# Surround a string with hashes

I've already made this in Python, but it seems that it could be shortened a lot:

txt = input("Type something.. ")
c = "#"
b = " "
print(c * (len(txt) + 4))
print(c, b * len(txt), c)
print(c, txt, c)
print(c, b * len(txt), c)
print(c * (len(txt) + 4))


So if the user types:

Hello World


The program prints:

###############
#             #
# Hello World #
#             #
###############


Fewest bytes wins—and of course, the answer can be written in any language.

• The same challenge as editor golf. Sep 13, 2015 at 12:44
• Not so different: codegolf.stackexchange.com/questions/57442/show-tree-rings-age/… Sep 13, 2015 at 13:40
• The input string won't contain linebreaks, right? Sep 13, 2015 at 13:51
• @edc65 I disagree, this challenge is hugely different Sep 13, 2015 at 13:55
• I would recommend to wait at least a week before accepting an answer. While it doesn't really matter if you are planning to update the accepted answer if a shorter submission comes in, there will be people complaining about an early accepted answer, or even downvote it. There will also be some people who won't be interested in posting an answer if there is already an accepted one. Sep 13, 2015 at 17:36

# CJam, 22 20 bytes

qa{4/3*' +f+W%z}8/N*


Test it here.

## Explanation

How do you wrap a 2D grid of characters in one layer of spaces (or any other character)? Correct: four times, you append a space to each line and then rotate the grid by 90 degrees. That's exactly what I'm doing here with eight rotations: four for spaces, four for #:

qa      e# Read the input and wrap it in an array, to make it a 2D grid.
{       e# Execute this block for each value from 0 to 7.
4/3*  e#   Divide by 4, multiply by 3. Gives 0 for the first four iterations and
e#   and 3 for the other four.
' +   e#   Add the result to a space character (space + 3 == #).
f+    e#   Append this to each line of the grid.
W%z   e#   Reverse the lines, then transpose the grid - together these rotate it.
}8/
N*      e# Join the lines of the grid by newlines.

• What a neat solution! Sep 13, 2015 at 19:26
• I love the explanation :-D Sep 13, 2015 at 20:01

## vim, 28 27 keystrokes

I# <esc>A #<esc>Y4PVr#G.kwv$3hr kk.  Assumes input is provided as a single line of text in the currently open file. Explanation: I# <esc> put a "#" and space at the beginning of the line A #<esc> put a space and "#" at the end of the line Y4P copy the line 4 times Vr# replace the entirety of the first line with "#"s G. do the same for the last line kwv$3hr<space>  replace middle of the fourth line with spaces
kk.             do the same for the second line


This can also be run as a "program" like so:

