Question

我正在尝试编写一个允许我将输入的字母转换为数字的函数，反之亦然，到目前为止，我一直在编译错误。我想保持这个小而有效，而不必做一个广泛的if，else if语句。

到目前为止，这是我的代码：

public class CaesarCipher {

    /*
     * create function that converts a letter to a number
     * ex. a -> 0, b -> 1, etc...
     */
    static char letterToNumber (char firstLetter){
        if (firstLetter < 'a' || firstLetter > 'z') {
            throw new IllegalArgumentException("Only lower-case ASCII letters are valid");
    }
    return (char) (firstLetter - 'a');
    }

    /*
     * function to allow a user input a number and converts to a letter
     * 0->a and 1->b, etc...
     */
    static int numberToLetter (int firstNumer){
        if (firstNumber < '0' || firstNumber > '25'){   
        }
        return firstNumber;
    }
    public static void main(String[] args) {
        char a = 0;
        // TODO Auto-generated method stub
        System.out.println (letterToNumber (a)); //suppose to compile to convert a -> the number 0
        System.out.println(numberToLetter (1)); //compile to convert 1 -> the letter b
    }

}

Answer 1

你缺少＆＃39; a＆＃39; +在numberToLetter的return语句中

return 'a' + firstNumber;

无法将字母转换为数字并反转

1 个答案: