# Is it a whole number?

A number is whole if it is a non-negative integer with no decimal part. So 0 and 8 and 233494.0 are whole, while 1.1 and 0.001 and 233494.999 are not.

## Input

A floating-point number in the default base/encoding of your language.

For example, the default integer representation for Binary Lambda Calculus would be Church numerals. But the default integer representation for Python is base 10 decimal, not Unary.

## Output

A truthy value if the input is whole, a falsy value if it is not.

Note that if your language only supports decimal precision to, say, 8 places, 1.000000002 can be considered whole.

Input and output may be done via any standard I/O methods.

# Test cases

Input        -> Output
332          -> true
33.2         -> false
128239847    -> true
0.128239847  -> false
0            -> true
0.000000000  -> true
1.111111111  -> false
-3.1415926   -> false
-3           -> false


# Scoring

As with , the shortest submission wins. Good luck!

• @StephenLeppik It's no easier than the vanilla parity challenge, or the Hello World, or the Truth Machine. (In fact, it's harder than most of those.) Nov 18, 2017 at 3:28
• May we take input as two numbers, representing a fraction? Nov 18, 2017 at 3:32
• @LyricLy No, that would either be much easier for some languages or unnecessary for others. Nov 18, 2017 at 3:35
• Note that whole numbers are non-negative integers. Please update your first sentence to reflect this. And you can add in negative numbers test cases if you wish to show how whole numbers are not negative and falsy outputs for negative numbers are valid. Nov 18, 2017 at 3:56
• @ThomasWard, that wikipedia article seems to not fully agree with you: "Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, but in other writings, that term is used instead for the integers (including negative integers)". I read the title as asking about integers. Nov 20, 2017 at 9:02

# Lua, 38 bytes

function f(n)print(n>=0 and n%1==0)end


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## REXX, 27 bytes

arg n
say trunc(n)=n & n>=0


# Kotlin, 17 bytes

{it%1==.0&&it>=0}


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# C + ecpp, 23 bytes

#def $x !fmod(x,1)&x>=0  Defines an operator $ that takes one right operand x and evaluates to 1 if x is whole and 0 if it is not.

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# C on Linux, 31 bytes

int f(float v){return!rintf(v);}


The correct prototype for rintf is float rintf(float). However, in IEEE floating-point, 0.0f has the binary representation of all zeros. We (mis)use this feature to get away without using a prototype; the zero/non-zero value is then inverted into 1 or 0.

• or #define f(v)!rintf(v) Nov 19, 2017 at 8:18
• You should be able to use the implicit int typing if you drop the int. Nov 20, 2017 at 4:55
• @technosaurus - Why not accept double and save a byte? #define f(v)!rint(v) Nov 20, 2017 at 14:50
• Actually, I don't think this answer is correct at all - should it be v==rint(v)? Or perhaps !fmod(v,1) Nov 20, 2017 at 14:52
• @TobySpeight, it would have to be !fmodf(v,1.0) for the return to be a float (int-sized), and 1.0 to force the parameter to a float, both needed because there is no prototype.
– CSM
Nov 21, 2017 at 21:53

# Pyth, 4 bytes

Not a difficult question, but I thought I’d give it a try. Also did not think that zero was whole, but it saves a byte.

qs.a


Explanation:

 s.a      Integer of the absolute value of the implicit input
q         Is it the same as the original implicit input?


Test Suite

Version that doesn’t find 0 as a whole number:

&qs.a


Ands the original answer with the input, so 0 falsifies the answer.

Test Suite

Returns a falsy or truthy value that isn’t always a Boolean.

# Julia, 13 bytes

f(x)=x in 0:x


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# D, 3633 30 bytes

T f(T)(T n){return(n<0)==n%1;}


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I will not be beaten in D!

# Taxi, 553 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 5 l 2 l.Pickup a passenger going to Trunkers.Pickup a passenger going to Equal's Corner.Go to Trunkers:s 1 l.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:w 1 l.Switch to plan "b" if no one is waiting.'1' is waiting at Writer's Depot.[b]'0' is waiting at Writer's Depot.Go to Writer's Depot:n 1 l 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 r 2 r 1 l.


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Ungolfed:

Go to Post Office: west 1st left 1st right 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: north 5th left 2nd left.
Pickup a passenger going to Trunkers.
Pickup a passenger going to Equal's Corner.
Go to Trunkers: south 1st left.
Pickup a passenger going to Equal's Corner.
Go to Equal's Corner: west 1st left.
Switch to plan "b" if no one is waiting.
'1' is waiting at Writer's Depot.
[b]
'0' is waiting at Writer's Depot.
Go to Writer's Depot: north 1st left 1st right.
Pickup a passenger going to Post Office.
Go to Post Office: north 1st right 2nd right 1st left.


