How to check if an integer is in a given range?

Question

Hoping for something more elegant than

if (i>0 && i<100) 

Answer 1

You could add spacing ;)

if (i > 0 && i < 100) 

Answer 2

For those using commons lang an option is to use Range:

Range<Integer> myRange = Range.between(100, 500);
if (myRange.contains(200)){
    // do something
}

Also see: how to construct a apache commons 3.1 Range<Integer> object

Answer 3

I think

if (0 < i && i < 100) 

is more elegant. Looks like maths equation.

If you are looking for something special you can try:

Math.max(0, i) == Math.min(i, 100)

at least it uses library.

Answer 4

ValueRange range = java.time.temporal.ValueRange.of(minValue, maxValue);
range.isValidIntValue(x);

• it returns true if minValue <= x <= MaxValue - i.e. within the range
• it returns false if x < minValue or x > maxValue - i.e. out of range

Use with if condition as shown below:

int value = 10;
if (ValueRange.of(0, 100).isValidIntValue(value)) {
    System.out.println("Value is with in the Range.");
} else {
    System.out.println("Value is out of the Range.");
}

The below program checks, if any of the passed integer value in the hasTeen method is within the range of 13 (inclusive) to 19 (inclusive).

---

import java.time.temporal.ValueRange;
  
public class TeenNumberChecker {

    public static void main(String[] args) {
        System.out.println(hasTeen(9, 99, 19));
        System.out.println(hasTeen(23, 15, 42));
        System.out.println(hasTeen(22, 23, 34));
    }

    public static boolean hasTeen(int firstNumber, int secondNumber, int thirdNumber) {

        ValueRange range = ValueRange.of(13, 19);
        System.out.println("*********Int validation Start ***********");
        System.out.println(range.isIntValue());
        System.out.println(range.isValidIntValue(firstNumber));
        System.out.println(range.isValidIntValue(secondNumber));
        System.out.println(range.isValidIntValue(thirdNumber));
        System.out.println(range.isValidValue(thirdNumber));
        System.out.println("**********Int validation End**************");

        if (range.isValidIntValue(firstNumber) || range.isValidIntValue(secondNumber) || range.isValidIntValue(thirdNumber)) {
            return true;
        } else
            return false;
    }
}

OUTPUT

true as 19 is part of range

true as 15 is part of range

false as all three value passed out of range

Answer 5

I think its already elegant way for comparing range. But, this approach cause you to write extra unit tests to satisfy all && cases.

So, you can use any of the below approach to avoid writing extra unit tests.

Using Java 8 Streams:

if(IntStream.rangeClosed(0,100).boxed().collect(Collectors.toList()).contains(i))

Using Math class:

if(Math.max(0, i) == Math.min(i, 100))

Personally I recommend the second approach because it won't end up creating an Array of the size equal to the range you want to check.

Answer 6

I don't see how that's not elegant, but if you repeat the expression often, then it's a good idea to put it into a method, e.g.

class MathUtil
{
   public static boolean betweenExclusive(int x, int min, int max)
   {
       return x>min && x<max;    
   }
}

This is particularly true if you mix exclusive and inclusive comparisons. The method name can help avoid typos, such as using < when <= should have been used. The method can also take care of ensuring that min < max etc..

Answer 7

If you're looking for something more original than

if (i > 0 && i < 100)

you can try this

import static java.lang.Integer.compare;
...
if(compare(i, 0) > compare(i, 100))

Answer 8

This guy made a nice Range class.

Its use however will not yield nice code as it's a generic class. You'd have to type something like:

if (new Range<Integer>(0, 100).contains(i))

or (somewhat better if you implement first):

class IntRange extends Range<Integer>
....
if (new IntRange(0,100).contains(i))

Semantically both are IMHO nicer than what Java offers by default, but the memory overhead, performance degradation and more typing overall are hadly worth it. Personally, I like mdma's approach better.

Answer 9

Google's Java Library Guava also implements Range:

import com.google.common.collect.Range;

Range<Integer> open = Range.open(1, 5);
System.out.println(open.contains(1)); // false
System.out.println(open.contains(3)); // true
System.out.println(open.contains(5)); // false

Range<Integer> closed = Range.closed(1, 5);
System.out.println(closed.contains(1)); // true
System.out.println(closed.contains(3)); // true
System.out.println(closed.contains(5)); // true

Range<Integer> openClosed = Range.openClosed(1, 5);
System.out.println(openClosed.contains(1)); // false
System.out.println(openClosed.contains(3)); // true
System.out.println(openClosed.contains(5)); // true

Answer 10

if ( 0 < i && i < 100)  

if ( 'a' <= c && c <= 'z' )

Answer 11

Use this code :

if (lowerBound <= val && val < upperBound)
or
if (lowerBound <= val && val <= upperBound)

Answer 12

if you are using Spring data you can also use the Range object from Spring.

range = new org.springframework.data.domain.Range(3, 8);
range.contains(5) // will return true.

Answer 13

That's how you check is an integer is in a range. Greater than the lower bound, less than the upper bound. Trying to be clever with subtraction will likely not do what you want.

Answer 14

Just my 2 cts

    // Exclusive    
    public boolean isOrdered(int n1, int n2, int n3) {
        return n1 < n2 && n2 < n3 ;
    }

call it like to see it is one of 1 to 99

    if (isOrdered(0,i,100)) 

Answer 15

Try:

if (i>0 && i<100) {} 

it will work at least ;)

Answer 16

if (i in 0..100) {}

please try this for efficient code ;)

Answer 17

If you just want to test whether the value of i is in the range [0..100), and you want the boolean result, then

i >= 0 && i < 100

is fine and will give you true or false. If you are willing to throw an exception if the value is out of range, you can use the built-in checkIndex method

Objects.checkIndex(i, 100)

which will throw IndexOutOfBoundsException if out of range. Granted, it is not a general solution, and is really only prettier if your context is one in which your range check is being done for checking array bounds, but it has the advantage of not needed any third party libraries, which is nice for small programs.

Answer 18

Try this if strategy. Seems solid.

int num = 5;
    
    //random values; can be changed
    int a = 3;
    int b = 7;
    
    if (num >= 3 && num <= 7) {
        //Your Code Here
    }

Answer 19

Wrote this program to check the value of 3 integer parameters against a range.

I have created a separate method to check the range of the each value using if/else-if statements.

I hope this helps someone.

public class TeenNumberChecker {

    public static void main(String[] args) {
        
        boolean isTeen = hasTeen();

    }

    public static boolean hasTeen(int valueOne, int valueTwo, int valueThree) {
        
        
        if (valueOne > 13 && valueOne < 19) {
            
            return true;
            
        } else if (valueTwo > 13 && valueTwo < 19) {
            
            return true;
            
        } else if (valueThree > 13 && valueThree < 19) {
            
            return true;
            
        }else {
            
            return false;
            
        }

    }
}

Answer 20

if(i <= 0 || i >=100)

It will work.

Answer 21

If you are confident that numbers are stored in 2's complement form:

return  ((x-low) <= (high-low));

A more general and safer solution would be:

return ((x-high)*(x-low) <= 0);

Answer 22

 Range<Long> timeRange = Range.create(model.getFrom(), model.getTo());
    if(timeRange.contains(systemtime)){
        Toast.makeText(context, "green!!", Toast.LENGTH_SHORT).show();
    }