Generate ;# code

Question

Related to: Make a ;# interpreter

In the above linked challenge the task was to create an interpreter for the esoteric language ;#.

The ;# language

The language has exactly two commands: ; and # (all other characters are ignored by the interpreter):

;: Increment the accumulator

#: Modulo the accumulator by 127, print the corresponding ASCII character and reset the accumulator to 0.

Challenge

Because I am lazy but still want to test some more testcases, I need a program or function which converts plain text to ;# code.

Input

The input is a string, taken either as argument or through stdin. It will only contain printable ASCII characters and newlines.

Output

The output is the generated ;# program by returning, or printing to stdout. As long as the program is valid, it may contain excess characters other than # and ; as all other characters are ignored.

Examples

Input: Hello, World!
Output: ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

Input: ABC
Output: ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

Input: ;#
Output: ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

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Answer 1

Husk, 13 10 bytes

ṁȯ`:'#R';c

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Ungolfed/Explained

ṁȯ          -- map function and concatenate result
         c  --   ASCII order of element times
      R';   --   repeat ';' and
  `:'#      --   append '#'

Answer 2

Python 3, 51 bytes

[print(''.join((';'*ord(i)+'#')for i in input()))] 

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Answer 3

q/kdb+, 24 bytes

Solution:

{,["#"sv(7h$x)#'";"]"#"}

Explanation:

q is interpreted right-to left. We cast the input to it's ASCII values, thus "Hello, World!" becomes 72 101 108 108 111 44 32 87 111 114 108 100 33. We then take this many of the ; character, which gives us a list of lists. These lists are joined together with #, then we a # on the end to terminate the string.

{,["#"sv(7h$x)#'";"]"#"} / the solution
{                      } / lambda function
 ,[                ]"#"  / concatenate with "#"
              #'         / apply each-left take (#) each-right
                ";"      / the right (the character ;)
        (7h$x)           / the left (input cast to ASCII values)
   "#"sv                 / join these lists back together with "#"

Bonus:

Could save 1 byte moving the cast outside the function, but that's a bit underhand:

{,["#"sv x#'";"]"#"}7h$

Taking inspiration from the APL solution and the above yields a 22 byte solution:

{raze(x#'";"),'"#"}7h$

which becomes 20 bytes in k

{,/(x#'";"),'"#"}7h$

Answer 4

Haskell, 50 32 bytes

This is a direct port of my Husk answer:

((\n->(';'<$['\1'..n])++"#")=<<)

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Thanks @Laikoni for saving me 18 bytes!

Answer 5

Charcoal, 8 bytes

ＦＳ⁺×;℅ι#

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Explanation (for those too lazy to click the link)

Ｆ         For
 Ｓ        each character in next input as string
   ⁺    #  Append "#"
    × ℅ι   Multiply (repeat) by the character code of i (loop variable)
     ;     ";"

Answer 6

Common Lisp, 71 bytes

(lambda(s)(map'list(lambda(x)(format t"~v,,,v<~>#"(char-code x)#\;))s))

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Answer 7

SNOBOL4 (CSNOBOL4), 90 bytes

	N =INPUT
T	N LEN(1) . L REM . N :F(END)
	&ALPHABET L @X
	OUTPUT =DUPL(';',X) '#' :(T)
END

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	N =INPUT			;* read input
T	N LEN(1) . L REM . N :F(END)	;* set L to the first character, N to the REMainder, and
					;* if there is no match (N is empty), goto END
	&ALPHABET L @X			;* @ssign X with the value of the match of L inside &ALPHABET (all ASCII chars)
	OUTPUT =DUPL(';',X) '#' :(T)	;* set OUTPUT to ';' duplicated X times concatenated with '#', and a newline
					;* then goto T
END

Answer 8

Gol><>, 11 bytes

iEHr{R`;`#r

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How it works

iEHr{R`;`#r

i            Input n as char
 EH          If EOF, print stack contents as chars (from top) and exit

             Stack content                     [x y z... n]
   r{        Reverse the whole stack except n  [...z y x n]
     R`;     Pop n and push that many ';'s     [...z y x ;...;]
        `#   Push '#'                          [...z y x ;...; #]
          r  Reverse stack                     [# ;...; x y z...]
             Repeat indefinitely

The r...r structure makes the invariant that the stack content is the desired string reversed.

Answer 9

J, 23 Bytes

Nothing special.

,([:,&'#'#&';')"0 u:inv

Explanation:

,([:,&'#'#&';')"0 u:inv
                  u:inv   | Convert string to code points
 (            )"0         | For each item:
  [:,&'#'#&';'            | Repeat ';' n times and append #
,                         | Ravel the resulting character array

Note that the result contains lots of whitespace, since arrays have to be rectangular.

