Question

当BigInteger大于Integer.MAX_VALUE时，我想要抛出异常。它不允许我为exponent案例抛出该异常。当biginteger值太大而无法传递给BigInteger.pow（）方法时，我不确定如何让它抛出异常。

提前致谢。

这是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

如果指数大于ArithmeticException("Negative Exponent Error")，它的编写方式应该抛出Integer.MAX_VALUE。你尝试的时候会发生什么？

Answer 2

你以错误的顺序弹出堆栈。指数将位于堆栈的顶部，而不是在尾数下。你在减法，除法和模数方面遇到同样的问题，并且以相同的方式进行加法和乘法也不会有什么坏处。在每种情况下，它应该是b = stack.pop（）;然后a = stack.pop（）。如果您将堆栈声明为Stack stack = new Stack（），则不需要所有这些类型转换。

带有BigInteger的BigInteger.pow（）

2 个答案: