變數在Flutter中意外回傳null

我正在遍歷串列以獲取具有布林值false的專案。

類：

class Words {
  String alphabet;
  String word;
  String wordtype;
  String image;
  bool used;

  Words({
    required this.alphabet,
    required this.word,
    required this.wordtype,
    required this.image,
    required this.used,
  });
}

以下是正在使用的示例資料

List<Words> initData = [
    Words(
        alphabet: 'a',
        word: 'apple',
        wordtype: 'fruit',
        image: 'assets/res/apple.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'ant',
        wordtype: 'insect',
        image: 'assets/res/ant.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'axe',
        wordtype: 'object',
        image: 'assets/res/axe.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'arrow',
        wordtype: 'object',
        image: 'assets/res/arrow.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'airplane',
        wordtype: 'object',
        image: 'assets/res/airplane.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'anchor',
        wordtype: 'object',
        image: 'assets/res/anchor.jpg',
        used: false),
    Words(
        alphabet: 'a',
        word: 'alligator',
        wordtype: 'animal',
        image: 'assets/res/alligator.jpg',
        used: false),
  ];

呼叫小部件

class WordsWidget extends StatelessWidget {
  final homeCtrl = Get.find<HomeController>();
  WordsWidget({Key? key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    print(homeCtrl.nextWord());
    return Container();
  }
}

以下函式迭代，如果所有單詞都已使用，則將它們重置并再次迭代。

職能：

  nextWord() {
    Words nextword =
        Words(alphabet: '', word: '', wordtype: '', image: '', used: false);
    bool found = false;
    for (var i = 0; i < words.length; i  ) {
      if (words[i].used == false && found != true) {
        nextword = words[i];
        words[i].used = true;
        found = true;
      }
    }
    if (found == false) {
      print("resetting");
      resetWords();
    } else {
      print("Found = true so now returning");
      print(nextword.word);
      return nextword.word;
    }
  }

  void resetWords() {
    for (var a = 0; a < words.length; a  ) {
      words[a].used = false;
    }
    nextWord();
  }

以上作業正常，直到它重置。重置后，即使上面的列印回傳顯示了該值，但是當我在呼叫小部件時列印相同的值時，它回傳 null。以下是說明問題的輸出。

flutter: Found = true so now returning
2    flutter: arrow
flutter: Found = true so now returning
2    flutter: airplane
flutter: Found = true so now returning
2    flutter: anchor
flutter: Found = true so now returning
2    flutter: alligator
flutter: resetting
flutter: Found = true so now returning
flutter: apple
flutter: null

正如你從上面看到的，在 alligator 這個詞之后，它會重置它自己并為所有詞設定 used = false。然后它在迭代中搜索下一個單詞，它找到了單詞 apple，但呼叫該函式的小部件仍然回傳 null。

如果有人能強調我做錯了什么，我將不勝感激。謝謝你。

uj5u.com熱心網友回復：

您當前的代碼有兩個主要問題

1. words如果您的串列為空，它將陷入無限遞回（因此是例外） 。

2. 如果您沒有在for回圈中找到可用的單詞，則不會回傳任何內容...

nextWord() {

  // if there are no available words, don't need to do anything
  if (words.length == 0) {
    return null;
  }

  for (int i = 0; i < words.length; i  ) {
    //when you found an unused word, set the flag and return the word
    if (!words[i].used) {
      words[i].used = true;
      return words[i].word;
    }
  }

  //in case you didn't find anything, return the result of resetWords
  return resetWords();
}


resetWords() {
  //reset the flags
  for (int i = 0; i < words.length; i  ) {
    words[i].used = false;
  }

  //and call nextWord again, and return its result
  return nextWord();
}

標籤：扑 镖 迭代

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