带有 BigInteger 的 BigInteger.pow()

Question

当 BigInteger 大于 Integer.MAX_VALUE 时，我试图抛出异常。它不允许我为指数情况抛出异常。我不确定当 biginteger 值太大而无法传递到 BigInteger.pow() 方法时如何让它抛出异常。

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

按照其编写方式，如果指数大于 Integer.MAX_VALUE，则应抛出 ArithmeticException("Negative Exponent Error")。当你尝试时会发生什么？

Answer 2

您以错误的顺序弹出堆栈。指数将位于堆栈顶部，而不是尾数下方。在减法、除法和模数中也有同样的问题，在加法和乘法中采用同样的方法也没有什么坏处。在任何情况下都应该是 b = stack.pop();然后 a = stack.pop()。如果将堆栈声明为 Stack stack = new Stack()，则不需要所有这些类型转换。