How can I represent a range in Java?

Question

Let's say an integer should be within the range: [0...2147483647]

I want to check whether an integer variable falls within this range. I know it can be accomplished by a simple if-else statement, but is there a more efficient way to check whether it's within the range?

I'd rather not do this:

if (foo >= 0 && foo <= 2147483647) 
{
    // do something
}

Answer 1

Apache Commons Lang has a Range class for doing arbitrary ranges.

Range<Integer> test = Range.between(1, 3);
System.out.println(test.contains(2));
System.out.println(test.contains(4));

Guava Range has similar API.

If you are just wanting to check if a number fits into a long value or an int value, you could try using it through BigDecimal. There are methods for longValueExact and intValueExact that throw exceptions if the value is too big for those precisions.

Answer 2

You could create a class to represent this

public class Range
{
    private int low;
    private int high;

    public Range(int low, int high){
        this.low = low;
        this.high = high;
    }

    public boolean contains(int number){
        return (number >= low && number <= high);
    }
}

Sample usage:

Range range = new Range(0, 2147483647);

if (range.contains(foo)) {
    //do something
}

Answer 3

I know this is quite an old question, but with Java 8's Streams you can get a range of ints like this:

// gives an IntStream of integers from 0 through Integer.MAX_VALUE
IntStream.rangeClosed(0, Integer.MAX_VALUE); 

Then you can do something like this:

if (IntStream.rangeClosed(0, Integer.MAX_VALUE).matchAny(n -> n == A)) {
    // do something
} else {
    // do something else 
}

Answer 4

You could use java.time.temporal.ValueRange which accepts long and would also work with int:

int a = 2147;

//Use java 8 java.time.temporal.ValueRange. The range defined
//is inclusive of both min and max 
ValueRange range = ValueRange.of(0, 2147483647);

if(range.isValidValue(a)) {
    System.out.println("in range");
}else {
    System.out.println("not in range");
}

Answer 5

If you are checking against a lot of intervals, I suggest using an interval tree.