76
\$\begingroup\$

COBOL is a very old language, at the time of writing it is 58 years old. It is so old, in fact, that it has a very interesting quirk: the first six characters of each line are comments.

Why is this, you ask? Well, those 6 characters were intended to be used as line numbers, back in the day where programs weren't completely digital and typed out on a computer.

In addition, the seventh character could only be part of a very small set (it is usually * to comment out the line or a space to separate the line number from the code)

But what if you're on a more digital system, and you just want the raw program?

The comment system

There are two types of comments in COBOL: line comments and the aforementioned "line number" comments.

Uncommenting line numbers is simple: just take the first seven (six plus a single space) characters off each line.

000000 apple
000001 banana
celery donuts

would become:

apple
banana
donuts

Line comments make it a bit more difficult. A line comment is started with an asterisk * placed in the seventh character position on the line, like so:

000323* this is a comment

This is not a line comment:

*00000 this isn't a comment

To uncomment a line comment, just remove the whole line.

An example commented "program":

000000 blah blah
000001* apples
000002 oranges?
000003* yeah, oranges.
000*04 love me some oranges

The uncommented version:

blah blah
oranges?
love me some oranges

In other words, to uncomment a string, remove the first six characters of each line, then return all but the first character of every line that does not begin with a star.

The challenge

Create a program or function that takes a commented program and returns its uncommented variant.

Clarifications

  • Asterisks (*) will never be found anywhere outside the first seven characters on a line (we're not asking you to verify syntax)
  • Each line will always have at least 7 characters.
  • You may assume the seventh character is always an asterisk or a space.
  • Input or output may be a matrix or list.
  • Only printable ASCII characters (plus newline) must be handled.
  • You may output with a trailing newline. You may also assume that the input will have a trailing newline, if you so choose.

Scoring

Since this is , the answer with the least bytes wins!

DISCLAIMER: I do not actually know COBOL and do not claim to. If any of the claims about COBOL I have made in this question are incorrect, I take no responsibility.

\$\endgroup\$
18
  • 27
    \$\begingroup\$ Line numbers are not comments. They are a column. Terminology please. \$\endgroup\$
    – user207421
    Commented Aug 24, 2017 at 5:47
  • 2
    \$\begingroup\$ Your examples all have a space after the *. Is this a coincidence? \$\endgroup\$
    – Neil
    Commented Aug 24, 2017 at 10:13
  • 8
    \$\begingroup\$ Old does not automatically imply bad. I have worked in an Agile COBOL shop. They could do things on the AS/400 we could not do in Java. \$\endgroup\$ Commented Aug 24, 2017 at 12:06
  • 4
    \$\begingroup\$ Can there be a space in the first 6 characters? \$\endgroup\$
    – user4768
    Commented Aug 24, 2017 at 15:30
  • 3
    \$\begingroup\$ Column 7 in addition to a space, could also contain not only * (asterix) a comment but also - (hyphen) indicating a continuation line and / (forward slash) form feed \$\endgroup\$ Commented Aug 27, 2017 at 6:31

62 Answers 62

112
\$\begingroup\$

COBOL (GnuCOBOL), 191 + 17 = 208 bytes

I "learned" COBOL for this answer, so it's probably not fully golfed.

This is a full program, taking input on what I presume to be standard input and writing to what I presume to be standard output. Perhaps one day I'll return to this and (1) determine whether COBOL has functions and, if so, (2) see whether a function solution would be shorter.

Byte count includes program and compiler flags (-free and -frelax-syntax).

program-id.c.select i assign keyboard line sequential.fd i. 1 l pic X(80). 88 e value 0.open input i perform until e read i end set e to true end-read if not e and l(7:1)<>'*'display l(8:73).

Try It Online

Ungolfed program

program-id. c.

select i assign to keyboard organization line sequential.

fd i.
    1 l pic X(80).
    88 e value 0.

open input i
perform until e
    read i
        end set e to true
    end-read
    if not e and l(7:1) <> '*'
        display l(8:73).

Limitations

The output is, technically speaking, not correct. From my cursory research, it seems the only practical way to store a string in COBOL is in a fixed-size buffer. I have chosen a buffer size of 80 characters, since this is the line length limit for fixed-format programs. This presents two limitations:

  • Lines longer than 80 characters are truncated.
  • Lines shorter than 80 characters are right-padded with spaces.

I'm guessing this is acceptable since, well, it's COBOL. If not, I'd be willing to look into alternatives.

Acknowledgments

  • -166 bytes thanks to Edward H
  • -2 bytes thanks to hornj
\$\endgroup\$
7
  • 14
    \$\begingroup\$ Asterisks (*) will never be found anywhere outside the first seven characters on a line ... and yet ... ;) \$\endgroup\$
    – Cœur
    Commented Aug 24, 2017 at 17:43
  • \$\begingroup\$ @Cœur Haha yes...but my solution doesn't use that assumption, so maybe it's appropriate! \$\endgroup\$
    – Jakob
    Commented Aug 24, 2017 at 17:50
  • 10
    \$\begingroup\$ You win one internet. \$\endgroup\$
    – Joshua
    Commented Aug 25, 2017 at 18:58
  • \$\begingroup\$ @Cœur except in a COMPUTE statement. \$\endgroup\$
    – ClickRick
    Commented Aug 25, 2017 at 19:21
  • 3
    \$\begingroup\$ Congratulations on your gold badge! \$\endgroup\$ Commented Dec 21, 2017 at 17:25
22
\$\begingroup\$

Python 2, 39 38 37 bytes

-1 byte thanks to LyricLy. -1 byte thanks to Mego.

lambda s:[i[7:]for i in s if'*'>i[6]]

Try it online!

I/O as lists of strings.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Save a byte by replacing != with <, since a space's code point is lower than an asterisk's one, and the seventh character will always be a space or an asterisk. \$\endgroup\$
    – LyricLy
    Commented Aug 24, 2017 at 2:34
  • \$\begingroup\$ So, the 7th character will always be a space or an asterisk? \$\endgroup\$ Commented Aug 24, 2017 at 2:35
  • \$\begingroup\$ Yes. You may assume the seventh character is always an asterisk or a space. \$\endgroup\$
    – LyricLy
    Commented Aug 24, 2017 at 2:37
  • 1
    \$\begingroup\$ Save 1 byte with if'*'!=i[6] \$\endgroup\$
    – user45941
    Commented Aug 24, 2017 at 2:43
13
\$\begingroup\$

Perl 5, 19 + 1 (-p) = 20 16 bytes

-4 bytes with Pavel's suggestions

s/.{6}( |.*)//s

Try it online!

\$\endgroup\$
3
  • 2
    \$\begingroup\$ You can save three bytes if you replace (\*.*$| ) with ( |.*) \$\endgroup\$
    – Pavel
    Commented Aug 24, 2017 at 6:10
  • \$\begingroup\$ Not as short as @Pavel's comment, but / /;$_=/\* /?$,:$' is another alternative \$\endgroup\$ Commented Aug 24, 2017 at 7:35
  • \$\begingroup\$ You also don't need the ^. \$\endgroup\$
    – Pavel
    Commented Aug 24, 2017 at 7:39
11
\$\begingroup\$

V, 13 11 10 bytes

Î6x<<
çª/d

Try it online!

Explanation

Î       ' On every line
  x     ' delete the first...
 6      ' 6 characters
   <<   ' and unindent the line (removes the leading space)
ç /     ' on every line
 ª      ' matching \*
   d    ' delete the line

Hexdump:

00000000: ce36 783c 3c0a e7aa 2f64                 .6x<<.../d
\$\endgroup\$
4
  • \$\begingroup\$ Couldn't you do 7x instead of 6x<<? \$\endgroup\$
    – DJMcMayhem
    Commented Aug 24, 2017 at 2:43
  • 1
    \$\begingroup\$ Then it deletes the * \$\endgroup\$
    – nmjcman101
    Commented Aug 24, 2017 at 2:43
  • \$\begingroup\$ Would it work to delete the lins with * first and then do Î7x ? (assuming a * can not not in the chars 0-5) \$\endgroup\$ Commented Aug 28, 2017 at 11:00
  • \$\begingroup\$ @12431234123412341234123 unfortunately no, because there can be a * in the first 6 characters. \$\endgroup\$
    – nmjcman101
    Commented Aug 28, 2017 at 14:37
9
\$\begingroup\$

Paradoc (v0.2.8+), 8 bytes (CP-1252)

µ6>(7#;x

Takes a list of lines, and results in a list of uncommented lines.

