# Comparing integers using Integer.compare vs subtraction

04 August 2015

TL;DR: Use Integer.compare to avoid overflow that can occur when subtracting two integers.

I took Oracle’s Java 8 MOOC: Lamdbas And Streams class. I had a question about using Integer.compare that arose from one of the homework questions. I am recording the question and answer here so I can find it easily in the future, instead of having it buried in a thread about a homework assignment using the Streams API. The original post is on the Oracle community forums.

My question was:

My answers were pretty similar to what has already been posted. However, for the sort in question 7 I did:

whereas most implementations in this thread have done something like:

Is there any advantage to using subtraction vs using the Integer.compare method? Or is just different ways of doing the same thing?

The reply from Stuart Marks-Oracle was:

For this particular problem, there is essentially no difference. However, there are cases when comparing integers, where writing a comparator that simply subtracts the two values can fail to produce the right answer. Consider the following:

This is clearly wrong. Can you see why? Overflow, that’s why! When comparing numbers of extreme magnitude, such as MAX_VALUE and -1, these are subtracted, giving a result with the wrong sign. That’s why the sorting fails in the case. The correct way to compare integers is, as you note, using Integer.compare():

There’s nothing magical about Integer.compare(). It’s basically a conditional expression that uses comparison operations instead of subtraction:

In the exercise, string lengths are always non-negative, so subtracting them cannot result in overflow. However, I always recommend using Integer.compare(), so that you don’t have to prove that overflow cannot occur.