Java chained inequality if (5

Question

Is there any operator or trick for such operation? Or is it necessary to use

if(5<i && i<10)

?

What is the meaning of `i<10` condition ? – Bhavik Ambani May 18 '12 at 19:04 — Bhavik Ambani, May 18 '12 at 19:04

score 24 · Answer 1 · answered May 18 '12 at 19:03

24

You cannot chain inequalities. You can, however, define a static boolean method isInRange(value, low, high) that will perform such a check.

Some other languages, like Python or Icon, allow this notation.

answered May 18 '12 at 19:03

Gyscos

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Perhaps the parameters should be ordered as `isInRange(low, value, high)` e.g. `isInRange(5, i, 10)`. – Salman A May 18 '12 at 19:07

score 13 · Accepted Answer · answered May 18 '12 at 19:03

13

There is no such thing in Java (unless you work with booleans).

The 5<iresults in a boolean on the left side on <10 which the compiler dislikes.

answered May 18 '12 at 19:03

Thorbjørn Ravn Andersen

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score 5 · Answer 3 · edited May 23 '17 at 12:24

5

I'm afraid chained inequalities are unsupported in Java at this time. Check this post for languages which do.

edited May 23 '17 at 12:24

Community

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answered May 18 '12 at 19:04

danielmhanover

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score 5 · Answer 4 · answered May 18 '12 at 19:07

5

You can use a single comparison but is more complicated than its worth usually.

if (i - (Integer.MIN_VALUE + 6) < Integer.MIN_VALUE + (10 - 6))

This uses underflow to adjust all the value 5 and below to be a large positive value.

The only reason you would use this is as a micro-optimisation.

answered May 18 '12 at 19:07

Peter Lawrey

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You wrote `6` twice, probably something OP wants to avoid. – Salman A May 18 '12 at 19:09

jihor · Answer 5 · 2018-01-31T18:53:14.400

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Range

Comparisons cannot be chained in Java.

The standard way for doing a range check is to use either Guava Ranges:

import com.google.common.collect.Range;
...
    if(Range.open(5, 10).contains(x)){

or Apache Commons Lang (it provides only closed ranges, though):

import org.apache.commons.lang3.Range;
...
    if(Range.between(5 + 1, 10 - 1).contains(x)){

edited Jan 31 '18 at 18:53

answered Jan 28 '18 at 17:40

jihor

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score 1 · Answer 6 · answered Jul 11 '14 at 14:05

a < x < b is equivalent to (x-a)(x-b)<0, or if x is an expression you only wanna calculate once,

(x - (a+b)/2) ^ 2 < (b-a)^2 / 4

Same deal with two nonstrict equalities. I don't think there's a way to deal with one strict and one nonstrict equality, however (unless x is an integer).

Java chained inequality if (5

6 Answers6

Range

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