Explanation:

μ        .. Map the following block over each line (the block is terminated
         .. by }, but that doesn't exist, so it's until EOF)
 6>      .. Slice everything after the first six characters
   (     .. Uncons, so now the stack has the 6th character on top
         .. and the rest of the line second
    7#   .. Count the multiplicity of factors of 7 in the character
         .. (treated as an integer, so '*' is 42 and ' ' is 32)
      ;  .. Pop the top element of the stack (the rest of the line)...
       x .. ...that many times (so, don't pop if the 6th character was a
         .. space, and do pop if it was an asterisk)

Hi, I wrote a golfing programming language. :)

I'm still developing this and added/tweaked a bunch of built-ins after trying to write this so that there are more reasonable ways to differentiate between a space and an asterisk than "7#", but I feel like that would make this noncompeting. It's fortunate that it still worked out (this only uses features from v0.2.8, which I committed three days ago).

\$\endgroup\$
4
  • \$\begingroup\$ "Hi, I wrote a golfing programming language." Was the version you're using released before or after this challenge was posted? \$\endgroup\$
    – Mast
    Commented Aug 24, 2017 at 19:18
  • 1
    \$\begingroup\$ It works on this version from three days ago: github.com/betaveros/paradoc/releases/tag/v0.2.8 \$\endgroup\$
    – betaveros
    Commented Aug 24, 2017 at 19:23
  • \$\begingroup\$ Right, you mentioned that, but somehow it didn't register explicitly... \$\endgroup\$
    – Mast
    Commented Aug 24, 2017 at 19:38
  • 2
    \$\begingroup\$ @Mast It doesn't really matter. \$\endgroup\$ Commented Aug 25, 2017 at 14:09
7
\$\begingroup\$

Octave, 23 bytes

@(s)s(s(:,7)~=42,8:end)

Try it online!

\$\endgroup\$
2
  • 4
    \$\begingroup\$ I never knew Octave could do strings like that... Take that, MATLAB. xD \$\endgroup\$
    – Sanchises
    Commented Aug 24, 2017 at 7:47
  • \$\begingroup\$ Since R2016b's introduction of string arrays, I'm pretty sure this would work in MATLAB too @Sanchises! I only have access to R2015b currently so can't confirm though. MATLAB can do it in 74 bytes, probably fewer @(s)cellfun(@(r)r(8:end),s(cellfun(@(r)r(7)~=42,s)),'uniformoutput',false), where s is a cell array, not a string array. Using regexp or something would likely be shorter, but the method I've written is comparable to the methodology in this answer just for old MATLAB \$\endgroup\$
    – Wolfie
    Commented Aug 25, 2017 at 15:51
6
\$\begingroup\$

Jelly, 11 9 bytes

ṫ€7Ḣ⁼¥Ðf⁶

Try it online!

Inputs and outputs as a list of lines.

-2 bytes thanks to @EriktheOutgolfer and @JonathanAllan

How it works

ṫ€7Ḣ=¥Ðf⁶
 €           On each line:
ṫ 7            Replace the line with line[7:]
      Ðf     Keep all lines that meet condition:
     ¥         Dyad:
   Ḣ             First Element (modifies line)
    =            Equals
        ⁶    Space
\$\endgroup\$
2
  • \$\begingroup\$ 7$€ can be €7 \$\endgroup\$ Commented Aug 24, 2017 at 9:29
  • \$\begingroup\$ take it down to 9 like so: ṫ€7Ḣ⁼¥Ðf⁶ \$\endgroup\$ Commented Aug 24, 2017 at 12:40
5
\$\begingroup\$

PowerShell, 32 bytes

$input-replace'^.{6}( |.*)'-ne''

Try it online!

