BigInteger.pow（）與BigInteger

Question

當BigInteger大於Integer.MAX_VALUE時，我正在綁定引發異常。它不會允許我爲指數情況拋出異常。我不確定如何在biginteger值太大而無法傳入BigInteger.pow（）方法時拋出異常。

在此先感謝。

這裏是toPostfix方法：

public BigInteger evalPostfix(String postfix){ 
    BigInteger a, b; 
    Stack stack = new Stack(); 

     for(int i=0; i<postfix.length(); i++){ 
      if(this.isOp(postfix.charAt(0))) 
       throw new ArithmeticException("Malformed Postfix Expression"); 
      switch(postfix.charAt(i)){ 
       case '+': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        stack.push(b.add(a)); 
        break; 
       case '-': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        stack.push(b.subtract(a)); 
        break; 
       case '*': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        stack.push(b.multiply(a)); 
        break; 
       case '/': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        if(a == BigInteger.valueOf(0)){ 
         throw new ArithmeticException("Cannot divide by 0"); 
        }else{ 
         stack.push(b.divide(a)); 
        } 
        break; 
       case '%': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        stack.push(b.mod(a)); 
        break; 
       case '^': 
        a = (BigInteger)stack.pop(); 
        b = (BigInteger)stack.pop(); 
        if(b.compareTo(BigInteger.valueOf(Integer.MAX_VALUE)) > 0) 
         throw new ArithmeticException("BigInteger value is too large"); 
        stack.push(a.pow(b.intValue())); 
        break; 
       default: 
        if(this.numbers.get(postfix.substring(i, i+1)) == null) 
         throw new NullPointerException(postfix.substring(i, i+1) + " is not mapped to any value"); 
        stack.push(this.numbers.get(postfix.substring(i,i+1))); 
      } 
     } 

    return (BigInteger)stack.pop(); 
} 

Answer 1

它是書面的方式，它應該拋出ArithmeticException("Negative Exponent Error")如果指數大於Integer.MAX_VALUE更大。當你嘗試時會發生什麼？

Answer 2

您按照錯誤的順序彈出堆棧。指數將位於堆棧的頂端，而不是尾數。你在減法，除法和模數方面有同樣的問題，也不會因爲加法和乘法的方式做同樣的事情。在任何情況下，它應該是b = stack.pop（）;那麼a = stack.pop（）。如果你將堆棧聲明爲Stack stack = new Stack（），那麼你不需要所有這些類型轉換。