Yes, of course I'm an adult!

Question

I think we've all done this as a kid: some websites require a minimum age of 18, so we just subtract a few years from the year of birth and voilà, we 'are' 18+.
In addition, for most rides at amusement parks the minimum height to enter is 1.40 meters (here in The Netherlands it as at least). Of course this can be cheated less easily than age, but you could wear shoes with thick heels, put your hair up, wear a hat, stand on your toes, etc.

Input:

Your program/function accepts a positive integer or decimal.

Output:

• Is the input an integer >= 18? Simply print the input.
• Is the input an integer 0-17? Print 18.
• Is the input a decimal >= 1.4? Simply print the input.
• Is the input a decimal 0.0-1.4? Print 1.4.

Challenge rules:

• Assume the input will always be in the range of 0-122 (oldest woman ever was 122) or 0.0-2.72 (tallest man ever was 2.72).
• You are allowed to take the input as a String, object, or anything else you prefer.
• The decimal inputs will never have more than three decimal places after the decimal point.
• 2 or 2. both aren't valid outputs for 2.0. You are free to output 2.00 or 2.000 instead of 2.0 however.
  Just like the input the output will never have more than three decimal places after the point.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code.
• Also, please add an explanation if necessary.

Test cases:

0      ->  18
1      ->  18
2      ->  18
12     ->  18
18     ->  18
43     ->  43
115    ->  115
122    ->  122

0.0    ->  1.4
1.04   ->  1.4
1.225  ->  1.4
1.399  ->  1.4
1.4    ->  1.4
1.74   ->  1.74
2.0    ->  2.0
2.72   ->  2.72

Answer 1

Python 2.7, 34 bytes

lambda x:max(x,[18,1.4]['.'in`x`])

Answer 2

JavaScript (ES6), 27 31

Input taken as a string. To check if the input value has decimals, it's appended to itself: if there is no decimal point the result is still a valid number, else it's not. To recognize a valid number (including 0), I use division as in javascript 1/n is numeric and not 0 for any numeric n (eventually the value is Infinity for n==0), else it's NaN

x=>x<(y=1/(x+x)?18:1.4)?y:x

Test

f=    
x=>x<(y=1/(x+x)?18:1.4)?y:x

;[
 ['0', '18' ],['1', '18' ],['2', '18' ],['12', '18' ],['18', '18' ],['43', '43' ],['115', '115'], ['122', '122' ]
,['0.0', '1.4'],['1.0', '1.4'],['1.04', '1.4'],['1.225', '1.4'],['1.399', '1.4'],['1.4', '1.4'],['1.74', '1.74'],['2.0', '2.0'],['2.72', '2.72']
].forEach(t=>{
  var i=t[0],k=t[1],r=f(i)
  console.log(i,k,r,k==r?'OK':'KO')
})

My prev (wrong) solution:

Taking input as a number, you can use remainder operator % to check if the number is integer.

x=>x<(y=x%1?1.4:18)?y:x

or

x=>Math.max(x,x%1?1.4:18)

But this does not work as the challenge request to discriminate between, say, 2 and 2.0 and that's the same number. So it's not possibile to get the input as a number

Answer 3

05AB1E, 13 11 bytes

Uses CP-1252 encoding.

ÐîQ18*14T/M

Explanation

Ð             # triplicate input
 î            # round up
  Q           # check for equality
   18*        # multiply 18 by this (18 if input is int, else 0)
      14T/    # push 14 / 10
          M   # take max of stack (input and 1.4 or 18)

Try it online

Answer 4

Java 8, 90 61 57 bytes

i->(i+"").contains(".")?(float)i<1.4?1.4:i:(int)i<18?18:i

-4 bytes returning Object instead of String; and some additional bytes converting Java 7 to 8.
-4 byte taking the input as Object instead of String as well.

Explanation:

Try it here.

i->                      // Method with Object as both parameter and return-type
  (i+"").contains(".")?  //  If the input as String contains a dot:
   (float)i<1.4?         //   If the input is a float below 1.4:
    1.4                  //    Return double 1.4
   :                     //   Else:
    i                    //    Return the input-float as is
  :(int)i<18?            //  Else if the input is an integer below 18:
   18                    //   Return integer 18
  :                      //  Else:
   i                     //   Return the input-integer as is

Answer 5

Pyth, 14 bytes

eS,@,18h.4}\.`

Try it online.

