我在用Java HashMap挣扎。我想将translationList作为带有字符串的数组返回。示例:word : "translated word"

Main类:

public static void main(String[] args) {
    Dictionary dictionary = new Dictionary();
    dictionary.add("apina", "monkey");
    dictionary.add("banaani", "banana");
    dictionary.add("cembalo", "harpsichord");

    ArrayList<String> translations = dictionary.translationList();
    for(String translation: translations) {
        System.out.println(translation);
    }
}


Dictionary类:

private HashMap<String, String> dictionary = new HashMap<String, String>();
public Dictionary(){};
public String translate(String word){
    if(dictionary.containsKey(word)){
        return dictionary.get(word);
    }
    return null;
}
public void add(String word, String translation){
    dictionary.put(word,translation);
}
public int amountOfWords() {
    return dictionary.size();
}
public ArrayList<String> translationList(){
    for ( String key : dictionary.keySet() ) {
        if(translationList().size()<dictionary.size()){
            translationList().add(key+" = "+dictionary.get(key));
        }
    }
    return translationList();
}


Java返回:

Exception in thread "main" java.lang.StackOverflowError
at java.base/java.util.HashMap$KeyIterator.<init>(HashMap.java:1515)
at java.base/java.util.HashMap$KeySet.iterator(HashMap.java:917)
at Dictionary.translationList(Dictionary.java:21)
at Dictionary.translationList(Dictionary.java:22)


感谢帮助 :)

最佳答案

您的translationList方法正在调用自身,从而导致无限递归调用和StackOverflowError

您可能打算编写类似以下内容的内容:

public List<String> translationList(){
    List<String> list = new ArrayList<>();
    for ( String key : dictionary.keySet() ) {
        list.add(key+" = "+dictionary.get(key));
    }
    return list;
}


要么

public List<String> translationList(){
    List<String> list = new ArrayList<>();
    for ( Map.Entry<String,String> entry : dictionary.entrySet() ) {
        list.add(entry.getKey()+" = "+entry.getValue());
    }
    return list;
}


您必须先创建一个ArrayList,然后向其中添加String。除此之外,即使没有递归调用,if (translationList().size()<dictionary.size())条件也没有意义。

10-04 11:07