79
\$\begingroup\$

Super simple challenge today, or is it?

I feel like we've heard a fair bit about double speak recently, well let's define it in a codable way...

Double speak is when each and every character in a string of text is immediately repeated. For example:

"DDoouubbllee  ssppeeaakk!!"

The Rules

  • Write code which accepts one argument, a string.
  • It will modify this string, duplicating every character.
  • Then it will return the double speak version of the string.
  • It's code golf, try to achieve this in the smallest number of bytes.
  • Please include a link to an online interpreter for your code.
  • Input strings will only contain characters in the printable ASCII range. Reference: http://www.asciitable.com/mobile/

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

var QUESTION_ID=188988;
var OVERRIDE_USER=53748;
var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(d){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+d+"&pagesize=100&order=asc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(d,e){return"https://api.stackexchange.com/2.2/answers/"+e.join(";")+"/comments?page="+d+"&pagesize=100&order=asc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){answers.push.apply(answers,d.items),answers_hash=[],answer_ids=[],d.items.forEach(function(e){e.comments=[];var f=+e.share_link.match(/\d+/);answer_ids.push(f),answers_hash[f]=e}),d.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){d.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),d.has_more?getComments():more_answers?getAnswers():process()}})}getAnswers();var SCORE_REG=function(){var d=String.raw`h\d`,e=String.raw`\-?\d+\.?\d*`,f=String.raw`[^\n<>]*`,g=String.raw`<s>${f}</s>|<strike>${f}</strike>|<del>${f}</del>`,h=String.raw`[^\n\d<>]*`,j=String.raw`<[^\n<>]+>`;return new RegExp(String.raw`<${d}>`+String.raw`\s*([^\n,]*[^\s,]),.*?`+String.raw`(${e})`+String.raw`(?=`+String.raw`${h}`+String.raw`(?:(?:${g}|${j})${h})*`+String.raw`</${d}>`+String.raw`)`)}(),OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(d){return d.owner.display_name}function process(){var d=[];answers.forEach(function(n){var o=n.body;n.comments.forEach(function(q){OVERRIDE_REG.test(q.body)&&(o="<h1>"+q.body.replace(OVERRIDE_REG,"")+"</h1>")});var p=o.match(SCORE_REG);p&&d.push({user:getAuthorName(n),size:+p[2],language:p[1],link:n.share_link})}),d.sort(function(n,o){var p=n.size,q=o.size;return p-q});var e={},f=1,g=null,h=1;d.forEach(function(n){n.size!=g&&(h=f),g=n.size,++f;var o=jQuery("#answer-template").html();o=o.replace("{{PLACE}}",h+".").replace("{{NAME}}",n.user).replace("{{LANGUAGE}}",n.language).replace("{{SIZE}}",n.size).replace("{{LINK}}",n.link),o=jQuery(o),jQuery("#answers").append(o);var p=n.language;p=jQuery("<i>"+n.language+"</i>").text().toLowerCase(),e[p]=e[p]||{lang:n.language,user:n.user,size:n.size,link:n.link,uniq:p}});var j=[];for(var k in e)e.hasOwnProperty(k)&&j.push(e[k]);j.sort(function(n,o){return n.uniq>o.uniq?1:n.uniq<o.uniq?-1:0});for(var l=0;l<j.length;++l){var m=jQuery("#language-template").html(),k=j[l];m=m.replace("{{LANGUAGE}}",k.lang).replace("{{NAME}}",k.user).replace("{{SIZE}}",k.size).replace("{{LINK}}",k.link),m=jQuery(m),jQuery("#languages").append(m)}}
body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, [Other information] N bytes

where N is the size of your submission. Other information may include flags set and if you've improved your score (usually a struck out number like <s>M</s>). N should be the right-most number in this heading, and everything before the first , is the name of the language you've used. The language name and the word bytes may be links.

