Write an interpreter for *

Question

The task is simple. Write an interpreter for the language *.

Here's a bigger link to the wiki.

There are only three valid * programs:

• * Prints "Hello World"
•  *  Prints a random number between 0 and 2,147,483,647
• *+* Runs forever.

The third case must be an infinite loop according to the specifications in this question

Input:

• The input can be taken via any acceptable input method by our standard I/O rules
• It will always be one of the above programs

Output:

• The first case should print exactly Hello World, with or without a trailing line break.
• For the second case, if your language's integer maximum is smaller than 2,147,483,647, use your language's integer maximum
• The first and second cases can print to any acceptable output by our standard I/O rules.
• The third case should not give any output.

Scoring:

As this is code-golf, the shortest answer, in bytes, wins.

Answer 1

*, 0 bytes

Since * has no way of reading input, the default rules allow specifying that the input must be given by concatenating it onto the program.

(... I think. There's an "at least twice as many upvotes as downvotes" condition that I don't have the rep to verify).

Answer 2

R, 69 bytes

switch(scan(,""),"*"="Hello, World!"," * "=sample(2^31,1)-1,repeat{})

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switch tries to match the named arguments and if there's no match, selects the first unnamed one after the first, which in this case is the infinite loop repeat{}.

Answer 3

Jelly,  21  20 bytes

ḊOSØ%HX’¤“½,⁾ẇṭ»¹Ḃ¿?

A monadic Link accepting a list of characters.

Try it online!

vL’... also works (see below).

How?

ḊOSØ%HX’¤“½,⁾ẇṭ»¹Ḃ¿? - Link: list of characters   e.g.: "*"        or  " * "    or  "*+*"
Ḋ                    - dequeue                          ""             "* "         "+*"
 O                   - to ordinals                      []             [42,32]      [43,42]
  S                  - sum                              0              74           85
                   ? - if...
                  ¿  - ...if-condition: while...
                 Ḃ   -    ...while-condition: modulo 2  0              0            1
                ¹    -    ...while-true-do: identity                                85
                     -                                  0              74           (looping)
        ¤            - ...then: nilad followed by link(s) as a nilad:
   Ø%                -    literal 2^32                                 2^32
     H               -    half                                         2^31
      X              -    random integer in [1,n]                      RND[1,2^31]
       ’             -    decrement                                    RND[0,2^31)
         “½,⁾ẇṭ»     - ...else: dictionary words        "Hello World"

---

Alternative

vL’... - Link: list of characters                 e.g.: "*"        or  " * "    or  "*+*"
 L     - length                                         1              3            3
v      - evaluate (left) as Jelly code with input (right)
       -                                                1^1            3^3          (3^3+3)^3
       -                                                1              27           27000
  ’    - decrement                                      0              26           26999
   ... - continue as above                              "Hello World"  RND[0,2^31)  (looping)

Answer 4

C (gcc), 66 63 bytes

Thanks to attinat for the -3 bytes.

I only have to check the second character: if the LSB is set, it's a + (thus the program is "*+*") and the program loops. After that, if it's a NUL, the program was "*" and we display Hello World; otherwise, it displays a random value (" * ", the only other option left.)

f(char*s){for(s++;*s&1;);printf(*s?"%d":"Hello World",rand());}

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Answer 5

Keg, -lp, -ir 30 26 25 24 20 19 bytes

!1=[_“H%c¡“| =[~.|{

-1 byte using flags

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Answer History

?!1=[_“H%c¡“| =[~.|{

Try it online!

Shortened Hello World to dictionary string

!1=[_Hello World| =[~.|{

Try it online!

I never ceased to be amazed at the power of Keg. Credits to user EdgyNerd for another byte saved.

Prior Versions

_!0=[Hello World|\*=[~.|{

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Credit to user A__ for the extra byte saved.

Old version

?!1=[_Hello World| =[__~|{

Essentially, takes the input program and:

• Checks to see if the input length is 1, printing "Hello World" if true
• Checks to see if the last character is a space, and prints a random number
• Otherwise runs an infinite loop

Then implicitly print the stack.

?                               #Get input from the user
 !1=                            #Compare the stack's length to 1
    [_Hello World           #Push "Hello, World!" to the stack
                     | =        #See if top item is a space
                        [__~|{  #If so, generate a random number, otherwise, infinite loop.

4 bytes saved due to the fact that hello world doesnt need punctuation.

