( ; Math.pow(finalBase, D) <= val; D++) {}

        // use char array to hold new digits
        char[] newNum = new char[D];

        double pwr; 
        for(int p = D-1; p >= 0; p--) {

            // calculate the digit for this power of newBase
            pwr = Math.pow(finalBase, p);

            decDigit = Math.floor(val / pwr);

            val -= decDigit*pwr;

            // store the digit character

            if(decDigit <= 9) {
                newNum[D-1-p] = (char) ('0' + (int)decDigit);
            }
            else {
                newNum[D-1-p] = (char) ('A' + (int)(decDigit - 10));
            }
        }       
        return new String(newNum);
    }
}

Answer 1

我注意到的问题是您从未调用 isValidInteger 函数。因此，当finalBase例如为1时，即无效输入，convertInteger函数中存在无限循环:

// determine number of digits in new base
        int D = 1;

        for( ; Math.pow(finalBase, D) <= val; D++) {}

我不确定这是否是唯一的问题，但这就是我发现的问题。