For example:

# [><>](http://esolangs.org/wiki/Fish), <s>162</s> 121 [bytes](https://esolangs.org/wiki/Fish#Instructions)
\$\endgroup\$
10
  • 3
    \$\begingroup\$ It will modify this string. Are you intentionally requiring pass-by-reference and modify in-place? And then return a copy or reference to that modified string? If so, languages like asm or C would need to accept an explicit-length string (pointer + length) where the length is either the current string length (with the buffer being twice that size), or it's the total size and you need to duplicate the low half. Thus you need to start from the end and work backwards, or allocate scratch space and then copy back. But there are answers in C and 8086 asm that totally violate all that. \$\endgroup\$ Aug 3 '19 at 1:49
  • 4
    \$\begingroup\$ @PeterCordes I do not care if it modifies the same object or builds a new one. \$\endgroup\$
    – AJFaraday
    Aug 3 '19 at 5:08
  • 3
    \$\begingroup\$ I'd suggest wording it as "modify (or produce a modified copy) of the string" to explicitly allow answers that do or don't modify in-place. Simplifying the wording to "return a string that's twice as long, with each character repeated" would be nice but then it's not clear if void foo(char *c, size_t len) is legal that takes one input/output buffer and a length, and doesn't have any return value, just a side-effect on the object it has a pointer to. \$\endgroup\$ Aug 3 '19 at 5:16
  • \$\begingroup\$ Can the string be empty? \$\endgroup\$ Aug 6 '19 at 19:58
  • 1
    \$\begingroup\$ @cschultz2048 it says the string will only contain printable ascii characters, so that implies that they’ll always be populated. I’d expect that any code for this challenge would leave an empty string empty... anyway, I don’t think it’s a test case that I’d use for this. \$\endgroup\$
    – AJFaraday
    Aug 6 '19 at 22:36

199 Answers 199

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\$\begingroup\$

Forth (gforth), 39 bytes

: f bounds ?do i 1 type i 1 type loop ;

Try it online!

Beats NieDzejkob's submission by one byte. If we don't care empty strings, we can safely remove the ? in ?do, making it 38 bytes.

A function that takes a string in the standard representation (addr, length), and prints the result to the console.

How it works

gforth documentation about bounds

: f ( addr len -- )
  bounds ?do        \ Iterate through char addresses..
    i 1 type        \   Print one char at the address
    i 1 type        \   One more time
  loop ;            \ End loop
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2
\$\begingroup\$

Backhand, 5 bytes

i}o: 

Try it online!

(note the trailing space) This should be the optimal answer, since all of io: are needed, 4 bytes needs at two movement modifiers to avoid getting stuck in two instruction loops, and no three byte permutation works.

Explanation:

i       Input
   :    Dupe
  o     Output
 }      Move back to output
  o     Output
   :    Unnecessary dupe
i       Beginning of loop again

Of course, for a suboptimal but ironic answer you could do:

vvii::oooo22jj

Try it online!

Which doubles up on every character.

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

SWI-Prolog, 87 bytes

I did not see any Prolog attempt, so I tried to make one.

d([A|X],[A,A|Y]):-d(X,Y). d([],[]). p(X,W):-string_codes(X,Y),d(Y,Z),string_codes(W,Z).

The predicate to be called is p:

?- p("Hello world!", X)

It produces (in SWI-Prolog online interpreter):

X = "HHeelllloo wwoorrlldd!!";

Explanation

The code uses two different predicates: the first one (d) works with a list and duplicates every element in it.

d([A|X], [A,A|Y]) :- d(X, Y).
d([],[]).

As most of Prolog's predicates, it uses recursion to implement iteration over elements: the first line is the recursive step, while the second is the base step. Duplication is obtained by forcing the head of the first argument to be equal to the "2-elements-head" of the second argument.

Unluckily, this is not enough to work with strings: in fact, predicate p serves as converter between strings and lists.

p(X,W) :- string_codes(X,Y), d(Y,Z), string_codes(W,Z).

Due to library names, this second predicate extends the solution by at least 22 unnecessary characters... So, if anyone could suggest me a way to get rid of those, or any general solution improvement, I'd be very grateful ^^

Note that you can still use d in a sort of "raw-mode", reducing the solution length to 35 bytes. However, the input is technically not a string.

?- d(['H','e','l','l','o',' ','w','o','r','l','d','!'],X)

Try it online!

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

R, 61 bytes

function(x){paste0(rep(strsplit(x,"")[[1]],e=2),collapse="")}

Try it online!

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

MIT-Scheme 10.*, 52 bytes

(lambda(x)(string-map(lambda(y)(make-string 2 y))x))

MIT-Scheme 10.* apparently has the new string-map which simplifies my earlier approach listed below. The only difference is that make-string is used to convert the character argument into a string and duplicate it.

74 bytes

This is the older solution for MIT-Scheme 9.* before string-map was introduced.