Try it online! Old version

Try it online! New version

Answer 6

Japt, 22/25 bytes

The first solution is for the original spec which had *<space> as the second programme and the other is for the updated spec which arbitrarily changed it to <space>*</space>, with thanks to EoI for the suggested "fix".

Both throw an overflow error upon entering the infinite loop of the third programme but, theoretically, with enough memory (which we may assume for the purposes of code-golf), would run forever.

Å?¢?ß:2pHÉ ö:`HÁM Wld

Try programme 1
Try programme 2
Try programme 3

Å?UÎ>S?ß:2pHÉ ö:`HÁM Wld

Try programme 1
Try programme 2
Try programme 3

Answer 7

Python 2, 103 93 89 87 bytes

I combined my earlier answer with Chas Browns's answer and got something a few bytes shorter.

The random number will be between 0 and 2**31-1 inclusive.

from random import*
i=input()
while'*'<i:1
print["Hello World",randrange(2**31)][i<'!']

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Previous versions:

103 bytes

from random import*
exec['print"Hello World"','while 1:1','print randint(0,2**31-1)'][cmp(input(),'*')]

93 bytes

from random import*
i=cmp(input(),'*')
while i>0:1
print["Hello World",randint(0,2**31-1)][i]

Answer 8

JavaScript (ES7), 66 bytes

s=>s[1]?s<'!'?Math.random()*2**31|0:eval(`for(;;);`):'Hello World'

Try it online! (Hello World)

Try it online! (random number)

Try it online! (infinite loop)

Answer 9

Jelly, 23 21 bytes

OS¹Ḃ¿ịØ%HX’;““½,⁾ẇṭ»¤

Try it online!

A monadic link taking a single argument and returning Hello World, a random 31 bit integer or looping infinitely as per the spec.

All options: * * *+*

Explanation

O                     | Convert to codepoints
 S                    | Sum
  ¹Ḃ¿                 | Loop the identity function while odd 
     ị              ¤ | Index into the following as a nilad:
      Ø%              | - 2 ** 32
        H             | - Halved
         X            | - Random integer in the range 1..2**31
          ’           | - Decrease by 1 
           ;          | - Concatenated to:
            ““½,⁾ẇṭ»  |   - "", "Hello World"

Answer 10

Befunge-93, 54 bytes

~"*"-_~1+#^_"dlroW olleH">:#,_@.%*2**:*::*88:*`0:?1#+<

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Annotated:

~"*"-                      _                ~1+                   #^_        "dlroW olleH">:#,_    @      .%*2**:*::*88:   *`0:             ?1#+<
Compare first      If equal, go right       Compare second       If equal,        Output          Exit    Modulo by 2^31   If less than      Add 1
character to       Otherwise, go left       character to       loop forever   "Hello World"                 and output     0, multiply     a random amount
'*'                and wrap around          -1 (EOF)                                                                         by 0            of times

The randomness is not uniform. At each increment there is a 50% chance to stop incrementing.

Answer 11

Python 2, 91 89 88 bytes

from random import*
def f(p):
 while'*'<p:p
 print['Hello World',getrandbits(31)][p<'!']

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2 bytes thanks to Jonathan Allan; 1 byte thx to ShadowRanger.

Answer 12

PowerShell, 60, 56 bytes

switch($args){*{'Hello World'}' * '{random}*+*{for(){}}}

Pretty dumb version, the only golfing technique here is omitting Get- in Get-Random.

UPD. Stripped down to 56 bytes by removing quotes, thanks to veskah!

• Try it online! (Hello World)
• Try it online! (random number)
• Try it online! (infinite loop)

Answer 13

///, 72 bytes

/+*/+*// */a random number between 0 and 0//*/"Hello World"/

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Here I go again, adding another cheating entry. Was I really going write my own pseudo-random generator just for this though... and in ///?!

Instead, I interpreted the Print instruction very literally, as a good minion should. Everything after Print is outputted:

* Prints "Hello World"

* → "Hello World", yes, even with the quotation marks. Otherwise, I'd admit to being aware of the quotation-marks-denote-literal-string convention, which wasn't explicitly stated in the question, and doesn't exist in the language with which I'm submitting. And then I wouldn't be able to cheat with the "Print a random number" part. So, I'm feigning ignorance... Shhh, don't tell!

​ * ​ Prints a random number between 0 and 2,147,483,647

* → a random number between 0 and 0. Lucky for me, although they didn't add the whole non-deterministic, different-number-every-time shtick in the rules, they did say "if our language's integer maximum is smaller than 2,147,483,647" we can use our language's "integer maximum" instead, so I did... There are no integer types in ///, only strings. So the integer maximum is 0, sort of...
Well, it shaves off a few bytes!

