17
\$\begingroup\$
  • Take a line of input (function arguments or stdin, etc.).
  • Remove comments in it, where a comment starts and ends with ".
  • Do not remove comments in string literals, which start and end with '. String literals do not contain escapes, because smalltalk has nice, easily lexable syntax ;).
  • Output this string.
  • The number of "s not inside strings and the number of 's not in comments in the input are both guaranteed to be even.
  • Because removing comments makes programs more golfy, this is : shortest code per language wins.

Examples

input => output

123 => 123
1"2"3 => 13
1""23 => 123
'12' => '12'
'1"2'3"" => '1"2'3
'1"2"3' => '1"2"3'
a"1'2'3"a => aa
"a"b"c" => b
'a'b'c' => 'a'b'c'
a'"'b'"'c => a'"'b'"'c
a"'"b"'"c => abc
Regex is boring as usual. Try something new, or answer in smalltalk!
\$\endgroup\$
3
  • \$\begingroup\$ sandboxed ;) \$\endgroup\$
    – Wezl
    Jun 10 at 13:10
  • 1
    \$\begingroup\$ This would be a good use-case for graphical programming languages for statecharts! Unfortunately, graphical programming languages are not good for golfing. \$\endgroup\$ Jun 10 at 13:28
  • 3
    \$\begingroup\$ @MartinRosenau Remember that this isn't a competition between languages. I'd still be interested in a solution using one of those languages, even if it isn't competitive with a dedicated golfing language. \$\endgroup\$
    – Wezl
    Jun 10 at 13:36

12 Answers 12

9
\$\begingroup\$

Smalltalk, 325 bytes

[:k|k inject:OrderedCollection new into:[:a :b|(a isEmpty not and:[a last=0])ifTrue:[(b=$')ifTrue:[a removeLast. a add:b. a]ifFalse:[a removeLast. a add:b. a add:0. a]]ifFalse:[(a isEmpty not and:[a last=$"])ifTrue:[(b=$")ifTrue:[a removeLast. a]ifFalse:[a]]ifFalse:[(b=$')ifTrue:[a add:b. a add:0. a]ifFalse:[a add:b. a]]]]]

Some hot garbage smalltalk code.

It essentially uses a stack, pushing a 0 for a placeholder for when it sees '. When it sees ", it keeps in on the top of the stack until it sees another ".

I couldn't get an Eval function (smalltalk docs are terrible), nor could I get regex to work (smalltalk docs are terrible).

You can use GNU Smalltalk to run this, in theory, although I only tested it on the online compiler (linked above).

Here's a (more) readable version that takes from stdin and writes to stdout. As you can tell smalltalk is a beautiful language...

Transcript show:(stdin nextLine inject:
    OrderedCollection new into:
        [:a :b|
        
        Transcript show: a printString; cr.

        (a isEmpty not and:[a last = 0])ifTrue: [
            (b = $') ifTrue: [
                a removeLast.
                a add: b.
                a
            ] ifFalse:[
                a removeLast.
                a add: b.
                a add: 0.
                a
            ]
        ] ifFalse: [
            (a isEmpty not and:[a last = $"]) ifTrue: [
                (b = $") ifTrue: [
                    a removeLast.
                    a
                ] ifFalse: [a]
            ] ifFalse: [
                (b = $') ifTrue: [
                    a add: b.
                    a add: 0.
                    a
                ] ifFalse: [
                    a add: b.
                    a
                ]
            ]
        ]
]).