(lambda(x)(list->string(append-map(lambda(k)(list k k))(string->list x))))

This is an anonymous lambda that takes the input as a string, converts it to a list so that it can be mapped (I don't know of another way to do the same), replicating each character and joining it together in one list, after which it is converted back into a string.

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

GolfScript, 4 bytes

{.}%

Try it online!

Takes an input, which it treats as a list of characters, and maps . (push top item of stack to stack) over them.

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

Pip, 4 bytes

aWVa

Try it online!

The operator WV ("weave") takes two iterables and interleaves their elements. In this case, weaving the input string with itself produces the desired output.

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

W j, 5 2 bytes

+M

If you want to specify your input, you need to split your input into a list of individual characters before you do your job. (If that disobeys the rules, I will implement a flag that does this job for the writer. (Flags don't count in bytecount.))

Say your input is abcdef. Find imps.py and change those variables:

read = [["a","b","c","d","e","f"]]

prog = '+M'

Explanation

Lots of implicit stuff here. Expanded program:

aaa+M

The + requires 2 inputs and M only provides a single input. Therefore two 'a''s are prepended.

The M only has 1 input required left, so a single 'a' is prepended.
a     % The Implicit input
    M % Map every item of this input with...
 aa   % This every item with this every item...
   +  % Concatenated together

% Implicit Output
j % Join the output without a separator

Wren, 24 bytes

Fn.new{|s|s.map{|n|n+n}}

Try it online!

For every item in the list, return every item concatenated with itself.

\$\endgroup\$
2
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Forth, (GForth) 55 bytes

: d pad 80 accept 0 do pad i + c@ dup emit emit loop ;

Usage: d [enter]

This text will be doubled. [enter]

Online Forth interpreter

'accept' takes a line of up to 80 chars into pad, then the do loop fetches chars from the pad, duplicates them and emits them to the terminal.

\$\endgroup\$
2
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Rexx (Regina), 41 bytes

I just realized the rules specifically state "It will modify this string, duplicating every character." Since the newline is part of the string, I will now leave it in the string and duplicate it, thus reducing my code size to 42 bytes (according to ls -l, TIO says 41).

b=''
DO 80
a=CHARIN()
b=b||a||a
END
SAY b

Try it online!

Old 60 byte version which strips newlines (or CRs depending on your OS.)

b=''
DO 80
a=CHARIN()
b=b||a||a
END
SAY LEFT(b,LENGTH(b)-2)

Alternate last line (same number of bytes):

SAY STRIP(b,,X2C('0a'))

Do 80 iterations, read a char into var a, concatenate two a's onto the end of b during each loop, output the string minus the two newlines or CRs (need to modify for AmigaOS/AROS because it uses CR-LF for line terminators.)

62 byte original version:

DO 80
a=CHARIN()
x=CHAROUT(,a)CHAROUT(,a)
END

Read up to 80 chars from stdin into string var a, write 'a' to stdout twice for each char read in.

\$\endgroup\$
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2
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Elixir, 26 bytes

"#{for<<c<-s>>,do: [c,c]}"

Try it online!

\$\endgroup\$
2
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PHP, 37 bytes

while($c=$s[$i++])$a.=$c.$c;return$a;

Try it online!

\$\endgroup\$
2
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FEU, 12 bytes

s/(.)/\1\1/g

Try it online!

Just a single regex substitution.

\$\endgroup\$
2
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Kite, 111 94 79 bytes

Really surprised that nobody has used Kite before on CGCC... it's a really easy to learn language.

method d(a)[(a|split("")<-method(a)[a+a;])<|method(a,b)[a|str*a|bool|int+b;];];

Test suite:

method d(a)[(a|split("")<-method(a)[a+a;])<|method(a,b)[a|str*a|bool|int+b;];];

(d("abc"))|print;

(d("Double speak"))|print;
\$\endgroup\$
2
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Marbelous, 26 bytes

..@0
..00
..]]\/
../\
@0../\..