echo 'Hello World' | vim - '+exe "norm I# \<esc>A #\<esc>Y4PVr#G.kwv$3hr kk."'  Which is a bit convoluted, but it works. • This is the best answer, because it works like my mind does. Sep 14, 2015 at 0:44 # pb - 89 bytes vw[Y!-1]{b^}w[B!0]{t[B]vvv>>b[T]^^^<}v>>>w[Y!4]{bv}w[X!0]{b^[Y]bv<}  This is the kind of challenge pb was made for! Not that it's competitive for this kind of challenge or anything. It's still a horrible golf language. However, challenges like this are a lot less of a pain to solve in pb than others are. Since pb treats its output as a 2D canvas and is able to write to any coords, anything involving positioning text/drawing around text (i.e. this challenge) is handled rather intuitively. Watch it run: This visualization was created with an in-development version of pbi, the pb interpreter. The line with the blue background is Y=-1, where input is stored when the program starts. The rectangle with the red background is the current location of the brush. The rectangles with yellow backgrounds are anywhere the ascii character 32 (a space) is explicitly written to the canvas. Any blank spaces without this background actually have the value 0, which is converted to a space. Here's the code with the comments I used while writing it, with some thematically relevant section headers ;) ################################ # # # Handle first column oddities # # # ################################ v # Start from Y=4 and go up (so we land on input afterwords) w[Y!-1]{ # While we're on the visible part of the canvas b^ # Write "#", then go up } ######################### # # # Insert text of output # # # ######################### w[B!0]{ # For each character of input t[B] # Save input char in T vvv>> # Down 3 + right 2 = where text part of output goes b[T]^^^< # Write T and go to next char } ############################### # # # Handle last column oddities # # # ############################### v>>> # Go to Y=0, X=(X of last text output's location + 2) w[Y!4]{ # Until we reach the last line of output bv # Draw "#", then go down } ########################### # # # Loop to finish Y=0, Y=4 # # # ########################### w[X!0]{ # Until we've gone all the way left b^[Y] # Print "#" at Y=4, go to Y=0 bv # Print "#" at Y=0, go to Y=4 < # Move left, printing until output is complete }  • Not that it's competitive... It's still a horrible golf language. I say that, but right now I'm only tied for last. I was expecting to be last outright. :D Sep 13, 2015 at 16:48 # brainfuck - 156 bytes ++++++++++>,[>>+++++[<+++++++>-],]<....[.---<<]>>+++>>+++.---.[.>>]<<.+++.[<]>.>>+++.---.<[.>>]<<<.>>.---[<]>.--->>+++.---[.>>]<<..+++.---[<]>[+++.>>]<<....  This is probably golfable. There's some places where I didn't know if it would be better to store a value somewhere for reuse or to remake it/go get it from elsewhere on the tape. Instead of doing the work to figure it out, I didn't do that. :D With comments: ++++++++++> Place a 10 (\n) at the beginning of the tape ,[>>+++++[<+++++++>-],] Place a byte of input; place a 35 (#); repeat until end of input <.... Print the last cell (35; #) 4 times [.---<<] Print every noninput cell on the tape in reverse (one # for each byte of input; then \n) After printing each cell; decrease it by 32 >>+++ Increase the 7 back up to 10 >>+++.---. Increase a 32 (space) back up to 35 (#); print it; put it back to 32 and print again [.>>] Print every 32 on the tape (one for each byte of input) <<.+++. Print the last space again; increase it by 3 and print the resulting # [<]>. Go to the beginning of the tape and print the \n >>+++.---. Increase a 32 (space) back up to 35 (#); print it; put it back to 32 and print again <[.>>] Print every byte of input <<<.>>.--- Print a space; then print a 35 (#} that was left behind earlier; Set it back to 32 after [<]>.--- Go to the beginning of the tape and print the \n; decrease by 3 >>+++.--- Set a space to #; print it; set it back to space [.>>] Print all spaces <<..+++.--- Print a space twice more; set it to #; print it again; set it back to space [<]> Go to the "newline" (currently a 7 instead of 10) [+++.>>] Increase by 3; print; do the same for all spaces (to set them to # before printing) <<.... Print the last # 4 more times  # K, 21 bytes 4(|+"#",)/4(|+" ",)/,  Enlist the string, Add a space to all four sides of a string, then add an octothorpe to each side of the string. In action:  4(|+"#",)/4(|+" ",)/,"Hello." ("##########" "# #" "# Hello. #" "# #" "##########")  Works in oK, Kona and k5. There are quite a few variations within one character of length which remove the redundancy in the above, but none seem to break even when we only have to perform the "wrap" operation twice: {x{|+x," #"y}/&4 4},: {x{|+x,y}/,/4#'" #"},: {x{|+x,y}/" #"@&4 4},: {|+x,y}/[;" #"@&4 4],:  # Python 3, 88 bytes Thanks @WorldSEnder s=" ";n=s+input()+s b=len(n) h="#";x=h*(b+2);y=h+s*b+h;z="\n" print(x+z+y+z+h+n+h+z+y+z+x)  ## Example I/O: This is a test ################## # # # This is a test # # # ##################  • s=" " at the beginning saves 1 byte overall I think. Sep 13, 2015 at 18:53 • Why using ; instead of new lines btw? Both should be one byte, right? Sep 13, 2015 at 21:56 • @JeromeJ I know it doesn't make a difference, but it looks shorter if you put it on one line ;) Sep 14, 2015 at 5:53 • 72 bytes Aug 11, 2020 at 21:23 # Python 2, 67 bytes s=input() n=len(s)+2 y='#\n#' x='#'*-~n+y+' '*n+y print x,s,x[::-1]  Try it online! # Pyth, 31 bytes Js[K\#*d+2lzKb*K+4lz)_Jjd[KzK)J  Thanks to people in comments giving hints on how to golf further, I really don't know the language well as you can (probably) tell. • Couple of hints: "#" is equal to \#. If you have to concatenate a lot of objects you're better of using j""[ followed by a list of objects, which is equal to jk[. – orlp Sep 13, 2015 at 13:24 • Some more hints. s[ is equal to jk[ for arrays of strings. Also, you can also assign K on the fly, like jk[K\#, and just drop the initial assign. Assigning lz to J doesn't help if only used twice, so save J for something else. Here, if you swap Z for J you can remove the =. Finally, you can assign J on the fly. The code then looks like this: _Js[K\#*d+2lzKb*K+4lz)jd[KzK)J Sep 13, 2015 at 16:56 # Perl, 43 76 bytes Transform each text input line as specified: s/.*/($x=("#"x(4+($z=length))))."\n".($y="#"." "x(2+$z)."#\n")."#$& #\n$y$x"/e


