CODEWITHSUNDEEP
javamath
KRB
KRB

Reputation: 5175

How do you find if a number is within a range in Java? Problems with Math.abs(num1-num2) <= inRange

I've seen in another question that the solution to finding if your number is in a range was,

Math.abs(num1-num2) <= inRange

Where inRange is the number you are trying to figure out if it is in range between num2 and num1.

Where this formula breaks for me is when I insert these numbers.

Math.abs(25-(-25)) <= -5

I'm trying to find if -5 is in between -25 and 25. This equation is false even though the answer is true, -5 falls between -25 and 25.

Please clarify for me!

Upvotes: 9

Views: 69836

Answers (6)

user2991413
user2991413

Reputation: 551

kotlin has .. operator

if (center.x in 0..MaxX) {
                   //Do stuff
}
if (center.y !in 0..MaxY) {
                    //Do stuff
}

Upvotes: 0

John
John

Reputation: 1

Old question, however the Math.abs method can be clearer depending what you're working on, and is still worth showing:

int x = 5;
int bounds = 25;
if(Math.abs(x) <= bounds) {
    //run if x is anywhere between -25 and 25
}

Upvotes: 0

Steve J
Steve J

Reputation: 2674

For bonus points, there is a new Range class (used with helper class Ranges) introduced in Guava 10.x:

import com.google.common.collect.Range;
import com.google.common.collect.Ranges;

public class RangeTest {

    Range<Integer> range = Ranges.closed(-25, +25);

    public boolean rangeTest(Integer candidate) {
        return range.contains(candidate);
    }

}


public class TestMain {
    static RangeTest rangeTest = new RangeTest();

    public static void doTest(Integer candidate) {
        System.out.println(candidate + " in -25..+25: "
                + rangeTest.rangeTest(candidate));
    }

    public static void main(String[] args) {
        doTest(-26);
        doTest(-25);
        doTest(-24);
        doTest(-1);
        doTest(-0);
        doTest(+1);
        doTest(+24);
        doTest(+25);
        doTest(+26);
    }

}

Output:

-26 in -25..+25: false
-25 in -25..+25: true
-24 in -25..+25: true
-1 in -25..+25: true
0 in -25..+25: true
1 in -25..+25: true
24 in -25..+25: true
25 in -25..+25: true
26 in -25..+25: false

The Range class supports open and closed ranges, ranges from -INF to +INF, and all sorts of range-related operations like membership, intersection, and span.

Upvotes: 8

Tu Tran
Tu Tran

Reputation: 1977

Below expression will check x is between a and b:

Math.abs(x - a) + Math.abs(b - x) == Math.abs(b - a)

Upvotes: 2

Saeed Amiri
Saeed Amiri

Reputation: 22555

Just do:

bool isInRange = Math.min(num1,num2) <= inRange 
                && Math.max(num1,num2) >= inRange;

Your current approach just checks number ranges. in fact smallest and largest number distance.

Upvotes: 7

Jon Skeet
Jon Skeet

Reputation: 1500665

I don't see any reason to use Math.abs at all. I'd use:

if (lowerBound <= value && value < upperBound)

or

if (lowerBound <= value && value <= upperBound)

if you want the upper bound to be inclusive too.

Indeed, the Math.abs() approach seems entirely broken - I strongly suspect that you misunderstood the question where it was posed as a solution.

Upvotes: 29

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