Marbelous is a language based on marble machines

  • @n (n from 0 to Z) is a portal which teleport the marble to another portal with the same value
  • 00-FF initiate a marble with this value
  • ]] read one byte of input into a parsing marble
  • \/ is a trash can
  • /\ create a duplicate passing marble to it's left and right
  • .. is a noop
  • marbles going out of the machine from the bottom are implicitly outputed

interpretor

\$\endgroup\$
2
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Rockstar, 82 73 bytes

listen to S
cut S
X's0
while S-X
let S at X be*2
let X be+1

join S
say S

Try it here (Code will need to be pasted in)

\$\endgroup\$
2
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Python 3, 67 64 bytes

t=input()
for i in map(lambda x:x[0]+x[1],zip(t,t)):print(end=i)

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ print("".join(map("".join,zip(*[input()]*2)))) \$\endgroup\$
    – Danis
    Jan 25 at 19:00
  • \$\begingroup\$ j="".join;print(j(map(j,zip(*[input()]*2)))) \$\endgroup\$
    – Danis
    Jan 25 at 19:02
2
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Whispers v1, 38 bytes

> Input
>> L+L
>> Each 2 1
>> Output 3

Try it online!

Squeezed pseudo code:

>> Output ( Input.map( c -> c + c ) )

Explanation:

As usual in Whispers, the last line is executed first by default:

>> Output 3

This line outputs the result of line 3:

>> Each 2 1

Here we expect line 2 to be a function and line 1 to be an array. It returns the array specified in line 1, whose elements were modified by the function in line 2. Let's look at this array first:

> Input

Takes the first line of the input. So we get an array of characters.

Now line 2:

>> L+L

L is an argument from the Each statement in line 3. Since we are working with strings, L is returned concatenated with itself.

So the squashed pseudo-code is:

>> Output ( Input.map( c -> c + c ) )
\$\endgroup\$
2
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Haskell, 18 bytes

f s=s>>= \c->[c,c]

Try it online!

\$\endgroup\$
2
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Factor + sequences.repeating, 12 bytes

[ 2 repeat ]

Try it online!

\$\endgroup\$
2
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Numberwang, 6 bytes

248337

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ That's Numberwang! \$\endgroup\$
    – Giuseppe
    Apr 16 at 14:18
  • \$\begingroup\$ Wang even is this language? \$\endgroup\$
    – AJFaraday
    Apr 16 at 14:33
2
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[MashedPotatoes], 315 bytes

synchronized(0.0f){WHILEstd::ignore=`uniq-c`s/nullptr//gWENDgoto++i;(formatt"case__dict__of{_->OUTPUT=void(0)}def$ARGV(--help)usestrictqw/*read-eval*/;endgoto<>;gotoObject;defstd::ignore(--help)guardnullptrend(formatt"procFS{STDERR}{SETLOCAL()import$ARGV;}std::cout<<--help<<std::endl;goto0.0f;goto++i;")goto<>;")}

This is not short, by any means. But I've wanted to try and use MashedPotatoes for something, since the language is so odd. So it might be possible to shorten the code, but it's the sort of language where I'm relieved if I get something to work. :)

It's not available on TIO, so here's a link to the Esolang entry for the language. https://esolangs.org/wiki/MashedPotatoes There's a link the GitHub entry with the code on the page.

And there's one caveat about the interpreter... I submitted a PR to the repo to make it possible to handle EOF when reading character input, and this code relies on that enhancement. The maintainer merged it today (meaning 4/17/2021) so if you may need to refresh your copy if you pulled it previously.

I think a brief explanation of some language quirks are probably necessary first... There are 3 registers ^A, ^C, ^E, three stacks 0, 1, 2, two triggers A, B, and three labels 0, 1, 2 available.

The constants STDERR, $[, ~uniq -c~ (where I've used ~ to replace backticks since I cannot figure out how to make backticks work inside backticks) references registers ^A, ^C, ^E respectively the first time they appear in the code. Then they change to the next register the next time they are used. Meaning STDERR evaluates to the value of ^A the first time it appears in the code, to ^C the second time, etc...

Trigger A is referenced by the constants use strict qw/LABEL/;, SETLOCAL LABEL, import LABEL;, CFLAGS=LABEL, \emph{LABEL}, ~cat LABEL~ (again ~ used in place of a backtick here) cyclically. Meaning the first occurence of trigger A in the code will be use strict qw/LABEL/;, the second time will be SETLOCAL LABEL, etc... And LABEL is replaced the target label.

Similarly trigger ^B is references by the constants s/LABEL//g, OUTPUT = LABEL, guard LABEL, lambda: LABEL, """LABEL""", also in the same cyclical manner.

Label 0 is referenced by the symbols 0.0f, ++i, <>, Object cycically.

The cyclical reference names for label 1 are std::ignore, nullptr, __dict__, void(0).

And the label 2 references are $ARGV, *read-eval*, FS, ().

A better description can be found on the Esolangs page for the language.