Answer 6

PHP, 40 Bytes

modified by @user59178 Thank You

<?=max(is_int(0+$i=$argv[1])?18:1.4,$i);

PHP, 42 Bytes first Version

<?=max(strpos($i=$argv[1],".")?1.4:18,$i);

Answer 7

Perl, 29 27 bytes

Includes +2 for -lp

Give input on STDIN

adult.pl <<< 1.24

adult.pl:

#!/usr/bin/perl -lp
$_>($a=/\./?1.4:18)or*_=a

If you don't mind an extra newline if you really were a full adult then leaving out the l option for 26 bytes works too

Answer 8

EXCEL: 26 31 29 Bytes

=MAX(A1;IF(MOD(A1;1);1,4;18))

Formula can go anywhere except A1, the input cell.

Fixed errors, and replaced with Emigna's suggestion.

Thanks to Alexandru for saving me some bytes using truths

Answer 9

Brachylog, 14 bytes

#$:18ot|:1.4ot

Try it online!

Explanation

    #$             Input is an integer
      :18o         Sort the list [Input, 18]
          t        Take the last element
|              Or
    :1.4o          Sort the list [Input, 1.4]
         t         Take the last element

Answer 10

GNU sed, 40 + 1 = 41 bytes

(score +1 for use of -r flag to interpreter)

s/^.$|^1[^9]$/18/
/^0|1\.[0-3]/s/.*/1.4/

Annotated:

#!/bin/sed -rf

# First, anything that's a single digit or is '1' followed by a
# digit other than '9' is replaced with '18'.
s/^.$|^1[^9]$/18/

# Now, any line beginning with '0' or containing '1.0' to '1.3' is
# replaced with '1.4'.
/^0|1\.[0-3]/s/.*/1.4/

We take advantage of the constraints on input, so don't have to test the start of string when we see '1.' - we know there's only one digit before the decimal point.

Test result:

$ ./94832.sed <<END
> 0
> 1
> 2
> 12
> 18
> 43
> 122
> 
> 0.0
> 1.04
> 1.225
> 1.399
> 1.4
> 1.74
> 2.0
> 2.72
> END
18
18
18
18
18
43
122

1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

Answer 11

Haskell, 50 bytes

x#y=show$max x$read y 
f s|elem '.'s=1.4#s|1<2=18#s

Usage example: f "1.0" -> "1.6".

Haskell's strict type requires usage of strings as input and output. Howevers, read, max and show are polymorphic and handle all numeric types.

Answer 12

Java, 79 70 bytes

int f(int x){return x<18?18:x;}
float f(float x){return x<1.4?1.4f:x;}

Defines two methods with overloading, which use the conditional operator.

Call it like f(5) or f(1.4f).

Answer 13

Convex, 9 bytes

_1ÛI1.4?»

Try it online!

Answer 14

CJam, 18 14 bytes

Saved 4 bytes thanks to Martin Ender.

q~_`'.&1.4I?e>

Online interpreter

Answer 15

IBM/Lotus Notes Formula Language, 58 49 bytes

@If(@Like(@Text(a);"%.%");@If(a<1.4;1.4;a);@If(a<18;18;a))

Computed field formula where a is an editable numeric field.

EDIT

@If(@Like(@Text(a);"%.%");@Max(1.4;a);@Max(18;a))

Alternative inspired by @Mego

Answer 16

AWK - 29 bytes

($0<c=$0~/\./?1.4:18){$0=c}1

Usage:

awk '{c=$0~/\./?1.4:18}($0<c){$0=c}1' <<< number

Testing was performed with gawk on RHEL 6. I tried with all the test cases, unfortunately I don't have AWK on the machine that has internet access, so copy-paste is not possible.

Is there a more compact way to do this in AWK?

Answer 17

C#, 58 bytes

x=>x is int?(int)x>17?x:18:(float)x<1.4?"1.4":$"{x:.0##}";

No crazy string parsing needed for C#. Input is expected to be an int or float (unfortunately C# can't cast a double to float if the double is in an object). Output will be either int or string in an object.