Pipeline input comes in as an array of strings, -replace works on every string, and -ne '' (not equal to empty string) applied to an array, acts to filter out the blank lines.

\$\endgroup\$
5
\$\begingroup\$

Vim, 14 bytes

Ctrl-VG5ld:%g/\*/dEnter

Loading the input file as the buffer to edit, then enter the above commands. Output is the new buffer.

\$\endgroup\$
4
\$\begingroup\$

Pyth, 9 bytes

Note that this only works if at least 1 line is not a comment and at least 1 line is a comment. All the other solutions work in all cases.

-2 bytes thanks to @pizzakingme!

m>d7.m@b6

Try it here!

Explanation

m>d7.m@b6     - Full program with implicit input. Takes input as a list of Strings.

m>d7          - All but the first 7 letters of 
    .m   (Q)  - The input, filtered for its minimal value using the < operator on
      @b6     - the 7th character -- note that "*" is greater than " "
              - Implicitly Output the result.

Pyth, 11 bytes

tMfqhTdm>d6

Try it here!

Explanation

tMfqhTdm>d6 - Full Program with implicit input. Takes input as a list of Strings.

       m>d6 - Remove the first 6 characters of each line.
    hT      - Get the first character of each.
  fq  d     - Keep those that have the first character an asterisk.
tM          - Remove the first character of each.
            - Output Implicitly.

Pyth, 11 bytes

m>d7fqd@T6Q

Try it here!

Explanation

m>d7fq@T6dQ  - Full program. Takes input as a list of Strings.

      @T6    - The sixth character of each.
    fq   dQ  - Keep the lines that have a space as ^.
m>d7         - Crop the first 7 characters.
             - Output implicitly.

Pyth, 12 bytes

tMfnhT\*m>d6

Try it here!

Explanation

tMfnhT\*m>d6 - Full Program with implicit input. Takes input as a list of Strings.

        m>d6 - Remove the first 6 characters of each line.
    hT       - Get the first character of each.
  fn  \*     - Filter those that aren't equal to an asterisk.
tM           - Remove the first character of each.
             - Output Implicitly.

Pyth, 12 bytes

m>d7fn@T6\*Q

Try it here!

Explanation

m>d7fn@T6\*Q  - Full program. Takes input as a list of Strings.

      @T6     - Get the sixth character of each string
    fn   \*Q  - And filter those that aren't equal to an asterisk.
m>d7          - Crop the first 7 characters.
              - Output implicitly.
\$\endgroup\$
6
  • \$\begingroup\$ "Keep those that have the first character an asterisk." I think you meant "Keep those of which the first character is NOT an asterisk." \$\endgroup\$ Commented Aug 24, 2017 at 13:25
  • \$\begingroup\$ m>d7.m@b6 should work at 9 bytes, abusing that * is after space in lexicographical order \$\endgroup\$
    – Dave
    Commented Aug 24, 2017 at 13:27
  • \$\begingroup\$ I can edit it in with explanation / test link if you want! \$\endgroup\$
    – Dave
    Commented Aug 24, 2017 at 13:36
  • \$\begingroup\$ @pizzakingme I'd be glad if you would edit in, because I am on mobile. Thanks a lot and don't forget to credit yourself for the new solution! \$\endgroup\$
    – Mr. Xcoder
    Commented Aug 24, 2017 at 13:40
  • \$\begingroup\$ Does this work if all lines are line comments? (I'm not sure if you have to handle this case) \$\endgroup\$
    – betaveros
    Commented Aug 24, 2017 at 18:50
4
\$\begingroup\$

C (gcc), 53 48 46 bytes

x;main(y){for(y=&x;gets(y-6);x&2||puts(y+1));}

Try it online!