*+* Runs Forever

*+* → . There's no Print instruction in this one so I can interpret it like a normal human. Replacing a pattern with itself (or another string containing itself), is the standard way to loop forever, and it's as succinct as it gets!

Note

To golf it a bit, /// looks for +* & ​ *​ rather than the whole thing. This works just as well. For the infinite loop, nothing ever prints because the replacing never stops, so you'll never find out that it's prepending an infinitely long list of * to the beginning every time it replaces. As for printing 'a random number,' we just don't replace one of the spaces, leaving a trailing space at the end of the output. Luckily, this hurts no one and is not noticeable. Plus, how's an ignoramus like me to know how many of the trailing spaces at the end of the Print instruction, until the margin, were on purpose :p

Answer 14

Perl 5 -p, 43 39 bytes

$_=/ /?0|rand~0:/\+/?redo:"Hello World"

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Answer 15

C# (Visual C# Interactive Compiler), 71 bytes

s=>{for(;s=="*+*";);return"*"==s?"Hello World":new Random().Next()+"";}

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Answer 16

Ruby -n, 47 bytes

puts~/ /?rand(1<<31):~/\+/?loop{}:"Hello World"

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Answer 17

Wolfram Language (Mathematica), 65 bytes

#/.{"*"->"Hello World"," * "->RandomInteger[2^31-1],_:>0~Do~∞}&

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Answer 18

Java (JDK), 83 bytes

s->{for(s=s.intern();s=="*+*";);return"*"==s?"Hello World":(-1>>>1)*Math.random();}

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Answer 19

Brachylog, 26 23 bytes

l₃∈&hṢ∧2^₃₁-₁ṙw∨Ḥ⊇ᶠ³⁶tw

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Takes the program as a string through the input variable, and ignores the output variable. Heavily exploits the guarantee that the input is only ever one of the three valid programs: any length-three input will behave like either " * " or "*+*" depending on whether or not the first character is a space, and any other input will behave like "*".

l₃                         The input has length 3
  ∈                        and is an element of something,
   &h                      and the input's first element
     Ṣ                     is a space
  ∈                        (if not, try some other thing it's an element of),
      ∧2^₃₁-₁              so take 2,147,483,647 and
             ṙw            print a random number between 0 and it inclusive.
               ∨           If the input's length isn't 3,
                Ḥ⊇ᶠ³⁶tw    print the 36th subsequence of "Hello, World!".

Answer 20

PHP, 51 bytes

for(;'*'<$l=$argn[1];);echo$l?rand():'Hello World';

Try it online! (Hello World)

Try it online! (Random Number)

Try it online! (Infinite Loop)

Takes second character of input which can be '', '*' or '+'. In case of '+' the '*'<'+' will be true and the loop will be infinite, else, after the loop, "Hello World" or a random number is shown. The rand() automatically outputs a number between 0 and getrandmax() which uses defined RAND_MAX in standard C library and by default is 2147483647 on most platforms/environments, including TIO.

Answer 21

APL (Dyalog Unicode), 39 bytes^SBCS

Anonymous prefix lambda.

{'+'∊⍵:∇⍵⋄' '∊⍵:⌊2E31×?0⋄'Hello World'}

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{ "dfn"; ⍵ is the argument:

 '+'∊⍵: if plus is a member of the argument:

  ∇⍵ tail recurse on argument

 ' '∊⍵ if space is a member of the argument:

  ?0 random float (0–1)

  2E31× scale to (0–2³¹)

  ⌊ floor

 'Hello World' else return the string

Answer 22

05AB1E, 21 bytes

'*KgDi[ë<ižIL<Ω딟™‚ï

Try it online. (NOTE: The random buildin is pretty slow with big lists, so it might take awhile before the result is given.)

Explanation:

'*K           '# Remove all "*" from the (implicit) input
   g           # Get the length of what's remain (either 0, 1, or 2)
    D          # Duplicate this length
     i         # If the length is exactly 1:
      [        #  Start an infinite loop
     ë<i       # Else-if the length is 2:
        žI     #  Push builtin 2147483648
          L    #  Create a list in the range [1,2147483648]
           <   #  Decrease each by 1 to make the range [0,2147483647]
            Ω  #  Pop and push a random value from the list
               #  (after which the top of the stack is output implicitly as result)
     ë         # Else:
      ”Ÿ™‚ï    #  Push dictionary string "Hello World"
               #  (after which the top of the stack is output implicitly as result)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ”Ÿ™‚ï is "Hello World".