And next, the code...

synchronized (0.0f) {

This just assign the label 0, the value 1 since the ( is on line 1 of the code.

  WHILE std::ignore = `uniq -c` s/nullptr//g WEND

Increments the value of register ^E.

  goto ++i;

Shifts the values of the registers, effectively ^A -> ^E -> ^C -> ^A

  (format t "

Loops while ^C is greater then 0

    case __dict__  of { _ -> OUTPUT = void(0) }

Reads one character from STDIN into ^A

    def $ARGV(--help) use strict qw/*read-eval*/; end

Pushes the value of ^A to stack 0, since ---help evaluates to 0

    goto <>;
    goto Object;

Shift the registers twice.

    def std::ignore(--help) guard nullptr end

Pop the value of stack 0 into ^A, the result of the the last 4 actions is effectively ^C = ^A

    (format t "

Loop as long as ^C > 0, effectively if not EOF { ... } since EOF is codepoint 0.

      proc FS {STDERR} {

Code block for output operations

        SETLOCAL ()

Print the value of ^A, mapping the codepoint to a character

        import $ARGV;

Repeats the same, meaning print ^A a second time

      }

End of output code block

      std::cout << --help << std::endl;

Set ^A to zero.

      goto 0.0f; 
      goto ++i;

Shift the registers twice, effectively sets ^C to 0 to cause the loop to exit

    ")

End of the print loop, which will never run more then once since we explicitly set ^C to 0 to prevent that.

    goto <>;

Shift the registers, sets ^C to the last codepoint read from STDIN, will be 0 on EOF.

  ")

End of the outermost (format loop, will decrement ^C and compare against 0, so if the read got any character the loop will repeat.

}

End of synchronize code block.

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

Hexagony, 11 bytes

,)</;(/@>;~

Try it online!

Just some simple control flow on this one, could def be improved.

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

Zephyr, 89 bytes

input s
set i to 1
set o to""
while""<s[i...i]
set o to(o+s[i])+s[i]
inc i
repeat
print o

Try it online!

Explanation

Oh. My. Goodness.

I spent 20 minutes convinced this challenge wasn't possible in Zephyr. All the language design choices that I made 11 years ago came back to bite me:

  • There is no way to get the length of a string (not a design choice, I just didn't implement anything).
  • The print statement adds a trailing newline by default. Suppressing the trailing newline with ... adds a space instead.
  • The concatenation operator | also puts a space between the two values.

But a couple of things went right:

  • String slicing exists, and doesn't throw an error if the indices are out of bounds.
  • You can concatenate any value to a string using the + operator. In hindsight, I'm not sure why I thought this was a good feature, but there it is.

And so:

input string
set index to 1    # Zephyr uses 1-based indices
set output to ""
# Loop until index is out of bounds, at which point the string slice returns empty string
while "" < string[index...index]
    set output to (output + string[index]) + string[index]
    inc index
repeat
print output
\$\endgroup\$
2
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Julia 1.0, 20 18 bytes

!s=join([s...].^2)

Try it online!

\$\endgroup\$
1
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Python 2, 31 bytes

Realised I was beaten to it already.

lambda s:''.join(c*2for c in s)
\$\endgroup\$
1
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Ruby, 16 bytes

gsub /./m,'\0\0'

Requires -p.

Try it online!

Alternatively:

gsub(/./m){$&*2}

Try it online!

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

SNOBOL4 (CSNOBOL4), 69 bytes

	I =INPUT
S	I LEN(1) . X REM . I	:F(O)
	O =O X X	:(S)
O	OUTPUT =O
END

Try it online!

Prints with a single trailing newline.

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

kavod, 16 bytes

*>1+8?99.1-#<#0.

Try it online!

I picked up a random tarpit and decided this would be a nice challenge to solve

Explanation:

*     Take a byte of input and push to stack
>     Push to the register stack without popping from normal stack
1+    Add one to the input
8?    Go to the eighth character in the program if the top of the stack is nonzero (not EOF)
99.   Go to the 99th character in the program (terminating the program). This will be skipped over if input is not EOF
1-    Decrement TOS
#     Print with popping
<#    Get input back from register stack and print it
0.    Go back to the start of the program.
\$\endgroup\$
1
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Charcoal, 5 bytes

⭆S⁺ιι

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

 S      Input string
⭆       Map over characters and join
   ι    Current character
  ⁺     Plus
    ι   Current character
        Implicitly print
\$\endgroup\$
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