-5 bytes: It was very tricky to get this "whole program" down to the same size as gurka's function. It's now writing out of bounds (in both directions) of an array of wrong type and relies on little endian and 4 byte integers to find the asterisk ... but hey, it works ;)

-2 bytes: Well, if we already write to some "random" .bss location, why bother declaring an array at all! So here comes the string handling program that uses neither the char type nor an array.

\$\endgroup\$
2
  • \$\begingroup\$ Nice! And the *x&2 made me remember "You may assume the seventh character is always an asterisk or a space.", so I should be able to shave some bytes off my answer :-) \$\endgroup\$
    – simon
    Commented Aug 25, 2017 at 18:56
  • \$\begingroup\$ @gurka thanks :D -2, hehe \$\endgroup\$ Commented Aug 26, 2017 at 5:39
4
\$\begingroup\$

C, 63 59 55 48 47 46 bytes

Thanks to "an anonymous user" for getting rid off yet another byte.

Thanks to Felix Palmen for reminding me of "You may assume the seventh character is always an asterisk or a space.", which shaved off one more byte.

f(char**a){for(;*a;++a)(*a)[6]&2||puts(*a+7);}

Use like:

char** program = { "000000 apple", "000001 banana", "celery donuts", 0 };
f(program);

Try it online!

\$\endgroup\$
1
  • \$\begingroup\$ 44 bytes \$\endgroup\$
    – c--
    Commented Nov 7, 2022 at 17:23
4
\$\begingroup\$

SNOBOL4 (CSNOBOL4), 72 70 66 50 bytes

R	INPUT POS(6) (' '  REM . OUTPUT | '*') :S(R)
END

Try it online!

Pattern matching in SNOBOL is quite different from regex but the idea here is still the same: If a line matches "six characters and then an asterisk", remove it, otherwise, remove the first seven characters of the line and print the result.

This now actually takes better advantage of SNOBOL's conditional assignment operator.

The pattern is POS(6) (' ' REM . OUTPUT | '*') which is interpreted as:

Starting at position 6, match a space or an asterisk, and if you match a space, assign the rest of the line to OUTPUT.

\$\endgroup\$
3
\$\begingroup\$

Actually, 13 bytes

⌠6@tp' =*⌡M;░

Input and output is done as a list of strings.

Explanation:

⌠6@tp' =*⌡M;░
⌠6@tp' =*⌡M    for each line:
 6@t             discard the first 6 characters
    p            pop the first character of the remainder
     ' =         is it a space?
        *        multiply the string by the boolean - returns the string if true, and an empty string if false
           ;░  filter out empty strings

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Gaia, 9 bytes

6>¦'*«⁈ḥ¦

A function accepting a list of strings and returning a list of strings.

Try it online!

Explanation

6>¦        Remove the first 6 characters of each string
   '*«⁈    Filter out ones that start with *
       ḥ¦  Remove the initial space from each
\$\endgroup\$
3
  • \$\begingroup\$ I count ten characters, and since three are non-ASCII, do they not take more than a byte? \$\endgroup\$
    – WGroleau
    Commented Aug 24, 2017 at 6:35
  • \$\begingroup\$ @WGroleau and « are both 1 character. Out golfing languages that use non-ascii character (maybe except for Neim) use custom encodings, that allow all those non-ASCIIs to be counted as single bytes. Here is Gaia's code page. \$\endgroup\$
    – Mr. Xcoder
    Commented Aug 24, 2017 at 9:19
  • \$\begingroup\$ @Mr.Xcoder Neim has an encoding too. \$\endgroup\$ Commented Aug 24, 2017 at 9:58
3
\$\begingroup\$

Ruby, 39 38 36 29 23 22 20 + 1 = 21 bytes

$_[/.{6}( |.*
)/]=''

Try it online!

Uses -p flag.

Explanation:

The -p flag adds an implicit block around the code, so the code that actually gets run looks like this:

while gets
    $_[/.{6}( |.*
)/]=''

    puts $_
end

gets reads a line of text and stores its result in $_.

$_[/.../]='' removes the first occurence of the regex ... in $_.