For example:

echo surround a long string with pounds |
perl -ple's/.*/($x=("#"x(4+($z=length))))."\n".($y="#"." "x(2+$z)."#\n")."# $& #\n$y$x"/e' ###################################### # # # surround a long string with pounds # # # ######################################  Here’s how to see what it’s really doing: perl -MO=Deparse,-p,-q,-x9 -ple '($x=("#"x(4+($z=length))))."\n".($y="#"." "x(2+$z)."#\n")."#$& #\n$y$x";'
BEGIN { $/ = "\n";$\ = "\n"; }
LINE: while (defined(($_ = <ARGV>))) { chomp($_);
(((($x = ('#' x (4 + ($z = length($_))))) . "\n") . ($y = (('#' . (' ' x (2 + $z))) . "#\n"))) . (((('# ' .$&) . " #\n") . $y) .$x));
}
continue {
(print($_) or die((('-p destination: ' .$!) . "\n")));
}
-e syntax OK


So something more like this:

((
(($x = ('#' x (4 + ($z = length($_))))) . "\n") . ($y = (('#' . (' ' x (2 + $z))) . "#\n")) ) . (((('# ' .$&) . " #\n") . $y) .$x)
)

• There should be a row of white space padding before and after the string. Take a look at the example output from the question.... but nice to see you submitting... especially in Perl :D Sep 13, 2015 at 23:52
• @rayryeng It made me sad that nobody was submitting in Perl. Sep 13, 2015 at 23:55
• To be honest I was surprised. This certainly seems like a problem suited for it :). Sep 13, 2015 at 23:56
• @rayryeng Fixed in the next release. :) Sep 14, 2015 at 0:31
• Actually, now that I think about it, if you split everything into separate statements, you can simply move them outside the substitution and drop the e modifier...but if you do that, you might as well just drop the substitution altogether: $x="#"x(4+($z=y!!!c));$y="#".$"x(2+$z)."#\n";$_="$x\n$y# $_ #\n$y$x". Use actual newlines instead of \n and it's only 65 bytes, plus 2 for -lp. Sep 16, 2015 at 18:26 # Perl 5.14+, 57 56 bytes perl -lpe '$_=join"#
#",($_="$_ ",y// /cr,"#".y//#/cr)[2,1,0..2]'


54 bytes + 2 bytes for -lp (if input doesn't end in a newline, -l can be dropped to save one byte).