(almost missed the at least 1 decimal requirement, added that now)

Ungolfed:

/*Func<object, object> Lambda = */ x =>
    x is int // if parameter is an int
        ? (int)x > 17 // check if x is at least 18
            ? x // at least 18 so return x
            : 18 // less than 18 so return 18
        : (float)x < 1.4 // x is float, check if at least 1.4
            ? "1.4" // less than 1.4 so return 1.4
            : $"{x:.0##"} // at least 1.4 so return x and ensure at least 1 decimal place
;

Alternate implementation which is also 58 bytes.

x=>x is int?(int)x>17?x:18:$"{((float)x<1.4?1.4:x):.0##}";

Answer 18

Actually, 16 bytes

;:.7τ9τ($'.íuIkM

Try it online!

Explanation:

;:.7τ9τ($'.íuIkM
;                 dupe input
 :.7τ             1.4 (.7*2) - note that :1.4 is the same length, but an additional delimiter would be needed to separate it from the following 1
     9τ           18 (9*2)
       ($'.íu     1-based index of "." in string representation of input, 0 if not found
             I    1.4 if input contains a "." else 18
              kM  maximum of remaining values on stack 

Answer 19

Emacs Lisp, 37 bytes

(lambda(x)(max(if(floatp x)1.4 18)x))

Guesses from the "datatype" whether the integer or float version should be used. (floatp returns t for 1.0, but not for 1.) The parameter is a number as integer or float, i.e. it should satisfy numberp.

Answer 20

Haskell, 49 bytes

x#y=show.max x.fst<$>reads y;f s=head$18#s++1.4#s

Basically, this first attempts to read the input as an integer and then as a double if that fails. Then proceeds to compare it to the respective comparison baseline.

Answer 21

Jelly, 16 15 13 12 bytes

-1 thanks to caird coinheringaahing (take input as a string).

ċ”.ị1.4,18»V

TryItOnline
Or see all test cases, also at TryItOnline

How?

ċ”.ị1.4,18»V - Main link: string, S
 ”.          - '.' character
ċ            - count (occurrences of '.' in S)
    1.4,18   - pair literals 1.4 and 18:  [1.4,  18]
   ị         - index into (1-indexed)
           V - evaluate (S) as Jelly code
          »  - maximum

Answer 22

C#, 69 bytes

s=>s.Contains(".")?float.Parse(s)<1.4?"1.4":s:int.Parse(s)<18?"18":s;

Try it online!

Full program with test cases:

using System;

namespace YesImAnAdult
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<string,string>f= s=>s.Contains(".")?float.Parse(s)<1.4?"1.4":s:int.Parse(s)<18?"18":s;
            
            Console.WriteLine(f("0"));  //18
            Console.WriteLine(f("1"));  //18
            Console.WriteLine(f("2"));  //18
            Console.WriteLine(f("12")); //18
            Console.WriteLine(f("18")); //18
            Console.WriteLine(f("43")); //43
            Console.WriteLine(f("122"));    //122

            Console.WriteLine(f("0.0"));    //1.4
            Console.WriteLine(f("1.04"));   //1.4
            Console.WriteLine(f("1.225"));  //1.4
            Console.WriteLine(f("1.399"));  //1.4
            Console.WriteLine(f("1.4"));    //1.4
            Console.WriteLine(f("1.74"));   //1.74
            Console.WriteLine(f("2.0"));    //2.0
            Console.WriteLine(f("2.72"));   //2.72
        }
    }
}

A pretty straightforward solution. Note that on some systems float.Parse() might return incorrect results. Pass CultureInfo.InvariantCulture as the second argument according to this answer.

Answer 23

Dyalog APL, 14 bytes Rendered invalid by further specifications

⎕IO←0 which is default on many systems. Takes string as argument.

⍎⌈18 1.4⊃⍨'.'∘∊

⍎⌈ max of the evaluated argument and

18 1.4⊃⍨ {18,1.4} selected by

'.'∘∊ whether the argument contains a period

Answer 24

C++, 68 bytes

int A(int a){return a<18?18:a;}float A(float h){return h<1.4?1.4:h;}

This answer is actually 2 functions with the same name, and the compiler works out for me which to call, so it acts as one function without me having to take in one input and decide which it is. Since input on the floating point is guaranteed to be of the same precision as the output, I can return it safely without having to truncate it either.

Ungolfed + tests

#include <iostream>

int A(int a)
{
   return a < 18 ? 18 : a;
}

float A(float h)
{
   return h < 1.4 ? 1.4 : h;
}

int main()
{
  std::cout << 0 << " " << A(0) << "\n";
  std::cout << 19 << " " << A(19) << "\n";
  std::cout << 1.1 << " " << A(1.1f) << "\n";
  std::cout << 2.2 << " " << A(2.2f) << "\n";
}

Answer 25

C, 50 bytes:

#define A(x)(x/2+(x+1)/2-x?x<1.4?1.4:x:x<18?18:x)

The byte count includes the newline at the end of the macro definition.

Test:

#define A(x)(x/2+(x+1)/2-x?x<1.4?1.4:x:x<18?18:x)
#include <assert.h>
int main() {
  assert(A(0) == 18);
  assert(A(1) == 18);
  assert(A(2) == 18);
  assert(A(12) == 18);
  assert(A(18) == 18);
  assert(A(43) == 43);
  assert(A(115) == 115);
  assert(A(122) == 122);
  assert(A(0.0) == 1.4);
  assert(A(1.04) == 1.4);
  assert(A(1.225) == 1.4);
  assert(A(1.399) == 1.4);
  assert(A(1.4) == 1.4);
  assert(A(1.74) == 1.74);
  assert(A(2.0) == 2.0);
  assert(A(2.72) == 2.72);
}

Answer 26

C#, 71 bytes

object A(object i){return i is int?(int)i>18?i:18:(double)i>1.4?i:1.4;}

try it here

Answer 27

C, 119 111 105 100

m;f(char*s){float atof(),l=atof(s);for(m=s;*s&&*s++!=46;);puts(*s?l<1.4?"1.4":m:atoi(m)>18?m:"18");}

Tested with

main(c,v)char**v;{
    f("0");
    f("1");
    f("2");
    f("12");
    f("18");
    f("44");
    f("115");
    f("122");
    f("0.0");
    f("1.04");
    f("1.225");
    f("1.339");
    f("1.4");
    f("1.74");
    f("2.0");
    f("2.72");
}

Output

18
18
18
12
18
44
115
122
1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72

Answer 28

Vyxal ḋ, 13 12 bytes

₌⌊±[18|1.4]∴

Try it Online!

Takes input as string.

Answer 29

Batch, 102 bytes

@set/ps=
@if %s:.=%==%s% (if %s% lss 18 set s=18)else if %s:~0,1%%s:~2,1% lss 14 set s=1.4
@echo %s%

First determines whether the input is an integer by checking whether deleting all .s has any effect on the string. If it is then the value is readily compared against 18, otherwise the first and third characters are combined into a number which is compared against 14.

Answer 30

C#, 95 Bytes

Golfed:

string y(string p){int a;return int.TryParse(p,out a)?a>17?p:"18":double.Parse(p)<1.4?"1.4":p;}

Ungolfed:

class YesOfCourseImAnAdult
  {
    public string y(string p)
    {
      int a;
      return int.TryParse(p, out a) ? a > 17 ? p : "18"
       : double.Parse(p) < 1.4 ? "1.4" : p;
    }
  }

Test cases:

var codeGolf = new YesOfCourseImAnAdult();
Console.WriteLine(codeGolf.y("0"));
Console.WriteLine(codeGolf.y("1"));
Console.WriteLine(codeGolf.y("2"));
Console.WriteLine(codeGolf.y("12"));
Console.WriteLine(codeGolf.y("18"));
Console.WriteLine(codeGolf.y("43"));
Console.WriteLine(codeGolf.y("122"));

Console.WriteLine(codeGolf.y("0.0"));
Console.WriteLine(codeGolf.y("1.04"));
Console.WriteLine(codeGolf.y("1.225"));
Console.WriteLine(codeGolf.y("1.399"));
Console.WriteLine(codeGolf.y("1.4"));
Console.WriteLine(codeGolf.y("1.74"));
Console.WriteLine(codeGolf.y("2.0"));
Console.WriteLine(codeGolf.y("2.72"));

Output:

18
18
18
18
18
43
122

1.4
1.4
1.4
1.4
1.4
1.74
2.0
2.72