/.{6}( |.*\n)/ matches 6 of any character at the start of a line, followed by either a space or the rest of the line. Because the space appears first, it will try to remove only the first 6 characters and a space before attempting to remove the entire line.

$_ is then printed out, and this process is repeated for each line.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Method calls in Ruby don't need parentheses, removing them will save you a byte. \$\endgroup\$
    – m-chrzan
    Commented Aug 24, 2017 at 3:07
3
\$\begingroup\$

GNU Sed, 19 + 2 = 21 characters

Requires -E argument to sed to enable extended regular expressions.

/^.{6}\*/d;s/^.{7}/
\$\endgroup\$
2
  • \$\begingroup\$ you could do the same thing as the perl guy, s/^.{6}( |.*)//g \$\endgroup\$ Commented Aug 24, 2017 at 21:01
  • \$\begingroup\$ which version of sed ignores absense of replacement substring for s command? \$\endgroup\$ Commented Nov 8, 2022 at 3:41
3
\$\begingroup\$

Java 8, 40 bytes

Regular expressions: just about, but not quite, the wrong tool for the job. Lambda from String to String (assign to Function<String, String>). Input must have a trailing newline.

s->s.replaceAll("(?m)^.{6}( |.*\\n)","")

Try It Online

Acknowledgments

\$\endgroup\$
1
  • \$\begingroup\$ The right tool! :) \$\endgroup\$ Commented Aug 25, 2017 at 8:41
3
\$\begingroup\$

Haskell, 27 25 bytes

Laikoni's version is shorter than mine:

f n=[x|' ':x<-drop 6<$>n]

Try it online!

My version:

f n=[drop 7x|x<-n,x!!6<'*']

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ 25 bytes: f n=[x|' ':x<-drop 6<$>n]. \$\endgroup\$
    – Laikoni
    Commented Aug 24, 2017 at 17:24
  • \$\begingroup\$ @Laikoni That's nice!! I didn't know it was possible to match patterns in a generator of a list comprehension. \$\endgroup\$
    – jferard
    Commented Aug 24, 2017 at 19:43
3
\$\begingroup\$

><>, 57 53 Bytes

>i~i~i~i~i~i~i67*=\
<o$/?:$/?=a:;?(0:i<
\~$/~\ $
/  <o\?/

try it online

Explanation

>i~i~i~i~i~i~i67*=    Read in the first seven bytes of the line
 i~i~i~i~i~i~         Read, and discard 6 characters
             i        Read the seventh
              67*=    Check if the seventh character was an 
                      asterisk (and leave that value on the stack );

<o$/?:$/?=a:;?(0:i<    Read characters until a newline or eof
                 i     Read the next character of the line
            ;?(0:      If it's a -1, terminate the program
       /?=a:           If it's a newline, break out of the loop
   /?:$                If the seventh character was not an asterisk
<o$                    Output this character
\~$/                   otherwise discard it

   /~\ $    Having reached the end of the line, output
/  <o\?/    the newline only if it was not a comment

Edit: 53 bytes

>   i~i~i~i~i~i~i67*=\
/?=a<o/?$@:$@:$;?(0:i<
~   \~/

Basically the same stuff as before, but restructured slightly

As a side note: I'm disappointed no-one's done this in cobol yet.

\$\endgroup\$
3
\$\begingroup\$

R, 47 45 bytes

function(x)gsub("(?m)^.{6}( |.*\\n)","",x,,T)
\$\endgroup\$
6
  • \$\begingroup\$ If you take the input as a list of strings I think you can shorten the regex to "^.{6}( |.*$)" for -6. \$\endgroup\$ Commented Aug 24, 2017 at 11:37
  • \$\begingroup\$ @CriminallyVulgar Correct. In this case, I could also drop the pe=T argument. However, I am not sure whether input as a list of strings is allowed. \$\endgroup\$ Commented Aug 24, 2017 at 12:18
  • \$\begingroup\$ From the OP: Input or output may be a matrix or list. \$\endgroup\$ Commented Aug 24, 2017 at 13:05
  • \$\begingroup\$ @CriminallyVulgar The problem is the presence of empty strings in the output. \$\endgroup\$ Commented Aug 24, 2017 at 13:24
  • \$\begingroup\$ @SvenHohenstein Ah of course, I hadn't considered that. \$\endgroup\$ Commented Aug 24, 2017 at 14:57
3
\$\begingroup\$

CLC-INTERCAL, 116 bytes.

