Question

我正在为一堂课的Pig Latin翻译工作。我之前有我的计划工作;它从translateWords数组列表中打印出底部for循环。但最新版本不会在底部打印循环。我怀疑它是一个大括号，但我似乎无法找到它。

package stringTest;

import java.util.ArrayList;

public class StringTest {

    public static void main(String[] args) {
        String userString = "THIS IS A STRING";

        // We might need to trim the string first it throws an error if           theres
    // white space around word
    // Making string lowercase, first
    userString = userString.toLowerCase();

    // Splitting up string into individual words
    String[] stringArray = userString.split(" ");
    ArrayList<String> translatedWords = new ArrayList<String>();

    // going through each string with foreach loop
    for (String ss : stringArray) {
        System.out.println("prints here in intial for loop");
        // pulling out the words that start with a vowel
        // since they just get "way" at the end
        if (ss.charAt(0) == 'a' || ss.charAt(0) == 'e'
                || ss.charAt(0) == 'i' || ss.charAt(0) == 'o'
                || ss.charAt(0) == 'u') {
            ss = ss.concat("way");
            translatedWords.add(ss);

        }

        // If the words don't start with a vowel
        // trying to figure out how to cut them at first vowel and
        // concatenate to end
        else {
            for (int i = 0; i < ss.length();) {
                if (ss.charAt(i) == 'a' || ss.charAt(i) == 'e'
                        || ss.charAt(i) == 'i' || ss.charAt(i) == 'o'
                        || ss.charAt(i) == 'u') {
                    ss = ss.substring(i, ss.length());
                    String sss = ss.substring(0, i + 1);
                    String ss44 = ss.substring(i + 1);
                    String ss33 = ss44 + sss;
                    ss33 = ss33 + "ay";
                    translatedWords.add(ss33);
                    System.out.println(ss33);
                    System.out.println("Why won't this print");
                    break; 

                }
            }

        }

        for (String fs : translatedWords) {
            System.out.print(fs.toString() + " ");
        }

       }
    }
}

Answer 1

它不是一个大括号，但for循环无限运行

i永远不会在for语句中或在for

中增加

因此if条件将继续为字符串的第一个字符运行，如果它不是字符串的假元素，那么它将永远不会进入if语句

  //for (int i = 0; i < ss.length();) {//no i++ implemented
  for (int i = 0; i < ss.length();i++) {
            if (ss.charAt(i) == 'a' || ss.charAt(i) == 'e'
                    || ss.charAt(i) == 'i' || ss.charAt(i) == 'o'
                    || ss.charAt(i) == 'u') {
                ss = ss.substring(i, ss.length());
                String sss = ss.substring(0, i + 1);
                String ss44 = ss.substring(i + 1);
                String ss33 = ss44 + sss;
                ss33 = ss33 + "ay";
                translatedWords.add(ss33);
                System.out.println(ss33);
                System.out.println("Why won't this print");
                break; 

            }
        }

我在Eclipse上的Java程序停止打印到控制台，但之前它正常工作

1 个答案: