Check if a variable is between two numbers with Java

Question

I have a problem with this code:

if (90 >>= angle =<< 180)

The error explanation is:

The left-hand side

Answer 1

I see some errors in your code.
Your probably meant the mathematical term

90 <= angle <= 180, meaning angle in range 90-180.

if (angle >= 90 && angle <= 180) {

// do action
}

Answer 2

//If "x" is between "a" and "b";

.....

int m = (a+b)/2;

if(Math.abs(x-m) <= (Math.abs(a-m)))

{
(operations)
}

......

//have to use floating point conversions if the summ is not even;

Simple example :

//if x is between 10 and 20

if(Math.abs(x-15)<=5)

Answer 3

Assuming you are programming in Java, this works:

if (90 >= angle  &&  angle <= 180 )  {

(don't you mean 90 is less than angle? If so: 90 <= angle)

Answer 4

<<= is like +=, but for a left shift. x <<= 1 means x = x << 1. That's why 90 >>= angle doesn't parse. And, like others have said, Java doesn't have an elegant syntax for checking if a number is an an interval, so you have to do it the long way. It also can't do if (x == 0 || 1), and you're stuck writing it out the long way.

Answer 5

public static boolean between(int i, int minValueInclusive, int maxValueInclusive) {
    if (i >= minValueInclusive && i <= maxValueInclusive)
        return true;
    else
        return false;
}

https://alvinalexander.com/java/java-method-integer-is-between-a-range

Answer 6

are you writing java code for android? in that case you should write maybe

if (90 >= angle && angle <= 180) {

updating the code to a nicer style (like some suggested) you would get:

if (angle <= 90 && angle <= 180) {

now you see that the second check is unnecessary or maybe you mixed up < and > signs in the first check and wanted actually to have

if (angle >= 90 && angle <= 180) {