爲什麼此列表在Android Studio中不斷返回NullPointerException？

Question

因此，基本上這段代碼應該做的是拼寫檢查一個字作爲EditText字段的輸入，並在TextFields中顯示建議。

getAll()是調用所有需要執行的其他操作的函數。

在執行完所有功能後，Public member variable "list"包含所有建議。

最後一個功能sort()增加了Listlist中的所有建議。在onClick()中，我們試圖將它們打印到textfield。但是

t2.setText(list.get(1).toString()); 

返回NullPointerException錯誤。

在eclipse上獨立運行時，拼寫檢查代碼工作得很好。

這裏的錯誤日誌：

E/AndroidRuntime: FATAL EXCEPTION: main 
        Process: com.example.daipayan.myapplication, PID: 2667 
        java.lang.NullPointerException: Attempt to invoke interface method 'java.lang.Object java.util.List.get(int)' on a null object reference 
         at com.example.daipayan.myapplication.MainActivity$1.onClick(MainActivity.java:69) 
         at android.view.View.performClick(View.java:5637) 
         at android.view.View$PerformClick.run(View.java:22429) 
         at android.os.Handler.handleCallback(Handler.java:751) 
         at android.os.Handler.dispatchMessage(Handler.java:95) 
         at android.os.Looper.loop(Looper.java:154) 
         at android.app.ActivityThread.main(ActivityThread.java:6119) 
         at java.lang.reflect.Method.invoke(Native Method) 
         at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:886) 
         at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:776) 

，代碼：

public class MainActivity extends AppCompatActivity { 
     public static char keyboard[][] = {{'0','0','0','0','0','0','0','0','0','0','0','0'}, 
       {'0','Q','W','E','R','T','Y','U','I','O','P','0'}, //mapping the keyboard to a multidimensional array 
       {'0','0','A','S','D','F','G','H','J','K','L','0'}, 
       {'0','0','0','Z','X','C','V','B','N','M','0','0'}, 
       {'0','0','0','0','0','0','0','0','0','0','0','0'}}; 


public static String possibleArray[]; 
public static ArrayList<String> comb; 
public static ArrayList<String> filtered; 
public static Set<String> dictionary; 
public static HashMap<String , Integer> distances; 
public static List list; 
public static String incorrect_word; 


EditText input; 
TextView t1,t2,t3; 
Button bt1; 
@Override 
protected void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 

    input=(EditText)findViewById(R.id.editText); 
    bt1 = (Button)findViewById(R.id.button); 
    bt1.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View v) { 
      String inputText = input.getText().toString(); 

      String words[] = inputText.split("\\s"); 

      String lastWord = words[words.length-1]; 

      incorrect_word = "HELLP"; 
      try { 
       getAll(); 
      } catch (IOException e) { 
       e.printStackTrace(); 
      } 

      t1 = (TextView)findViewById(R.id.text1); 
      t2 = (TextView)findViewById(R.id.text2); 
      t3 = (TextView)findViewById(R.id.text3); 
      t1.setText(""+list.get(0)); 
      t2.setText(list.get(1).toString()); 
      t3.setText(list.get(2).toString()); 
     } 
    }); 


} 

public static void getAll() throws IOException { 
    loadDictionary(); 
    Log.d("getall","called"); 
    //System.out.println(checkWord(incorrect_word)); 

    // 
    //fetching the location of characters on the keyboard 
    // 
    int index[][] = new int[incorrect_word.length()][2]; 
    for (int i = 0 ; i < incorrect_word.length() ; i++){ 
     int xy[] = locate(incorrect_word.charAt(i)); 
     index[i][0]=xy[0]; 
     index[i][1]=xy[1]; 
    } 

    //print(index, incorrect_word.length(), 2); 

    // 
    //Fetching all the possible words with the 
    // 
    ArrayList<String> possible = characters(incorrect_word, index); 
    { 
     possibleArray = possible.toArray(new String[0]); 
     comb = new ArrayList<>(); 
     dfs("",0); 
    } 

    filtered = new ArrayList<String>(); 

    filterCombinations(); //filter out the words that aren't in the dictionary 

    Levenshtein(incorrect_word);//Calculate distances between words and the incorrect word 

    printHash();//Print the hash map consisting of the words and their respective distances 

    sort(); 
} 

public static int[] locate(char a){ 
    int i=1,j=0; 
    for(; i < 4 ; i++ ){ 
     for(j=1 ; j < 11; j++){ 
      if(keyboard[i][j]==a) 
       return new int[] {i,j}; 
     } 
    } 
    return new int[] {i,j}; 
} 

public static ArrayList<String> characters(String incorrect_word, int index[][]){ 
    ArrayList<String> possible = new ArrayList<String>(); 
    for (int i = 0 ; i < incorrect_word.length() ; i++){ 
     String character=""+incorrect_word.charAt(i); 
     int x=index[i][0] , y=index[i][1]; 