Answer 10

Java 10, 63 bytes

c->{var s="";for(var d:c){for(;d-->0;)s+=";";s+="#";}return s;}

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Ungolfed version:

c -> { // lambda taking a char[] as argument and returning a String
    var s = ""; // the output String
    for(var d : c) { // iterate over all the input chars
        for(; d-- > 0 ;) // as many times as the ASCII value of the current char
            s += ";";    // append ";" to the program to compute the character
        s += "#"; // append "#" to output the character
    }
    return s; // return the finished program
}

Answer 11

Ahead, 13 bytes

~W$:ki@j";#"~

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Answer 12

Noether, 21 bytes

I~sL(";"si/B*P"#"P!i)

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Explanation:

I~s                   - Get input and store in the variable s
   L(               ) - Loop until the top of stack equals the length of s
     ";"              - Push the string ";"
        si/           - Return the character of string s at position i
           B          - Return the ASCII code of the character
            *P        - Multiply ";" by the ASCII code and print
              "#"P    - Push the string "#" and print
                  !i  - Increment i

Answer 13

Pascal (FPC), 94 bytes

var c:char;i:word;begin repeat read(c);for i:=1to ord(c)do write(';');write('#')until eof end.

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Assumes non-empty input.

Answer 14

VBA (Excel) 82 Bytes

I really think this challenge should've been scored in the length of your program after it is converted into ;#. I would score 6769 characters (including newlines)

Function d(e)
  For i=1To Len(e)
    d=d &String(Asc(Mid(e,i,1)),";") &"#"
  Next
End Function

Note: the leading spaces aren't needed, just make it more read-able, not that I should be worried about that... just look at the output from this thing!!!

Un-golfed With Comments:

Function toSemiHash(translate)
    'For 1 to the length of the string
    For i = 1 To Len(translate)
        'Append a semicolon the char-code number of times along with a hash
        fromSemiHash = fromSemiHash & String(Asc(Mid(translate, i, 1)), ";") & "#"
    Next
End Function

Answer 15

PHP, 75 Bytes

This shorter then the parser I wrote for it.

Generator

function g($s){foreach(str_split($s)as$a)echo str_repeat(';',ord($a)).'#';}

You can test it like this:

$strings = [
    'Hello, World!',
    ';#',
    '!'
];

foreach($strings as $s){
    g($s); echo "\n\n";
}

function g($s){foreach(str_split($s)as$a){echo str_repeat(';',ord($a)).'#';}}

Parser (second attempt)

The one from my first attempt should work fine as well. This one is shorter, so well just use it to verify that the generator works correctly.

function f($s){foreach(split('#',$s)as$a)echo!$a?'':chr(substr_count($a,';')%127);}

You can test both like this:

//turn of warning message or set in PHP.ini
error_reporting(0);
$s = "Hello, World!";

function g($s){foreach(str_split($s)as$a)echo str_repeat(';',ord($a)).'#';}
function f($s){foreach(split('#',$s)as$a)echo!$a?'':chr(substr_count($a,';')%127);}

//output buffer the generator
ob_start();
    g($s);
$s=ob_get_clean();
f($s);

Output

 Hello, World!

Test both of them

A few things to note:

• Because neither one return information, we can capture the output by using output buffering.
  • We could also put it a file, use $argv[1] for input, remove the function calls, add the shebang #!/bin/env php and run it from the terminal. Which would cut it down to around something like 65. Then we could just pipe that into the other script. So :-P
• Because the parser is old, we have to use PHP < 7 for it. Don't worry our dev team is currently working hard on a patch to make it compatible with PHP7,(just change split to explode). They should have it done sometime in the spring after they quite code golfing and get back to work!

Lastly

My first attempt at a parser was recursive, so I thought it only fitting I make a generator that is also recursive. Unfortunately as with the first one, it's longer 89 bytes.

 function g($s,$i,$r){echo''==($a=$s[$i])?$r:g($s,++$i,$r.=str_repeat(';',ord($a)).'#');}

You can test it with the same code above:

error_reporting(0);
$s = "Hello, World!";

//function g($s){foreach(str_split($s)as$a)echo str_repeat(';',ord($a)).'#';}
function f($s){foreach(split('#',$s)as$a)echo!$a?'':chr(substr_count($a,';')%127);}

//recursive
function h($s,$i,$r){echo''==($a=$s[$i])?$r:h($s,++$i,$r.=str_repeat(';',ord($a)).'#');}

//output buffer the generator
ob_start();
   // g($s);
   h($s,0,'');
$s=ob_get_clean();
f($s);

Recursive Generator

Things that didn't make the cut

  //parser 98 bytes, strtok + do while
  function p($s){$f=strtok;$a=$f($s,'#');do{echo chr(substr_count($a,';')%127);}while($a=$f('#'));}

Answer 16

brainfuck, 48 bytes

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

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How it works

+++++++[>+++++>++++++++<<-]>>+++>       'write ; and # onto the tape
,[                                      'read input as ascii values and loop
    [-<.>]                              'decrement input 1 and write ; then repeat until 0
    <<.>>                               'write #
,]                                      'read input as ascii values and loop

Answer 17

J, 23 bytes

[:,[:('#',~#&';')"+3&u:

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[:,(35,~#&59)"+&.(3&u:)

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Posting since the existing J answer is incorrect.