Note: the regex solution almost works, using the same approach as Jakque, but it fails since smalltalk gives nil instead of an empty string:

Transcript show:(stdin nextLine replacingAllRegex: '".*?"|(''.*?''|.)' with: '%1')
\$\endgroup\$
4
  • \$\begingroup\$ By "the online compiler" you mean codingground? \$\endgroup\$
    – Wezl
    Jun 10 at 16:12
  • \$\begingroup\$ yes I do mean that \$\endgroup\$
    – Riolku
    Jun 10 at 16:17
  • \$\begingroup\$ I didn't know about character literals in smalltalk. Because of those, this challenge won't work on real smalltalk code :/ \$\endgroup\$
    – Wezl
    Jun 10 at 16:35
  • \$\begingroup\$ Transcript show: 'a"''"b"''"c' "works", but single quotes get escaped by placing... another single quote. Edit: nvm it doesn't work since it's not parsed as code, duh. \$\endgroup\$
    – Riolku
    Jun 10 at 16:45
6
\$\begingroup\$

APL (Dyalog Extended), 30 26 bytes (SBCS)

Anonymous prefix lambda.

{⍵/⍨≠\⍛⍱⍨<⌿≠\@1⊢'''"'∘.=⍵}

Try it online!

{}dfn; argument is :
  `a"1'2'3"a'1"2"3'a"'"b"'"c`

'''"'∘.=⍵ equality table for ' and " versus the argument
   [[0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0],
    [0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0]]

≠\@1⊢ running parity (lit. XOR scan) at the 1st row (lit XOR scan) indicating characters inside strings
   [[0,0,0,1,1,0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,1,0,0,0],
    [0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0]]

<⌿ second row, i.e. comment delimiters, and not first row, i.e. inside string, i.e. active comment delimiters (lit. vertical less-than reduce)
   [0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0]

≠\⍛⍱⍨ neither that (active comment delimiters) NOR its running parity (inside comments)
   [1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,1]

⍵/⍨ use that to filter the characters of the argument
   `aa'1"2"3'ac`

\$\endgroup\$
2
  • \$\begingroup\$ it's nice to see that the regex solution and the cool and refreshing solution are the same length in APL \$\endgroup\$
    – Wezl
    Jun 10 at 14:36
  • 1
    \$\begingroup\$ @Wzl Shorter, even. \$\endgroup\$
    – Adám
    Jun 10 at 15:57
6
\$\begingroup\$

Jelly, 48 bytes

iⱮ⁾'"o1:0¤>/
ṣ”"2œPj€”"µẹ”'2ị‘œṖµÇ?;ß}¥/$¹f⁾'"$?

Try it online!

This is definitely not the best approach, even considering Jelly is typically not great with strings.

iⱮ⁾'"o1:0¤>/  Helper Link
iⱮ⁾'"         Find the first ' and "; 0 if not found
     o        Logical OR with
      1:0¤    1 / 0 (infinity) - that way, "not found" is at the end, not the start
          >/  Is the first greater than the second?

