How to compare an Integer with a long?

Question

I am trying to check if a long integer is greater than the INTEGER.MAX value but it's not working. It's very straight forward so I am just wondering if there is some problem with comparing an Integer object with a long as I have done because otherwise I don't know what the problem is. At the moment when the nextTotal value exceeds INTEGER.MAX, it kicks into negative numbers instead of printing the error message.

public Integer initialValue=0;

int amount = Integer.parseInt(amountStr);
        System.out.println("Received from client: " + amount);

        long nextTotal=amount+initialValue;
            if((nextTotal>Integer.MAX_VALUE)||    (nextTotal<Integer.MIN_VALUE)){
                System.out.println("Error: Integer has not been added as the total exceeds the Maximum Integer value!");
                out.flush();
            }  

            else{
                initialValue+=amount;
                out.println(initialValue); //server response
                System.out.println("Sending total sum of integers to the client:"+initialValue);
                out.flush();
                }
            }

rgettman · Accepted Answer · 2014-03-24 18:42:30Z

2

The problem is that you have added two ints, but they haven't been promoted to long yet, so it overflows before being converted to a long, and of course an int can never be greater than Integer.MAX_VALUE. It will only get converted to a long upon assignment, which is after the addition.

Convert to a long before the addition, with a cast.

long nextTotal = (long) amount + initialValue;

answered Mar 24, 2014 at 18:42

rgettman

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1 Comment

kellzer Over a year ago

Thanks for the help-I figured it was something simple!

erickson · Accepted Answer · 2014-03-24 18:52:16Z

0

Since you don't really want to use a long value, I would normally do the check like this:

int amount = Integer.parseInt(amountStr);
if (amount > (Integer.MAX_VALUE - initialValue)) 
  throw new IllegalArgumentException("Amount exceeds maximum value");
initialValue += amount;

In other words, check whether the amount will overflow before adding, and throw an exception (or send an error message or whatever is appropriate for your application) instead of proceeding normally.

answered Mar 24, 2014 at 18:52

erickson

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