Answer 23

Pyth, 32 bytes

It/Jw\*#;?tlJOhC*4\ÿ"Hello World

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Explanation (Python-ish)

I                                   # if 
  /Jw\*                             #    (J:=input()).count("*"))
 t                                  #                             - 1:
       #;                           #     try: while True: pass;except: break;
         ?                          # if (ternary)
           lJ                       #    len(J):
             O                      #     randInt(0,                    )
               C                    #                int(     , 256)
                *4\ÿ                #                    4*"ÿ"
              h                     #                                + 1
                    "Hello World    # else: (implicitly) print "Hello World"

Answer 24

JavaScript, 62 Bytes, no infinite recursive

s=>s[1]?eval("while(s>'!');Math.random()*2**31"):'Hello World'

Answer 25

JavaScript (V8), 79 bytes

-2 bytes thanks VFDan

s=>eval([`"Hello World"`,`Math.random()*2**31|0`,`for(;;);`][s.lastIndexOf`*`])

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Answer 26

Pip, ³⁹ 35 bytes

IaQ'*P"Hello World"EIsQ@aPRRmE{T0a}

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Simple if else based program, prints random integer between 0 and 1000.

Answer 27

Vyxal, 16 bytes

×=[kh|tð=[k×ʀ℅|{

Try it Online! The 5 flag is just so that you don't have to wait as long for it to timeout. Even at the default 60 second timeout, this times out on * , but you can see that it works by replacing k× with a smaller number.

Currently, this beats all other answers.

Explanation:

                  # Implicit input
×=[               # If input = "*":
   kh             #   Push "Hello World"
     |tð=[        # Else if last character of input = " ":
          k×      #   Push 2147483648
            ʀ     #   Range [0, 2147483648]
             ℅    #   Random integer from range
              |   #     Else:
               {  #       While True:  (Infinite loop)
                  # Implicit output

Answer 28

Charcoal, 30 bytes

Ｗ№θ*Ｆ⁼θ*≔Hello Worldθ∨θＩ‽Ｘ²¦³¹

Try it online! Link is to verbose version of code. Abuses Charcoal's default input format which splits on spaces if there is only one line, thus the random number input actually looks like three inputs. Explanation:

Ｗ№θ*

Repeat while the first input contains a *.

Ｆ⁼θ*

If the first input is a * only...

≔Hello Worldθ

... then replace it with Hello World, thus causing the loop to terminate. *+* doesn't get replaced, resulting in an infinite loop.

∨θ

If the first input is not empty then output it.

Ｉ‽Ｘ²¦³¹

But if it is empty then output a random integer in the desired range.

Answer 29

Add++, 78 bytes

z:"Hello World"
`y
xR2147483647
x:?
a:"*"
b:" * "
c:"*+*"
Ix=a,Oz
Ix=b,O
Wx=c,

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Explanation

z:"Hello World"	; Set z = "Hello World"
`y		; Set y as the active variable
xR2147483647	; Set y to a random number between 0 and 2147483647
x:?		; Set x to the input
a:"*"		; Set a = "*"
b:" * "		; Set b = " * "
c:"*+*"		; Set c = "*+*"
Ix=a,		; If x == a then...
	Oz	;	...output z
Ix=b,		; If x == b then...
	O	;	...output y
Wx=c,		; While x == c then...
		;	...do nothing

Answer 30

Commodore BASIC (VIC-20, C64, TheC64Mini etc) - 170 tokenize BASIC bytes

 0a%=32767:goS9:b$=leF(b$,len(b$)-1):ifb$="*"tH?"hello world
 1ifb$=" * "tH?int(rN(ti)*a%)
 2ifb$="*+*"tHfOi=.to1:i=.:nE
 3end
 9b$="":fOi=.to1:geta$:i=-(a$=cH(13)):b$=b$+a$:?a$;:nE:reT

I think to do this more accurately, I'll have to delve into the weird world of 6502 assembly language, but this is a first draft.

First point, the INPUT keyword in Commodore BASIC ignores white spaces, so the sub-routine at line 9 is a quick-and-dirty way to accept keyboard entries including spaces.

Second point, Commodore BASIC integers have a range of 16-bit signed, so -32768 to +32767 source - so I've kept the random number generated to 0 - 32767 inclusive