'Absolute value of a number in Clojure

How can the absolute number of a value be calculated in Clojure?

(abs  1) => 1
(abs -1) => 1
(abs  0) => 0
clojure


Solution 1:[1]

As of Clojure 1.11, it is simply (abs -1) in both Clojure and Clojurescript.

In versions before, for double, float, long and int you can use the java.lang.Math method abs (Math/abs -1)

Take care it won't work for decimals, ratio's, bigint(eger)s and other Clojure numeric types. The official clojure contrib math library that tries guarantee working correctly with all of these is clojure.math.numeric-tower

Solution 2:[2]

you could always do 

(defn abs [n] (max n (- n)))

Solution 3:[3]

The deprecated clojure.contrib.math provides an abs function.

The source is:

(defn abs "(abs n) is the absolute value of n" [n]
  (cond
   (not (number? n)) (throw (IllegalArgumentException.
                             "abs requires a number"))
   (neg? n) (- n)
   :else n))

As @NielsK points out in the comments, clojure.math.numeric-tower is the successor project.

Solution 4:[4]

abs is now available in the core clojure library since Clojure 1.11 release.

Docs for abs https://clojure.github.io/clojure/branch-master/clojure.core-api.html#clojure.core/abs

Usage: (abs a)

Returns the absolute value of a.
If a is Long/MIN_VALUE => Long/MIN_VALUE
If a is a double and zero => +0.0
If a is a double and ##Inf or ##-Inf => ##Inf
If a is a double and ##NaN => ##NaN

Added in Clojure version 1.11

Sources

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Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Shlomi
Solution 3
Solution 4 oxalorg

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