ṣ”"2œPj€”"µẹ”'2ị‘œṖµÇ?;ß}¥/$¹f⁾'"$?  Main Link
                                  ?  If
                             f⁾'"$   There are any ' or "
                     ?     $         - If
                    Ç                - " was found first
ṣ”"                                    - Split on "
   2œP                                 - Split over and discard the second item
      j€”"                             - Join each of those with " (basically, this slices out
                                         the part between and including the first and second ",
                                         and leaves the part before and after as two sublists)
                                     - Otherwise
           ẹ”'                         - Find all indices of '
              2ị                       - Get the second of these
                ‘                      - And increment it
                 œṖ                    - And partition the list at that point (basically, this
                                         divides the list at the second ' without removing anything)
                         ¥/          - Then, reduce (apply to first and second as left and right) by
                      ;ß}            - Concatenating the left side with this function applied to the right link
                            ¹        Otherwise, if ' and " aren't found, just return the string itself
```
\$\endgroup\$
2
  • \$\begingroup\$ can we get an explanation of how it works? \$\endgroup\$ Jun 11 at 18:19
  • 2
    \$\begingroup\$ @NooneAtAll I usually don't do much work on answers anymore after they've been beaten by so much, but here you go. \$\endgroup\$
    – hyper-neutrino
    Jun 11 at 18:49
5
\$\begingroup\$

Jelly, 17 bytes

⁾'"iⱮnaoɗ\ŻṖoƊỊx@

Try it online!

A link taking a Jelly string argument and returning a Jelly string without the comments.

Explanation

⁾'"iⱮ              | Positions of each character in '" (so single quote -> 1, double quote -> 2, anything else -> 0)
         ɗ\        | Cumulative reduce (x,y)
     nao           | - (x != y) and (x or y)
              Ɗ    | Following as a monad:
           Ż       | - Prepend zero
            Ṗ      | - Remove last item from list
             o     | - Or
               Ị   | Less than or equal to 1
                x@ | Original input with 0 or 1 copies of each character as appropriate
\$\endgroup\$
3
  • \$\begingroup\$ sadly you missed the fgitw :/ \$\endgroup\$
    – Wezl
    Jun 10 at 19:56
  • \$\begingroup\$ @Wzl what do you mean? \$\endgroup\$ Jun 10 at 19:56
  • \$\begingroup\$ fastest gun in the west \$\endgroup\$
    – Razetime
    Jun 11 at 3:42
4
\$\begingroup\$

QuadR g, 18 12 bytes

‒6 thanks to Neil.

'.*'
".*"
&

Try it online!

This is equivalent to the Dyalog APL function '''.*''' '".*"'⎕R'&' ''⍠'Greedy'0

The g flag turns off greedy patterns, essentially making * mean *?. Then the two patterns simply replace '.*' strings and ".*" comments with & themselves and  nothing, respectively.

\$\endgroup\$
5
  • 3
    \$\begingroup\$ okay now we've gotten regex out of the way :P \$\endgroup\$
    – Wezl
    Jun 10 at 13:22
  • \$\begingroup\$ No non-greedy quantifiers? \$\endgroup\$
    – Neil
    Jun 10 at 19:29
  • \$\begingroup\$ @Neil Wouldn't make a difference, as '[^']*' matches exactly all the chars between ' and the next '. The important thing is that the two patterns run in parallel, with, for each starting position, preference for the first one. \$\endgroup\$
    – Adám
    Jun 10 at 19:57
  • \$\begingroup\$ Sure, but Retina lets me use '.*?' which is shorter. \$\endgroup\$
    – Neil
    Jun 10 at 20:52
  • \$\begingroup\$ @Neil D'uh, of course. Thank you! \$\endgroup\$
    – Adám
    Jun 10 at 20:56
4
\$\begingroup\$

QBasic, 118 113 112 bytes

LINE INPUT s$
FOR i=1TO LEN(s$)
a=ASC(MID$(s$,i))
k=c
c=c XOR(a=34)<q
q=q XOR(a=39)<c
IF c+k=0THEN?CHR$(a);
NEXT

Explanation

Looping over the ASCII code a of each character in a line of input s$, we track three boolean values:

  • c is true inside a comment, false otherwise
  • q is true inside quotes, false otherwise
  • k is the previous iteration's value of c (we need to track this so we can suppress both double quotes instead of just the first one)

The values are updated as follows:

  • If the current character is a double quote (a=34) and q is false, toggle c
  • If the current character is a single quote (a=39) and c is false, toggle q

If both c and k are false, we're not in a comment, so output the character.