I'm surprised it took this long to post such a basic challenge. I like them. Taxi is... still not good for golfing.

# Jelly, 3 bytes

Ḟ⁼A


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Uses Erik's strategy of floor(n) == abs(n).

# Jelly, 3 bytes

AḞ=


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# Pyt, 3 4 bytes

Đ⎶Å≡

    implicit input
Đ   duplicates top value
⎶   pushes closest int to top value
Å   abs(TOS)
≡   checks if equal
implicit output


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# SNOBOL4 (CSNOBOL4), 73 bytes

	DEFINE('W(N)D')
W	N SPAN('1234567890') '.' REM . D
W =EQ(D) 1	:(RETURN)


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An anonymous function; returns 1 for truthy and an empty string for falsey.

Takes N the input and matches it against the regex \\d.(\\d*)$, setting (\\d*) as D. Then if D==0 (''==0 in SNOBOL), sets W to 1. If not, it doesn't set W, which defaults to the empty string, and returns W. ## Javascript - 16 Bytes a=>a>=0&&a%1<=0 f=a=>a>=0&&a%1<=0 // ----------- Test Cases ------------- console.log(332, f(332)) console.log(33.2, f(33.2)) console.log(128239847, f(128239847)) console.log(0.128239847, f(0.128239847)) console.log(0, f(0)) console.log(0.000000000, f(0.000000000)) console.log(1.111111111, f(1.111111111)) console.log(-3.1415926, f(-3.1415926)) console.log(-3, f(-3)) • Fails on 3 and -3.1415926 Mar 5, 2018 at 18:14 # TI-BASIC, 5 bytes Takes input and gives output through Ans. Ans=abs(int(Ans  # JavaScript (SpiderMonkey), 24 bytes s=>s<0?0:s==Math.ceil(s)  Try it online! A great thanks to @Jo King for converting the code from 29bytes to 24. • @JoKing thanks for pointing out. Nov 19, 2018 at 11:44 • false can be 0 and I'm pretty sure === can be == – Jo King Nov 19, 2018 at 12:00 • @JoKing yes. that way I will reduce 1 byte. Nov 19, 2018 at 12:11 • and four bytes for false to 0.. or if you're really insistent on booleans rather than falsey values, at least do 0>1 – Jo King Nov 19, 2018 at 12:21 • @JoKing the output requires false. If think that way then 0 is not significant option. Nov 19, 2018 at 12:24 # MathGolf, 3 bytes i±=  Try it online! ## Explanation Same as a bunch of other answers. i Read input as integer ± Convert to absolute value = Is it equal to the implicit input?  # Python 2, 5960 57 bytes lambda s:re.match(r"^\+?[0-9]+(\.0*)?$",s)>None
import re


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Takes input as a string.

Regex is as follows:

• starts with + or a number
• has AT LEAST 1 digit
• either:
• has a period, but only zeroes follow the period
• has a period, but nothing after the period
• terminates without a period

Python 2's re.match objects are greater than None objects, so it will return True if the regex matched and False otherwise.

Edit: no longer has false positives on negative numbers (not sure what i was thinking...) and correctly handles trailing periods. Thanks, Deadcode!

• This is incorrect. The problem defines a whole number as a non-negative integer, so you need \+? (or nothing) not [+-]?. Also, shouldn't it be 0* not 0+? Ending with a decimal point is generally valid. Feb 11, 2019 at 0:52
• @Deadcode Well, crap. You're right. Thanks for pointing this out! Feb 12, 2019 at 15:37

# Gol><>, 17 bytes

I:0(q0h:S(-zq1!0h


A simple solution, but I will be golfing this alot more

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• 7 bytes, with error/0 as falsey, 1 as truthy. If you don't like that than you can at least get rid of no-ops (like q1!0 lol) and switch to modulo-1 for 10 bytes
– Jo King
Feb 12, 2019 at 21:47
• @JoKing Those look really good, I honestly didn't realize that the '0' wasn't doing anything! I feel like such an idiot, in a little while I'll add your programs on (attributed to you of course) Feb 12, 2019 at 22:05
• @KrystosTheOverlord in a little while? :P Mar 7, 2019 at 6:47

# cQuents, 4 bytes

?$-1  Try it online! ## Explanation ? Given input n, output true if n is in the sequence and false if it is not Each term in the sequence equals$        current index
-1                    - 1


The -1 in to include 0 - if the challenge just wanted positive integers, ?\$ would be enough.

# Runic Enchantments, 6 bytes

i:kn=@


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Works on inputs up to 65535. Compares the input with the input cast to character and back to number. Inputs larger than 65535 cast to char exceed the byte size and roll over.

# Stax, 2 bytes

1%


Run and debug it

Modulus by 1.

# ARBLE, 9 7 bytes

nt(a%1)


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