Is it a whole number?

Question

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.

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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.

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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

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Scoring

As with code-golf, the shortest submission wins. Good luck!

Answer 1

Haskell, 27 16 bytes

This function checks whether x is contained in the list of nonnegative integers that are not greater than x.

f x=elem x[0..x]

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Answer 2

APL (Dyalog), 3 bytes

⌊≡|

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Note that f← has been prepended in the TIO link due to technical limitations, but it's not normally needed.

Answer 3

Pyth, 4 bytes

qs.a

Test suite

Is the number equal (q) to the floor (s) of its absolute value (.a)?

Answer 4

Wolfram Language (Mathematica), 17 15 14 bytes

Saved 1 byte thanks to Not a tree!

#>=0==#~Mod~1&

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First Mathematica answer \o/

Mathematica, 15 bytes

Saved 2 bytes thanks to @user202729!

#==⌊Abs@#⌋&

Answer 5

Husk, 3 bytes

£ΘN

Try it online! The third test case times out in TIO, so I chopped off a couple of digits. I tried to run it locally, but killed it after a couple of minutes since it was using over 6GB of memory and my computer started to stutter. It should theoretically finish at some point...

Explanation

This corresponds to the challenge description pretty directly.

£ΘN  Implicit input: a number.
  N  The infinite list [1,2,3,4...
 Θ   Prepend 0: [0,1,2,3,4...
£    Is the input an element of this list?

Answer 6

Python 2, 18 bytes

lambda n:n%1==0<=n

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Answer 7

///, 94 bytes, input hard-coded

/=/\/x://:/\/\///-0/-:.0/.:.|/|:-|/|:0=1=2=3=4=5=6=7=8=9=|x/|:||/true:x:/-:/.:/|/+:++/false/||

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Input between the two terminating vertical lines (||)

Strips -00...0 and .00...0, converts all remaining digits to xs, then tests whether the remaining number still has xs after . or a - not followed by ..

Could save up to 7 bytes depending on what's counted as truthy and falsey since this language doesn't have native truthy/falsey values, currently is outputting true and false but could change to, for example, T and F for 87 bytes if that's allowed.

Answer 8

Python 2 and Python 3, 21 18 bytes

lambda n:n>=0==n%1

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3 bytes saved thanks to Mr. XCoder.

Answer 9

Octave, 15 bytes

@(x)any(x==0:x)

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This one is based on the approach used in @flawr's Haskell answer.

While it brings the byte count down to 15, it is shamefully inefficient (no offence intended), creating a list of every integer from 0 to x and seeing if x is contained within.

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Octave, 18 bytes

@(x)fix(x)==abs(x)

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Takes input as a double precision number. Returns true if whole.

Checks if the input when rounded is equal to the magnitude of the input. This will only be the case when the number is positive and whole.

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Octave, 18 bytes

@(x)~mod(x,+(x>0))

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An alternate solution for 18 bytes. Sadly the mod function in Octave won't implicitly convert a bool to a double, so the +( ) is needed around the greater than comparison. Otherwise this solution would have been shorter.

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Octave, 18 bytes

@(x)x>=0&~mod(x,1)

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And another one... I can't seem to get lower than 18 bytes. All because of having to allow for 0 to be true with the >= instead of just >.

Answer 10

C++ (gcc), 33 bytes

int f(float x){return(uint)x==x;}

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Answer 11

Aceto, 6 bytes

rfdi±=p

r grabs input
f converts it to a float
d and i duplicates it and converts it to an integer
± pushes the absolute value of it (b/c can't be negative)
= checks if they are equal
p prints out the result

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Answer 12

Pyth, 6 bytes

&gQZsI

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Answer 13

QBIC, 17 bytes

?(:=int(a))*(a>0)

Explanation

?             PRINT
 (:=int(a))     if the input (: assigns the cmd line param to 'a') equals itself 
                cast to int, -1 else 0
 *              times
 (a>0)          if the input > 0, -1 else 0

If either check fails, this returns 0. If both are true, it returns -1 x -1 = 1

Answer 14

C (gcc), 27 28 27 25 bytes

^{-2 thanks to PrincePolka}

#define g(f)!fmod(f,f>=0)

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Defines a macro "function" g that takes a parameter f (of any type). Then checks if casting f mod 1 is zero, and if f is non-negative.

