两个长值之和返回负值

Question

我已经编写了完成后缀评估的代码，但当我尝试处理以下情况时，我继续得到一个负值，我知道这是不正确的。为什么我得到这个负数，如何得到正确的输出18000000000000000000？我已经在下面发布了我的代码，任何帮助将不胜感激。

     public static Number postfixEvaluate(String e){
    Long number1;
    Long number2;
    Number result = new Long(0);

    Stack<Number> stack = new Stack();

    String[] tokens = e.split(" ");

        for(int j = 0; j < tokens.length; j++){
            String token = tokens[j];
            //System.out.println(tokens[j]);
        if (!"+".equals(token) && !"*".equals(token) && !"-".equals(token) && !"/".equals(token) && !"".equals(token))  {
            stack.push(Long.parseLong(token)); 

    }  else if ("".equals(token)) {
        System.out.println(token);
    } else {
            String Operator = tokens[j];
            number2 = (Long) stack.pop();
            System.out.println(number2);
            number1 = (Long) stack.pop();
            System.out.println(number1);
            if (Operator.equals("/")){
                result = number1 / number2;
                System.out.println(result);
            }
            else if(Operator.equals("*")){
                result = number1 * number2;
                System.out.println(result);
            }
            else if(Operator.equals("+")){
                result = Long.sum(number1, number2);
                System.out.println("Addition of: " + number1 + "+ " + number2 + "= " + result);
            }
            else if(Operator.equals("-")){
                result = number1 - number2;
            System.out.println(result);
            }
            else System.out.println("Illeagal symbol");
            stack.push(result);
        }
        }

                stack.pop();


    //s.pop();
    System.out.println("Postfix Evauation = " + result);

        return result;
   }

我的输入和输出：

   Input: 9000000000000000123 9000000000000000987 +
   Expected Output: 18000000000000000000
   Current Output: -446744073709550506

Answer 1

您得到负值，因为您超过了long可以容纳的最大值。当您超过最大值时，它将再次从最小值开始，直至超过最大值的数字。当您尝试为long变量分配一个小于最小值的数字时，也是如此。要了解这一点，您可以查看以下程序的输出：

public class Main {
    public static void main(String[] args) {
        System.out.println("Long.MAX_VALUE: "+Long.MAX_VALUE);
        System.out.println("Long.MIN_VALUE: "+Long.MIN_VALUE);
        long x = Long.MAX_VALUE + 1;
        long y = Long.MIN_VALUE - 1;
        System.out.println("Long.MAX_VALUE + 1: "+x);//will be assigned the value of Long.MIN_VALUE
        System.out.println("Long.MIN_VALUE - 1: "+y);//will be assigned the value of Long.MAX_VALUE

        x = Long.MAX_VALUE + 2;
        y = Long.MIN_VALUE - 2;
        System.out.println("Long.MAX_VALUE + 2: "+x);//will be assigned the value of Long.MIN_VALUE + 1
        System.out.println("Long.MIN_VALUE - 2: "+y);//will be assigned the value of Long.MAX_VALUE - 1

    }
}

输出：

9223372036854775807
-9223372036854775808
-9223372036854775808
9223372036854775807

根据您的要求，您需要BigInteger例如

BigInteger bi = new BigInteger("18000000000000000000");