     String a = (keyboard[x-1][y] != '0' ? character+=keyboard[x-1][y]: "0");//Upper Right 
     a = (keyboard[x-1][y-1] != '0' ? character+=keyboard[x-1][y-1]: "0");//Upper Left 
     a = (keyboard[x][y-1] != '0' ? character+=keyboard[x][y-1]: "0");//left 
     a = (keyboard[x][y+1] != '0' ? character+=keyboard[x][y+1]: "0");//right 
     a = (keyboard[x+1][y] != '0' ? character+=keyboard[x+1][y]: "0");//Lower Left 
     a = (keyboard[x+1][y+1] != '0' ? character+=keyboard[x+1][y+1]: "0");//Lower Right 



     possible.add(character); 
    } 
    return possible; 


} 

public static void dfs(String x,int i) { 
    if(i == possibleArray.length) { // there is no more string that can be generated 
     if(checkWord(x)) 
      comb.add(x); // save the found string 
     return; 
    } 
    for(int j=0;j<possibleArray[i].length();j++) // for each character in the current string 
     dfs(x+possibleArray[i].charAt(j),i+1); // take the current character and move to the next string 
} 


public static void loadDictionary() throws IOException{ 
    BufferedReader dictReader = new BufferedReader(new FileReader("dictionary.txt")); 
    dictionary = new HashSet<>(); 
    while (dictReader.ready()) 
    { 
     String dictInput = dictReader.readLine() ; 
     String [] dict = dictInput.split("\\s"); 

     for(int i = 0; i < dict.length;i++) 
     { 
      // key and value are identical 
      dictionary.add(dict[i]); 
     } 
    } 
    dictReader.close(); 
} 

public static Boolean checkWord(String str){ 
    String toCheck = str.toLowerCase(); 
    if(dictionary.contains(toCheck)){ 
     return true; 
    } 
    else 
     return false; 
} 

public static void filterCombinations(){  for(String x:comb) 
     if (checkWord(x)) 
      filtered.add(x); 

} 
public static void printHash(){ 
    Iterator iterator = distances.keySet().iterator(); 

    while (iterator.hasNext()) { 
     String key = iterator.next().toString(); 
     String value = distances.get(key).toString(); 

     System.out.println(key + " " + value); 
    } 
} 
public static void Levenshtein(String incorrect_word){ 
    distances = new HashMap<String, Integer>(); 
    for(String x:filtered) 
     distances.put(x,distance(incorrect_word , x)); 

} 
public static int distance(String a, String b) { 
    a = a.toLowerCase(); 
    b = b.toLowerCase(); 
    // i == 0 
    int [] costs = new int [b.length() + 1]; 
    for (int j = 0; j < costs.length; j++) 
     costs[j] = j; 
    for (int i = 1; i <= a.length(); i++) { 
     // j == 0; nw = lev(i - 1, j) 
     costs[0] = i; 
     int nw = i - 1; 
     for (int j = 1; j <= b.length(); j++) { 
      int cj = Math.min(1 + Math.min(costs[j], costs[j - 1]), a.charAt(i - 1) == b.charAt(j - 1) ? nw : nw + 1); 
      nw = costs[j]; 
      costs[j] = cj; 
     } 
    } 
    return costs[b.length()]; 
} 

public static void sort(){ 

    list=new ArrayList(distances.entrySet()); 
    Collections.sort(list,new Comparator(){ 
     public int compare(Object obj1, Object obj2){ 
      return ((Comparable)((Entry)(obj1)).getValue()).compareTo(((Entry)(obj2)).getValue()); 
     } 
    }); 
    if(list.size() > 3) 
     list.subList(4, list.size()).clear(); 
    Log.e("list","the values " + list.get(0) + list.get(1)+list.get(2)); 
} 

} 

Answer 1

根據你有兩個元素，你得到第三個元素，因此拋出空指針異常的t3.setText(list.get(2).toString());

名單