DO;1<-#6DO,1<-#999DOCOMEFROM.1DOWRITEIN;1+,1(91)DO.1<-,1SUB#2(1)DO.2<-.1~#1DDOCOMEFROM.2DO,1SUB#3<-#0(95)DOREADOUT,1

Try it online, although you need to copy and paste!

Assumption

  • Each line should have <=80 characters except trailing newline character(s), although this program can handle up to 339 characters per line (or more, depending on what kind of characters are used).
  • Does not matter whether input has trailing newline or not.

Explaination

DONOTE stores first six characters of line
DO;1<-#6
DONOTE stores rest of characters of line except trailing newline
DO,1<-#999
DONOTE label one
DOCOMEFROM.1
DOWRITEIN;1+,1
DONOTE goto label one if tail one is asterisk
DONOTE first two items of tail one shall be
DONOTE 91 and 95 if space, or
DONOTE 91 and 95 if asterisk
(91)DO.1<-,1SUB#2
DONOTE goto label two if line is not empty
DONOTE obtw every item of tail will be zero
DONOTE if empty line
(1)DO.2<-.1~#1
DONOTE die here
D
DONOTE label two
DOCOMEFROM.2
DONOTE erase the space
DONOTE obtw zeros on tail are ignored
DONOTE when reading out
DONOTE see source code for more info
DO,1SUB#3<-#0
DONOTE goto label 1 after reading out
(95)DOREADOUT,1
\$\endgroup\$
3
\$\begingroup\$

APL, 23 bytes

  • 23 bytes assuming SBCS. In UTF-8 it'll be 42 bytes

This is my first ever attempt at a code golf challenge, feel free to improve, i'm sure i could have saved one or two characters if i wasn't so adamant on making this code 'tacit'

This assumes the input is a list of lines and return a list of uncommented lines, this makes everything a bit easier since APL was explicitly made to work on list!

(('*'≠7∘⌷)⊢⍤/(8≤⍳8)∘⊂)¨

Go ahead and try it! (Thanks @Sʨɠɠan)

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Welcome to Code Golf! It's typically allowed to take I/O how you did; see Default for Code Golf: Input/Output methods \$\endgroup\$ Commented Nov 8, 2022 at 0:50
  • 1
    \$\begingroup\$ You can count it as 23 bytes, as there's an SBCS (single-byte character set) to cover the set of characters: github.com/abrudz/SBCS \$\endgroup\$
    – naffetS
    Commented Nov 8, 2022 at 1:15
  • 1
    \$\begingroup\$ Here's a link so others can test your code: try it online! \$\endgroup\$
    – naffetS
    Commented Nov 8, 2022 at 1:15
  • \$\begingroup\$ Thanks a lot for the feedback! \$\endgroup\$ Commented Nov 8, 2022 at 1:49
2
\$\begingroup\$

Pyke, 9 bytes

36 4C 3E 23 64 DE 29 6D 74

Try it here!

Readable:

6L>#d.^)mt

Try it here!

6L>        -   [i[6:] for i in input]
   #d.^)   -  filter(i.startswith(" ") for  i in ^)
        mt - [i[-1:] for i in ^]
\$\endgroup\$
2
\$\begingroup\$

Retina, 23 15 bytes

5 bytes saved thanks to nmjcman101
1 byte saved thanks to Neil

m`^.{6}( |.*¶)

Try it online!

\$\endgroup\$
0
2
\$\begingroup\$

JavaScript, 44 34 bytes

Crossed-out 44 is still regular 44.

6 bytes saved thanks to tsh

a=>a.replace(/^.{6}( |.*\n)/gm,'')

Try it online!