How it works

[:,[:('#',~#&';')"+3&u:
                   3&u:  Convert to ASCII value
     (          )"+      Apply to each number...
           #&';'           Convert to that many ';'s
      '#',~                and add '#' to the right
   [:                    Treat the above as monadic fork
  ,                      Ravel; flatten the multidimensional array
[:                       Treat the above as monadic fork

[:,(35,~#&59)"+&.(3&u:)
              f&.g       Apply g, f, then inverse of g...
                 (3&u:)    g: convert to ASCII value
   (        )"+            f: apply to each number...
        #&59                 convert to that many 59's (charcode of ';')
    35,~                     and add 35 to the right (charcode of '#')
[:,                      Ravel the result

Both use the "+ trick to avoid numeric literal problem.

Answer 18

Whitespace, 82 bytes

[N
S S N
_Create_Label_OUTER_LOOP][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_STDIN_as_character][T   T   T   _Retrieve][N
S S S N
_Create_Label_INNER_LOOP][S N
S _Duplicate][N
T   S T N
_If_0_Jump_to_Label_HASTAG][S S S T N
_Push_1][T  S S T   _Subtract][S S S T  T   T   S T T   N
_Push_59_;][T   N
S S _Print_as_character][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T   N
_Create_Label_HASHTAG][S S S T  S S S T T   N
_Push_35_#][T   N
S S _Print_as_character][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Start OUTER_LOOP:
  Integer c = read character from STDIN
  Start INNER_LOOP:
    If(c == 0):
      Jump to function HASHTAG
    c = c - 1
    Print ";" to STDOUT
    Go to next iteration of INNER_LOOP

function HASHTAG:
  Print "#" to STDOUT
  Go to next iteration of OUTER_LOOP

Note: All characters read from STDIN in Whitespace are automatically stored in the heap as their code-point values.

Answer 19

Backhand, 27 bytes

}o" i# :" }]@_!}|{:" [; o".

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I'm sure a shorter solution exists, given this has 6 no-ops. This prints a null byte to begin with

Explanation:

}o       Print a null byte
    i  :  }]  ! |{     Get a character of input and check if it is EOF
                   "  ; o"   If it is not, print a ;
               }| :  [       Decrement the counter and check again
                   Once the counter hits zero exit the loop
            @_    If the value was -1 (EOF) exit
           ]  !   Otherwise execute some garbage
 o" i#  "         Print a # and get the next character of input

Answer 20

GolfScript, 10 bytes

{";"*"#"}%

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Answer 21

Burlesque, 13 bytes

m{'#j**';j.*}

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m{      # For each char in string (implicit concat)
  '#    # Push #
  j**   # Find ord(char)
  ';j.* # That many ;
 }

Answer 22

W j d, 7 bytes

The input format is [['a','b',..]]

◘Pp½h#û

Explanation

Uncompressed:

C';*'#+M

       M % For every item in the input:
C        % Convert to a character
 ';*     % Repeat ; ASCII codepoint times
    '#+  % Append a #

Flag:j   % Join without newline
```

Answer 23

Raku, 44 bytes

say $*IN.slurp.ords.map(';'x*).join('#'),'#'

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Answer 24

AWK, 52 49 + 8 = 57 bytes

-lordchr command-line parameter to import the ORDCHR extension.

BEGIN{RS=".|"}{for(;i++<ord(RT);)s=s";"}i=$0=s"#"

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This code roams the input one character at a time, and prints as many ; as needed, and a # between characters. To print everything in one line, just put ORS="" inside the brackets of the BEGIN pattern.

Answer 25

Lua, 57 bytes

(...):gsub('.',load'print((";"):rep((...):byte()).."#")')

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Essentially grabs the byte value of each character and repeats ; that many times, followed by a # and a newline.

Answer 26

Thunno 2 J, 7 bytes

C';×'#+

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Explanation

C';×'#+  # Implicit input
C        # Cast to ordinals
 ';×     # Repeat ";" that many times
    '#+  # Append "#" to each
         # Implicit output, joined

Answer 27

C, 51 bytes

c;main(){for(;c||~(c=getchar());puts("#;"+!!--c));}

Outputs lots of extra newlines, and an extra ; at the end, but neither of those affect the output of the ;# program, so it should be fine.

Using puts was suggested by ais523 in a comment: Make a ;# transpiler.

~(x) is a shorter way to write (x)!=-1.

"#;"+!!--c is a pointer arithmetic equivalent of --c?";":"#;".