Find the absolute value of a number without built-in functions [closed]

Question

The challenge is to take any real number as input and output or return the absolute value. You may not use any built-in functions other than those dealing with input and output/returning. This is code golf, so the shortest code wins. This is my first question here, so bear with me if I've left something obvious out of the challenge.

As per Quincunx's suggestion, I am limiting input to anywhere between -9E99 and 9E99.

Also, the only functions/operators you can use are input, output, return, +, -, *, /, %, ^ +=, -=, *=, /=, >, <, ==, >=, <=, !=, square, and square root or their equivalents

Answer 1

J, 2 bytes

**

Usage:

   f =: **
   f 9.3
9.3
   f -9.3
9.3

Explanation:

This uses the * verb in both its monadic and dyadic forms. The monadic form returns -1 if it's given a negative number, 0 if it's given 0 and 1 if it's given a positive number. The dyadic form is just plain old multiplication. Putting them in a function literal turns them into a hook which gets evaluated like this: y * (*y).

Answer 2

GolfScript, 4 characters

I believe I win. ;)

Works with integer or decimal numbers of any length.

'-'/

Try it online

These programs do the same and are of equal length:

'-'-

'-'%

GolfScript (old version), 16 13

My first GS program! (that actually does something)

Doesn't work with decimal numbers because GolfScript doesn't have floating point.

~:$0<{0$-}$if

Answer 3

Perl: 5 characters

s/-//

The example:

perl -e '$_=-82.923; s/-//; print' # will print 82.923 or return it unless use 'print'

Answer 4

C - 21

#define x(a) a<0?-a:a

This is a preprocessor macro, but does the same thing as Quincunx's Java solution when a real number is used as input.

Answer 5

J - 7 3

Max of number and inverse (3)

>.-

When assigned to a function: take the maximum between the negative and the number, using a hook so: (f g) y = y f g y

f=:>.-
f _4 5 _1 0
   4 5 1 0

root of the square (5 or 4)

]&.*: _4 5 _1 0
    4 5 1 0
NB. or if an expression is good enough:
%:*: _4 5 _1 0
    4 5 1 0

Negate if number smaller than its negative (7)

-^:(<-) 

Takes the inverse (the inverse is bigger than the number itself) times.

Would loosely translate to:

if -num > num then
    num= -num
end

Answer 6

Brainfuck, 40 chars

+++++[>+++++++++<-],[->->+<<]>[>.>],[.,]

Or 35 chars if wrapping is allowed.

-[+>+[+<]>+],+[->->+<<]>[>-.>],[.,]

Answer 7

Mathematica 10 8

Method 1 (8 chars)

Take the square root of a number squared.

Sqrt@#²&

Examples

Sqrt@#²&[-.0176987]
Sqrt@#²&[.0176987]

.0176987
.0176987

---

Method 2 (8 chars)

Sign returns -1 if negative, 0 if zero, 1 if positive.

# Sign@#&

Examples

# Sign@#&[-4.3]
# Sign@#&[4.3]

4.3
4.3

---

4 chars?

As the following picture shows, it is possible to legitimately reduce the function, but I haven't been able to replicate the input on SO.

Answer 8

APL, 5

×∘×⍨⎕

A function would be 4 chars

×∘×⍨

Explanation

×, when used with one argument, is the signum function; when used with two arguments, is multiplication.
∘ is the compose(combination) of two functions.

×∘× is a function that takes two arguments and return the left argument times the signum of the right arugment.
⍨ means "use right argument as both left and right argument".
⎕ takes input from screen.

Answer 9

C: 13 characters

I would assume using overloaded operators are not permitted? Just because something is abstracted and doesn't follow the standard formatting of a function call doesn't mean it isn't a function.

For example: (**)n <==> abs(n) <==> '-'/

Anyways, here is my code: With explanation:

n+: ==> +(true - false) ==> +(1-0) ==> +1

n-: ==> -(false - true) ==> -(0-1) ==> -(-1) ==> +1

a=n*(n>0 - (n<0))

Answer 10

Python CLI, 20

lambda x:[x,-x][x<0]

Answer 11

Haskell, 14 characters

sqrt.flip(^)2$

Answer 12

J, 13

Without using build-in functions (like signum and the like):

f =: -`]@.(0&<)

f -1.253
1.253
f 0.91235
0.91235

Answer 13

Python 2.7 (16 charcters)

This one is the shortest here in python.

(input()**2)**.5

Let n = input(), then

Answer 14

EXCEL, 10:

=SQRT(n^2)

-- n is a defined cell name with the input value.

Answer 15

TI-Basic, 3 bytes

Outputs square root of input squared.

√(Ans²

Alternates

√(AnsAns          3 bytes
Ans²^.5           5 bytes
max(Ans,-Ans      5 bytes
Anscos(angle(Ans  5 bytes
-min(Ans,-Ans     6 bytes
If Ans<0:-Ans     7 bytes
Ans-2Ans(Ans<0    8 bytes
Ans(2(Ans>0)-1    10 bytes
Ans(-1+2(Ans>0    10 bytes

Answer 16

Python - 31 28 26 24 23 18

Uses boothby's idea of a lambda function, saving 5 characters (or 3 if I need to assign it to a variable):

lambda x:(x*x)**.5

---

Old Methods

Uses a generator function to save 1 char, so it is necessary to print the value with some function that uses a generator/iterator, like for i in a(b):print(i)

def a(b):yield(b*b)**.5

Saved 2 3 characters by squaring and unsquaring.

