Java：我一辈子都想不通为什么我会在这里收到 NullPointerException

Question

我正在创建一个基于控制台的 Evil Hangman 程序。不是在游戏开始时选择单词，而是通过选择包含用户输入的字母的单词逐渐缩小单词列表。

这是一个很常见的任务，所以我相信你以前听说过类似的事情。

这是规范，如果您需要更多说明。

Specification

编译所需的文件和客户端：

File

HangmanMain

这个特定类中有问题的方法是 record() 方法，这是迄今为止在这个赋值中最难编程的方法。 我观察到两个主要问题：

a) 有以下 NullPointerException 我无法弄清楚原因。它发生在将单词添加到列表中的行。据我所知，单词和列表都应该存在并且都已初始化。这不是列表不包含单词或列表未正确初始化的问题。它似乎只发生在第二次用户输入之后。

Exception in thread "main" java.lang.NullPointerException
at HangmanManager.record(HangmanManager.java:66)
at HangmanMain.playGame(HangmanMain.java:59)
at HangmanMain.main(HangmanMain.java:39)

b) NullPointerException 可能是在我之前尝试修复另一个问题时引起的。在 NullPointerException 发生之前，我在确定程序选择的单词没有遵循预期输出的原因时遇到了问题。不管我怎么做，程序在遍历结束时选择的单词总是“aa”。即使我故意选择“a”作为猜测，它应该有效地过滤掉“aa”作为选择，它仍然出现。

这是我目前的程序。

import java.util.*;
public class HangmanManager {
private String pattern;
private int length;
private int max;
private SortedSet<Character> guessesMade;
private Set<String> words;
private Set<String> currentWords;
private Map<String, Set<String>> patternMap;

public HangmanManager(List<String> dictionary, int length, int max){
    if (length < 1 || max < 0){
        throw new IllegalArgumentException();
    }
    this.length = length;
    this.max = max;
    words = new TreeSet<String>();
    for (String word : dictionary){
        if (word.length() == length){
            words.add(word);
        }
    }
    currentWords = new TreeSet<String>();
    guessesMade = new TreeSet<Character>();
    patternMap = new TreeMap<String, Set<String>>();
    pattern = "";
    for (int i = 0; i < length; i++){
        pattern += "- ";
    }

}
public Set<String> words(){
    return words;
}
public int guessesLeft(){
    return max - guessesMade.size();
}
public SortedSet<Character> guesses(){
    return guessesMade;
}
public String pattern(){
    if (words.isEmpty()){
        throw new IllegalArgumentException("There are no words.");
    }
    return pattern;
}

public int record(char guess){
    if (guessesLeft() < 1 || words.isEmpty()){
        throw new IllegalStateException();  
    }
    if (!words.isEmpty() && guessesMade.contains(guess)) { 
        throw new IllegalArgumentException();
    }
    guessesMade.add(guess);
    int largestOccurences = 0;
    for (String word: words){
        System.out.println(word);
        System.out.println(words.size());
        if (patternMap.containsKey(pattern)){   
            System.out.println("patternMap contains pattern");
            largestOccurences = generatePattern(word, guess);
            currentWords.add(word);
            currentWords = patternMap.get(pattern);
            patternMap.put(pattern, currentWords);          
        } else {
            currentWords.add(word);
            patternMap.put(pattern, currentWords);
        }
    }
    words = findFamily();
    return largestOccurences;
}
private Set<String> findFamily(){
    int maxSize = 0;
    Map <String, Integer> patternCount = new TreeMap<String, Integer>();
    for (String key : patternMap.keySet()){
        patternCount.put(key, patternMap.get(key).size());
            if (patternMap.get(key).size() > maxSize){
                maxSize = patternMap.get(key).size();
                pattern = key;
            } else if (patternMap.get(key).size() == maxSize){
                if (key.length() >= pattern.length()){
                    pattern = key;
                    maxSize = patternMap.get(key).size();
                }
            }
        }
    System.out.println("Current pattern: " + pattern);
    return patternMap.get(pattern);
}

private int generatePattern(String s, char guess) {
    int count = 0;
    pattern = "";
        for (int i = 0; i < length; i++){
            if (s.charAt(i) == guess){
                pattern += guess + " ";
                count++;
            } else {
                pattern += "- ";
            }
        }
    return count;
}
}

Answer 1

您的问题在于patternMap。

 if (patternMap.containsKey(pattern))
    {   
        System.out.println("patternMap contains pattern");
        largestOccurences = generatePattern(word, guess);
        //=> generatePattern changes the global variable pattern, so it may 
        //not be contained in patternMap anymore

        currentWords.add(word);
        currentWords = patternMap.get(pattern);//since pattern was changed, 
                                               //patternMap return null and  
                                         //you get NullPointerException on 
                                         //the next iteration of the loop

        patternMap.put(pattern, currentWords);          
    }  

当您调用currentWords = patternMap.get(pattern); 时，您需要确保新更改的pattern 存在于patternMap 中。否则，你会得到null。

Answer 2

在分配值之前检查 null.. currentWords = patternMap.get(pattern);

将其替换为 if(patternMap.get(pattern)!=null) currentWords = patternMap.get(pattern);

检查一下...如果您有任何此类分配，请进行空值检查..

祝你好运！