Question

大家好，我一直在为我的单词益智游戏制作一个测试类，然后输出就是将对象参考号打印到对象上。任何人都有解决方案来打印对象的return语句。

输出：

生成器统计：从长度为3的单词生成的单词拼图 Puzzle.WordPuzzleGenerator @ c68c3Puzzle.WordPuzzleGenerator @ b2002fPuzzle.WordPuzzleGenerator @ 2a4983Puzzle.WordPuzzleGenerator @ 406199Puzzle.WordPuzzleGenerator @ c7b00cPuzzle.WordPuzzleGenerator @ 1f6f296Puzzle.WordPuzzleGenerator @ 1df5a8fPuzzle.WordPuzzleGenerator @ b2a2d8Puzzle.WordPuzzleGenerator @ 1e13d52Puzzle.WordPuzzleGenerator @ 80fa6f

测试类

public class Test_WordPuzzleGenerator {
    public static void main(String[] args) throws FileNotFoundException {

        int sizeTest1 = 3;
        System.out
                .println("Generator stats: word-puzzles generated from words of length "
                        + sizeTest1);

        for (int i = 0; i < 10; i++) {
            WordPuzzleGenerator puzzle = new WordPuzzleGenerator(sizeTest1);
            System.out.print(puzzle);
        }

        int sizeTest2 = 3;
        System.out
                .println("Generator stats: word-puzzles generated from words of length "
                        + sizeTest2);
        for (int i = 0; i < 10; i++) {
            new WordPuzzleGenerator(sizeTest2);
        }
    }
}

主程序：

public class WordPuzzleGenerator {
    static ArrayList<String> wordList = new ArrayList<String>();

    public WordPuzzleGenerator(int size) throws FileNotFoundException {
        ArrayList<String> puzzleListY = new ArrayList<String>();
        ArrayList<String> puzzleListX = new ArrayList<String>();
        String randomXWord;
        String letterSize = "" + size;
        makeLetterWordList(letterSize);
        boolean finished = false;
        while ( !finished ) {
            finished = true;
            puzzleListX.clear();
            puzzleListY.clear();
            for (int i = 0; i < size; i++) {
                int randomYWord = randomInteger(wordList.size());
                String item = wordList.get(randomYWord);
                puzzleListY.add(item);
            }
            for (int i = 0; i < puzzleListY.size(); i++) {
                StringBuilder sb = new StringBuilder();
                for (int j = 0; j < puzzleListY.size(); j++) {
                    sb.append(puzzleListY.get(j).charAt(i));
                }
                randomXWord = sb.toString();
                if (!wordList.contains(randomXWord) && !puzzleListY.contains(randomXWord)) {
                    finished = false;
                    break;
                }
                puzzleListX.add(randomXWord);

            }




        }
        toString(puzzleListX, puzzleListY);
    }

    public static int randomInteger(int size) {
        Random rand = new Random();
        int randomNum = rand.nextInt(size);
        return randomNum;
    }



    public static void makeLetterWordList(String letterSize) throws FileNotFoundException {

        Scanner letterScanner = new Scanner( new File (letterSize + "LetterWords.txt"));
        wordList.clear();
        while (letterScanner.hasNext()){
            wordList.add(letterScanner.next());
        }
        letterScanner.close();
    }

    public static String toString(ArrayList<String> ArrayList1, ArrayList<String> ArrayList2){
        StringBuilder group1 = new StringBuilder();
        for (int i = 0; i < ArrayList1.size(); i++) {
            group1.append(ArrayList1.get(i) + " ");
        }
        String wordsInString1 = group1.toString();

        StringBuilder group2 = new StringBuilder();
        for (int i = 0; i < ArrayList2.size(); i++) {
            group2.append(ArrayList2.get(i) + " ");
        }
        String wordsInString2 = group2.toString();

        return String.format("\t( %s) ( %s)", wordsInString1, wordsInString2);
    }
}

Answer 1

您的WordPuzzleGenerator课程未覆盖Object的{{1}}。相反，它包含一个具有不同签名的静态toString方法。

您需要在toString课程中使用此签名的方法：

WordPuzzleGenerator

再次看一下，它会使你的@Override public String toString() { ... }只有静态方法且没有实例成员，因此不清楚你期望WordPuzzleGenerator返回什么，或换句话说 - 它不清楚是什么{{ 1}}预计会打印。

编辑：

如果您希望toString打印在构造函数中创建的列表，您应该将它们设为实例成员：

System.out.print(puzzle);

然后你可以像这样覆盖toString：

toString()

Answer 2

您必须覆盖对象的toString方法，因为您的对象是从java对象继承的

@Override
public String toString(){
\\mystring build up...
return mystring;

注意覆盖注释，这就是诀窍;）快乐的编码！

Answer 3

尝试覆盖＆＃39; toString＆＃39;你班上的方法如下：

@Override
public String toString()
{
    //your code
}

在测试类中打印toString（）方法输出

3 个答案: