卡在开关循环中答案

【问题标题】：Stuck in a switch loop卡在开关循环中
【发布时间】：2015-05-27 18:10:13
【问题描述】：

我写了一个作业来尝试计算字符串中每种元音的数量。

它编译得很好，但似乎在 switch 语句上循环，尽管谷歌搜索了大约一个小时，但我看不出我做错了什么。

请帮忙！ =]

 import java.util.Scanner;

 public class assignment3b
 {
public static void main(String[] args)
{
    int alpha=0, epsilon=0, india=0, oscar=0, uniform=0, position=0, length;

    String input;

    char letter;

    Scanner scan = new Scanner(System.in);

    System.out.println("Welcome to the vowel parse-o-matic");
    System.out.println("\nThis program will count all lower case vowels in whatever you type.");
    System.out.print("\nPlease enter the word you'd like to have parsed : ");

        input = scan.next();

        System.out.println("\n\nThe word " + input + " has: "); // reprints the word before stripping spaces

        input = input.replaceAll("\\s+","");                // Removes whitespace so they don't get counted.

    while (position < input.length());
    {
        letter = input.charAt(position);

        switch (letter)
        {
            case 'a':
                alpha = alpha + 1;
                position = position +1;
                break;
            case 'e':
                epsilon = epsilon + 1;
                position = position +1;
                break;
            case 'i':
                india = india + 1;
                position = position +1;
                break;
            case 'o':
                oscar = oscar + 1;
                position = position +1;
                break;
            case 'u':
                uniform = uniform + 1;
                position = position +1;
                break;
            default:
                position++;
                break;  
        }

        System.out.println("a's = " + alpha);
        System.out.println("e's = " + epsilon);
        System.out.println("i's = " + india);
        System.out.println("o's = " + oscar);
        System.out.println("u's = " + uniform);
        System.out.println("\nOther characters = " + (input.length() - alpha -epsilon - india -oscar - uniform));
    }
}

}

【问题讨论】：

您的 printlns 都在 while 循环之后的 {} 中，这可能就是为什么...
附带说明，为了帮助我的强迫症，请使用position++ 或position = position + 1，但请不要同时使用。这让我哭了。
另外，如果您要增加位置，则无需在每种情况下增加位置。而且根本不需要位置，你可以在String.toCharArray上快速枚举。
旁注：北约拼音字母表，这似乎启发了您自己的灵感，使用带有 f 的 alfa，以及 echo 而不是 epsilon
switch 不会“循环”。这基本上是表达长 if/elseif/elseif/.../else 树的好方法。

标签： java loops switch-statement

【解决方案1】：

去掉分号

while (position < input.length());
                                 ^

这会阻止position 递增

【讨论】：

【解决方案2】：

import java.util.Scanner;

    public class assignment3b
    {
        public static void main(String[] args)
        {
            int alpha = 0, epsilon = 0, india = 0, oscar = 0, uniform = 0, position = 0, length;

            String input;

            char letter;

            Scanner scan = new Scanner(System.in);

            System.out.println("Welcome to the vowel parse-o-matic");
            System.out.println("\nThis program will count all lower case vowels in whatever you type.");
            System.out.print("\nPlease enter the word you'd like to have parsed : ");

            input = scan.next();

            System.out.println("\n\nThe word " + input + " has: "); // reprints the word before
                                                                    // stripping spaces

            input = input.replaceAll("\\s+", ""); // Removes whitespace so they don't get counted.
            int len = input.length();

            while (position < len)

            {
                letter = input.charAt(position);

                switch (letter)
                {
                    case 'a':
                        alpha = alpha + 1;
                        position = position + 1;
                        break;
                    case 'e':
                        epsilon = epsilon + 1;
                        position = position + 1;
                        break;
                    case 'i':
                        india = india + 1;
                        position = position + 1;
                        break;
                    case 'o':
                        oscar = oscar + 1;
                        position = position + 1;
                        break;
                    case 'u':
                        uniform = uniform + 1;
                        position = position + 1;
                        break;
                    default:
                        position++;
                        break;
                }
            }
                System.out.println("a's = " + alpha);
                System.out.println("e's = " + epsilon);
                System.out.println("i's = " + india);
                System.out.println("o's = " + oscar);
                System.out.println("u's = " + uniform);
                System.out.println("\nOther characters = "
                        + (input.length() - alpha - epsilon - india - oscar - uniform));

        }
    }

试试这个。

正如其他人提到的，您有一个不属于它的分号。而且我更正了你的逻辑并将打印语句移到循环之外。

希望这是您所期望的最终结果

Welcome to the vowel parse-o-matic

This program will count all lower case vowels in whatever you type.

Please enter the word you'd like to have parsed : striker


The word striker has: 
a's = 0
e's = 1
i's = 1
o's = 0
u's = 0

Other characters = 5

【讨论】：

【解决方案3】：

你有一个额外的分号，它不属于它。

while (position < input.length());

这个分号不是语法错误。它描述了一个带有空循环内容的 while 块。它翻译为：

while (position < input.length()) {}

所以，去掉分号就可以了。

【讨论】：