\$\endgroup\$
6
  • \$\begingroup\$ s=>s.replace(/^.{6}( |\*.*\s)?/mg,'') \$\endgroup\$
    – tsh
    Commented Aug 24, 2017 at 2:44
  • \$\begingroup\$ s.match(/(?<=^.{6} ).*/mg) ESNext (Nonstandard, Stage3) Chrome62+ \$\endgroup\$
    – tsh
    Commented Aug 24, 2017 at 2:52
  • \$\begingroup\$ @tsh. Until there is a stable interpreter which allows it, I suppose it doesn't count as a valid programming language. \$\endgroup\$
    – user72349
    Commented Aug 24, 2017 at 3:02
  • \$\begingroup\$ It doesn't seem like this gives the correct output if the last line is a comment line. \$\endgroup\$
    – LyricLy
    Commented Aug 24, 2017 at 3:42
  • \$\begingroup\$ @LyricLy. It is because I assumed that input will always contain a trailing new line. You can see that it works is there is an empty line after the input. If I should not assume it, then fix will cost 1 byte (adding ? after \n). \$\endgroup\$
    – user72349
    Commented Aug 24, 2017 at 3:55
2
\$\begingroup\$

JavaScript (ES6), 48 bytes

s=>s.map(c=>c[6]<"*"?console.log(c.substr(7)):1)

Try it online!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ This is neither a function, nor a full program as it assumes the input is stored in z, which isn't allowed here. You could however turn it into an anonymous arrow function, in order to make it valid. \$\endgroup\$ Commented Aug 27, 2017 at 11:35
  • 1
    \$\begingroup\$ @cairdcoinheringaahing you're absolutely right. Updated the solution - not sure what the rules are regarding the ( and ) surrounding the fn, added them to be sure. \$\endgroup\$
    – sgtdck
    Commented Aug 27, 2017 at 13:03
  • 1
    \$\begingroup\$ You don't need the () around the function, but otherwise it looks fine. \$\endgroup\$ Commented Aug 27, 2017 at 13:06
2
\$\begingroup\$

C#, 160 145 90 89 bytes

t=>{var o="";foreach(var s in i.Split('\n'))if(s[6]!=42)o+=s.Substring(7)+"\n";return o;}

Thanks to Pavel & auhmaan for reducing the size.

\$\endgroup\$
8
  • \$\begingroup\$ Welcome to PPCG! I would suggest adding a try it online link to your answer so that others can test your code. Besides that, great first (well, second) answer! \$\endgroup\$
    – LyricLy
    Commented Aug 24, 2017 at 3:52
  • \$\begingroup\$ You can make this shorter by writing a lambda in the form t=>{...} \$\endgroup\$
    – Pavel
    Commented Aug 24, 2017 at 5:46
  • \$\begingroup\$ @LyricLy I tried doing that, actually, but for some reason, this doesn't work there. It works perfectly fine in a VS console app, though. \$\endgroup\$
    – snorepion
    Commented Aug 24, 2017 at 13:07
  • \$\begingroup\$ @Pavel Like so? I'm not sure if I did it completely correctly; I've never needed to use a lambda expression before. \$\endgroup\$
    – snorepion
    Commented Aug 24, 2017 at 13:08
  • \$\begingroup\$ Yes, exactly. You can test it by assigning it to a func<string, string>. \$\endgroup\$
    – Pavel
    Commented Aug 24, 2017 at 16:36
2
\$\begingroup\$

Python 3, 62 bytes (no regex)

def f(s):
 for w in s.split('\n'):
  if w[6]!='*':print(t[7:])

It works!

>>> s="""000000 blah blah
000001* apples
000002 oranges?
000003* yeah, oranges.
000*04 love me some oranges"""
>>> f(s)
blah blah
oranges?
love me some oranges
\$\endgroup\$
2
\$\begingroup\$

sed, 11 bytes + 3 bytes option -nr = 14 bytes.

s/^.{6} //p

Try it online!

sed, 16 bytes

#n
s/^.\{6\} //p

Try it online!

#n on top of line is for -n, which is GNU-extension POSIX-compatible.

\$\endgroup\$

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