ArrayList在AsyncTask中不起作用

Question

我已經創建了這個在doInBackground中未正確工作的AsyncTask。爲了更好地理解，我指出了一些事情：

1. getFunctionVal(String function, double Val);這個方法返回一個點（例：getFunctionVal("x+3", 2.0) = 5.0）計算的函數的值。
2. funzione和derivata是兩個字符串（它們是方程式）。

問題：我不明白爲什麼soluzioni.size()始終爲0，同樣的事情發生與risultati.size()。當我在UI中使用OnClickListener時，沒有發生這種情況。

public double b = -20; 
    public double a = -20; 

    public class TaskAsincrono extends AsyncTask<Void, Void, Boolean> { 

    private Context mContext; 
    private View view; 
    public List<Double> risultati = new ArrayList<>(); 
    public List<Double> soluzioni = new ArrayList<>(); 
    private ProgressDialog dialog; 

    public TaskAsincrono (Context context, View v){ 
     mContext = context; 
     view = v; 
    } 

    @Override 
    protected void onPreExecute(){ 
     dialog = ProgressDialog.show(view.getContext(), "", "Attendi...", false, true); 
    } 

    @Override 
    protected void onProgressUpdate(Void[] values) { 

    }; 

    @Override 
    protected Boolean doInBackground(Void... params) { 

     int step = 500; 
     double k = (b-a)/step; 

     List<Double> valoriIntervallo = new ArrayList<>(); 
     List<Double> valoriFunzione = new ArrayList<>(); 

     //Generates number for intervals 
     for(int i = 0; i < step; i++) { 
      double ak = a + k; 
      double fak = getFunctionVal(funzione, ak); 
      a += k; 
      valoriIntervallo.add(ak); 
      valoriFunzione.add(fak); 
     } 

     //show the number intervals 
     for(int j = 0; j < valoriIntervallo.size()-1; j++) { 

      if (Math.signum(valoriFunzione.get(j)) != Math.signum(valoriFunzione.get(j+1))) { 
       risultati.add(valoriIntervallo.get(j)); 
       risultati.add(valoriIntervallo.get(j+1)); 
      } 

     } 

      k = 0; 

      for (int i = 0; i < (risultati.size()); i += 2) { 

       k++; 
       double a = risultati.get(i); 
       double b = risultati.get(i + 1); 
       double res = (b + a)/2; 

       for (int j = 1; j < 10; j++) { 
        res = res - (getFunctionVal(funzione, res)/getFunctionVal(derivata, res)); 
       } 

       soluzioni.add(res); 

      } 

     return true; 
    } 

    @Override 
    protected void onPostExecute(final Boolean success) { 

     if(dialog.isShowing()) { dialog.dismiss(); } 

     TableLayout tl = (TableLayout) view.findViewById(R.id.tableLayoutTangenti); 

     for (int o = 0; o < 1; o++) { 

      TableRow tr = new TableRow(mContext); 
      tr.setLayoutParams(new TableRow.LayoutParams(TableRow.LayoutParams.MATCH_PARENT, TableRow.LayoutParams.WRAP_CONTENT)); 

      TextView coefficente = new TextView(mContext); 
      coefficente.setText("x" + String.valueOf((o+1)) + " = "); 
      coefficente.setGravity(Gravity.CENTER); 
      coefficente.setLayoutParams(new TableRow.LayoutParams(0, TableRow.LayoutParams.WRAP_CONTENT, 0.2f)); 

      EditText decimale = new EditText(mContext); 
      decimale.setGravity(Gravity.CENTER); 
      decimale.setText(String.valueOf(risultati.size())); 
      decimale.setFocusable(false); 
      decimale.setFocusableInTouchMode(false); 
      decimale.setClickable(false); 
      decimale.setLayoutParams(new TableRow.LayoutParams(0, TableRow.LayoutParams.WRAP_CONTENT, 0.8f)); 

      tr.addView(coefficente); 
      tr.addView(decimale); 
      tl.addView(tr, new TableLayout.LayoutParams(TableLayout.LayoutParams.MATCH_PARENT, TableLayout.LayoutParams.WRAP_CONTENT)); 
     } 
    } 

    @Override 
    protected void onCancelled() { 
     mAuthTask = null; 
    } 
} 

如果你問爲什麼我使用的AsyncTask，那是因爲我需要有step = 50000我做不到「壓力」的UI這麼多。

Answer 1

k是0. funzione是恆定的。假設getFunctionVal是一個純函數，valoriFunzione包含500次相同的值。

這意味着Math.signum(valoriFunzione.get(j)) != Math.signum(valoriFunzione.get(j+1))總是錯誤的。

這就是爲什麼risultati什麼都沒有。

Answer 2

檢查條件if（Math.signum（valoriFunzione.get（j））！= Math.signum（valoriFunzione.get（j + 1）））的計算結果爲true。