(There's a fun trick in the update statements: In QBasic, truthy is -1, so instead of a=39AND c=0, we can get the same result from (a=39)<c. The inequality will be true only when a=39 is -1 and c is 0.)

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

JavaScript (Node.js), 41 bytes

s=>s.replace(/('|").*?\1/g,_=>_<'%'?'':_)

Try it online!

Trivial, thank EliteDaMyth for -3 bytes

JavaScript (Node.js), 88 86 bytes, No RegEx

f=([c,...s],y=c=='"')=>c?[[c+s.splice(0,s.indexOf(c=="'"|y?c:s)+1).join``][+y]]+f(s):s

Try it online!

If s.indexOf(s) happens to be non-negative, then s==c and f([c,c])==c+c

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

C, 92 91 bytes

c;s;main(){while((c=getchar())+1)s=s?s-2?putchar(c):0,s*(c!=44-5*s):c-34?putchar(c)==39:2;}

All my submissions so far have been in esolangs, so even though C might not be anything groundbreaking, it's at least new to me. :)

Try It Online!

Explanation

c;s;                          // for current char and state
main() {
  while ((c=getchar())+1)     // while input is not EOF:
    s=s?            // if state is not 0 (0=default, 1=string literal, 2=comment)
      s-2?putchar(c):0,       // output the current char if state is not 2
      s*(c!=44-5*s) // set s to 0 if the relevant character is found
                    // (39 ['] if state = 1, 34 ["] if state = 2)
     :              // else (state = 0)
      c-34?         // if the current char is not ["]
        putchar(c)==39 // output current char, and if it is ['], set s to 1
       :      // else (current char is ["])
        2     // set s to 2
    ;}        // end expression, statement, and main function
\$\endgroup\$
2
  • 2
    \$\begingroup\$ Looks great! Would you mind adding a "readable" version as well? \$\endgroup\$
    – MechMK1
    Jun 11 at 14:06
  • 1
    \$\begingroup\$ No problem, although I'm not sure how 'readable' I can make it. \$\endgroup\$
    – Ross Long
    Jun 12 at 16:21
3
\$\begingroup\$

Python 3.8 (pre-release), no regex, 75 65 bytes

a=b=1
for i in input():c=b*i;a^="'"==c;b^='"'==a*i;print(end=c*b)

Try it online!

Much cleaner solution than my previous one

how it works :

a and b store the state of the string:

  • a is equal to 0 if the char is inside a literal and 1 otherwise
  • b is equal to 0 if the char is inside a comment and 1 otherwise
  • c store the char multiplied by b. If the char is inside a comment (b is equal to 0), c is equal to the empty string
  • a^= use of the bitwise operator xor to swich the state of a if the current char is ' and the char is not inside a comment
  • same goes for b but with "
  • print(end=c*b) print the char if b is not set to 0. end= is for avoiding the trailing new line

Python 3.8 (pre-release), regex, 52 bytes

lambda s:re.sub("\".*?\"|('.*?')","\\1",s)
import re

Try it online!

Pretty basic: I replace comments by nothing and literals by themselves. Regex substitution are not overriding. Lazy operators do the rest

\$\endgroup\$
2
  • \$\begingroup\$ any particular reason for putting the import after the lambda? I guess it makes no difference; I've just never seen it before \$\endgroup\$
    – Quelklef
    Jun 12 at 0:49
  • \$\begingroup\$ It's for the TIO link. I can add f=\ in the header to name my lambda function. But yeah, it doesn't make any difference \$\endgroup\$
    – Jakque
    Jun 12 at 8:42
2
\$\begingroup\$

Retina 0.8.2, 16 bytes

('.*?')|".*?"
$1

Try it online! Link includes test cases. Explanation: Regex, obviously. Since by default matches can't overlap, the regex can match both strings and comments and won't get confused. It then remains to delete the comments without changing the strings.

\$\endgroup\$
1
\$\begingroup\$

Charcoal, 28 bytes

≔ωζFS«F¬⁼∨ζι"ι¿№"'ι≔⁻∨ζι∧ζιζ

Try it online! Link is to verbose version of code. Explanation:

≔ωζ

Keep track of the current quote character, initially , but can be " or '.

FS«

Loop over the input.

F¬⁼∨ζι"ι

If the current quote, or the current character if there is no current quote, is not ", then print the current character.

¿№"'ι

If the current character is a quote, then...

≔⁻∨ζι∧ζιζ

... turn the current quote on or off as appropriate.

  • If the current quote is empty, then we subtract nothing from the current character, so the current quote becomes the current character.
  • Otherwise, we subtract the current character from the current quote. This ends the current quote only once we find the matching character.
\$\endgroup\$
1
\$\begingroup\$

Haskell, 105 bytes

f('"':s)=g s
f(a@'\'':s)=a:h s
f(a:s)=a:f s
f[]=[]
g('"':s)=f s
g(a:s)=g s
h(a@'\'':s)=a:f s
h(a:s)=a:h s

Try it online!

\$\endgroup\$

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