Answer 15

C#, Java : 43 bytes

-1 byte thanks to Zacharý
-1 byte thanks to TheLetalCoder

int w(float n){return(n==(int)n&&n>=0)?1:0;}

C# has a special 33 bytes optimization that you can not do in java :

bool w(float n){return(int)n==n;}

For testing

C# code :

class Program {
    int w(float n){return(n==(int)n&&n>=0)?1:0;}

    static void Main(string[] args)
    {
        var p = new Program();
        float[] fTab = new float[]{
            332,33.2f,128239847,0.128239847f,0,0.0000f,1.1111111111f,-3.1415926f,-3
        };
        foreach (float f in fTab) {
            Console.WriteLine(string.Format("{0} = {1}", f, (p.w(f) != 0).ToString()));
        }
        Console.ReadKey();
    }
}

Java Code :

public class MainApp {
    int w(float n){return(n==(int)n&&n>=0)?1:0;}

    public static void main(String[]a) {
        MainApp m = new MainApp();
        float testArr[] = new float[]{
                332,33.2f,128239847,0.128239847f,0,0.0000f,1.1111111111f,-3.1415926f,-3
        };

        for (float v : testArr) {
            System.out.println(v + " = " + String.valueOf(m.w(v)!=0));
        }
    }
}

Answer 16

Google Sheets, 10 Bytes

Anonymous worksheet function that tales input from cell A1 and outputs to the calling cell.

=A1=Int(A1

Answer 17

Prolog (SWI), 24 bytes

f(X):-X>=0,round(X)=:=X.

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Answer 18

Ly, 35 47 bytes

ir"-"=[0u;]pr[["0"=![psp>l"."=[ppr!u;]p<2$]pp]]

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Ly has float support, but the only way to create a float currently is by performing division. There is no way to take a float as input, so I had to manually check the string instead.

Lost 13 bytes adding support for negative numbers.

Answer 19

JavaScript, 14

n=>!(n<0||n%1)

Answer 20

Perl 5, 11 +1(-p) bytes

$_=abs==int

The -l switch not counted because for tests display

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Answer 21

Perl 6,  15  13 bytes

{.narrow~~UInt}

Test it

{.Int==$_>=0}

Test it

Answer 22

Java (OpenJDK 8), 14 bytes

n->n%1==0&n>=0

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Answer 23

C#, 40 37 29 bytes

bool z(float x)=>x%1==0&x>=0;

Saved 8 bytes thanks to @Dennis_E!

Old Answer (37 bytes)

bool z(float x){return x%1==0&&x>=0;}

Old old answer (40 bytes):

bool z(float x){return (int)x==x&&x>=0;}

Answer 24

JavaScript, 17 15 bytes

_=>_<0?0:_==~~_

_=>_<0?0:_==(_|0)

Thanks to edc65 for catching my mistakes.

Answer 25

Unexpanded Sinclair ZX81, 20 bytes

 1 INPUT A
 2 PRINT ABS A=INT A

20 bytes because BASIC is tokenized.

Simply will output 1 (true) if number is positive and the value of the number entered equals its integer value. Outputs 0 either of those conditions are not met.

Answer 26

C, C++ : 38 37 bytes

-1 byte thanks to Zacharý

-1 byte thanks to ceilingcat

int w(float n){return(int)n==n&n>=0;}

For Testing

C : Try it online

C Code :

#include <stdio.h>
int main() {
    float f[] = { 332,33.2f,128239847,0.128239847f,0,0.0000f,1.1111111111f,-3.1415926f,-3 };
    int t;
    for ( t = 0; t < 9; ++t) {
        printf("%f = %s\n", f[t], w(f[t])?"true":"false");
    }
    return 0;
}

C++ Code :

#include <iostream>
int main() {
    std::initializer_list <std::pair<float,bool>> test{
        {332.f,true}, {33.2f,false}, {128239847.f,true}, {0.128239847f,false}, {0.f,true}, {0.000000f,true}, {1.111111f,false}, {-3.1415926f,false}, {-3.f,false}
    };

    for (const auto& a : test) {
        if (w(a.first) != a.second) {
            std::cout << "Error with " << a.first << '\n';
        }
    }
    return 0;
}

Answer 27

JavaScript (Node.js), 11 bytes

n=>n<0==n%1

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Answer 28

J, 7 4 bytes

Removed the unnecessary celing check after the solution of Erik The Outgolfer

<.=|

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Answer 29

Common Lisp, 36 32 bytes

(lambda(n)(=(max n 0)(floor n)))

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Transposition of marmeladze’s answer.

Answer 30

Symbolic Python, 21 bytes

_=_%(_==_)==(_!=_)<=_

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Uses a chained comparison:

• _%(_==_) == (_!=_) checks if n%1 == 0, only true if n has no decimal part.
• (_!=_) <= _ checks if 0 <= n, only true if the integer is non-negative.