---

def a(b):yield(b<0)*-2*b+b

Old one:

def a(b):yield b if b>0 else-b

Edit: saved two characters by factoring b in.

Answer 17

C, 19

Slightly more than some other C answers, but guaranteed branch-less. the f variable is a float. I hope the bitwise operator & is allowed.

*(int*)&f&=INT_MAX;

Inspired from http://devmaster.net/posts/9998/the-ultimate-fast-absolute-value

Answer 18

JavaScript 26 13

alert((b=prompt())<0?-b:b)

Reducing it further with fat arrow functions...

a=b=>b<0?-b:b

Et voila! Reduction by 50%! Only issue is that this now only works for Firefox 22 and above with thanks to the fat arrows...

Answer 19

Seriously, 8 7 4 3 bytes

Thanks @quintopia

My first Seriously answer, to a 2 yr old question!

Note the challenge is older than the language

,ª√

Seriously is a stack based language. What this does is it pushes the input on the stack, then multiplies if with itself. Then it takes the square root of the result. It is based on the fact that sqrt(x^2)=|x|.

Answer 20

Java - 36

This is the obvious solution.

double a(double b){return b<0?-b:b;}

Answer 21

Ruby, 15 characters

a=->b{b>0?b:-b}

Ruby, 22 chars (with I/O)

p (a=gets.to_f)>0?a:-a

Accepts input on stdin, outputs on stdout

Answer 22

Befunge 98 - 11 7 8

~:'-`!j,

Explained:

~  read character input
:  duplicate
'- fetch - and push it
`  compare duplicated value with it, 1 if first is greater, otherwise 0
!  if the number on the stack is 0, set it to 1, otherwise 0
j  jump the number of characters on the top of the stack, which is 1 if the input character is -
,  print the character on top of the stack

If I understand this correctly, then this works for any input, but only if constant looping is allowed. It simply prints everything but the minus sign. If infinite looping is not allowed, then this should work (13 10 chars (Thanks FireFly)):

~:'-`!j,#@

---

Old version (Befunge 93) - 11

&:0`2*1-*.@

It works like this:

& push input
: push a duplicate
0 push a zero
` pop duplicate and 0, if duplicate greater than 0, push 1 else 0
2 push a two
* pop the 1 or 0 and 2 , multiply the 1 or 0 by the 2
1 push a one
- pop last two values and subtract the second value on the stack by the first (ie a-1)
* pop last two values and multiply last two values on stack (ie + or - 1 * input)
. print
@ end program

Note: Only integers are valid numbers in Befunge.

---

If I need to support floating point input, then it is 18 chars:

&&\:0`2*1-*.".",.@

It is 17 chars in Befunge 98:

&&\:0`2*1-*.'.,.@

Note: these print a space before the decimal point

Answer 23

Clojure 13 18 chars

#(max %(- %))

i am not sure if max is allowed, but i have seen others use it so here it goes :)

after the debate in the comments below, it was decided that max is indeed allowed!

use it like that:

(#(max %(- %)) -2) ; returns 2

---

edit - it seems max isnt allowed, so lets resort to the trivial solution

a trivial solution:

#(if(< % 0)(- %)%)

or

#((if(< % 0)- +)%)

Answer 24

C# 6.0, 29 bytes

double m(double n)=>n<0?-n:n;

Answer 25

MATLAB, 10 bytes

@(a)a^2^.5

Pretty straight forward. We are allowed sqauare and square root, so the absolute value of a number (assuming it is not complex!) is simply square it then square root it.

Answer 26

Burlesque, 4 bytes

XXim

XX returns the digits of an integer and im converts the digits back to an integer. XX removes the sign.

You might also go with Jsn?* depending on the exact types.

Answer 27

Prolog, 22 bytes

Saves 7 bytes over printing as we were allowed to return the absolute value.

Code:

p(X,Y):-Y is(X^2)^0.5.

Explanation:

Input X is squared, taken the root of and returned as Y.

Example:

p(-9.0e99,X).
X = 9.0e+99

Answer 28

Milky Way 1.1.5, 13 bytes

':0e?{_^_;-}!

---

Explanation

'              # read input from the command line
 :             # push a duplicate of the TOS to the stack
  0            # push 0 to the stack
   e           # push the truth value of A > B where A and B are the top two stack elements
    ?{_ _  }   # if-else statement
       ^       # pop the TOS without outputting
         ;     # swap the top two stack elements
          -    # push the value of A - B where A and B are the top two stack elements
            !  # output the TOS

---

Usage

./mw <path-to-code> -i <input-integer>

Answer 29

PHP, 30

<?=($n=$argv[1])*(1-($n<0)*2);

Answer 30

Gema, 2 characters

Same as edem's Perl answer.

-=

Sample run:

bash-4.3$ gema -p '-=' <<< '-82.923'
82.923