Accepts input on STDIN:

$echo Hello World | perl -lpe '$_=join"#
#",($_="$_ ",y// /cr,"#".y//#/cr)[2,1,0..2]'
###############
#             #
# Hello World #
#             #
###############


## How it works

The core of the program is a list slice:

($_="$_ ",y// /cr,"#".y//#/cr)[2,1,0..2]'


This provides a compact way to store the three unique rows of the output (the first two rows of the bounding box are the same as the last two, only mirrored). For the input string foo, the results of the slice would be:

index   value
--------------
2    "######"
1    "     "
0    " foo "
1    "     "
2    "######"


Joining these values with #\n# gives us our box.

Note that Perl 5.14+ is required to use the non-destructive r modifier to the transliteration operator y///.

# JavaScript (ES6), 73

Heavily using template string, the 2 newlines are significant and counted.

Test running the snippet below in any EcmaScript 6 compliant browser (FireFox and latest Chrome, maybe Safari).

f=s=>(z=c=>*${c.repeat(s.length+2)}* )*+z +*${s} *
+z +z*

// Less golfed

U=s=>(
z=c=>'*' + c.repeat(s.length+2) + '*\n',
z('*') + z(' ') + '* ' + s + ' *\n' + z(' ') + z('*')
)

// TEST

O.innerHTML=f('Hello world!')
<pre id=O></pre>

This is quite shorter than my first try, derived from this other challenge:

f=s=>(q=(c,b,z=c.repeat(b.length))=>[z,...b,z].map(r=>c+r+c))('*',q(' ',[s])).join\n

• Doesn't run for me (Canary).
– user42643
Sep 13, 2015 at 15:05
• @minxomat what is Canary? Sep 13, 2015 at 15:10
• It works in Chrome 45.0.2454.85 m, Windows 7 Sep 13, 2015 at 15:12
• Canary is always the latest Chrome build. Doesn't work in my Chrome Stable either. Edit: Works in FireFox though.
– user42643
Sep 13, 2015 at 15:15
• (Is it legal to assign to the global z as a side-effect?)
– Neil
Sep 13, 2015 at 20:53

## Python 2, 74

s='# %s #'%input()
n=len(s)
b='\n#'+' '*(n-2)+'#\n'
print'#'*n+b+s+b+'#'*n


Takes input in quotes like "Hello World".

• The third line is the input encased in # _ #.
• The second and fourth lines b are # # with the right number of spaces, surrounded with newlines to either side to take care of all four newlines.
• The first and fifth lines are # multiplied to the length of the input

The lines are concatenated and printed.

# MATLAB, 93 91 bytes

Not the prettiest, but it gets the job done.

t=[32 input('','s') 32];m='#####'.';n=repmat('# ',numel(t),1)';disp([m [n;t;flipud(n)] m])


# Code Explanation

## Step #1

t=[32 input('','s') 32];


Read in a string from STDIN and place a leading and trailing single space inside it. 32 is the ASCII code for a space and reading in the input as a string type coalesces the 32s into spaces.

## Step #2

m='#####'.';


Declare a character array of 5 hash signs in a column vector.

## Step #3

n=repmat('# ',numel(t),1)'


Create a 2 row character matrix that is filled by hash signs first followed by white space after. The number of characters is the length of the input string plus 2 so that we can accommodate for the space before and after the string.

## Step #4

disp([m [n;t;flipud(n)] m])


We're going to piece everything together. We place the first column of 5 hashes, followed by the centre portion and followed by another column of 5 hashes. The centre portion consists of the 2 row character matrix created in Step #3, the input string itself which has a trailing and leading space, followed by the 2 row character matrix but reversed.

# Example Runs

>> t=[32 input('','s') 32];m='#####'.';n=repmat('# ',numel(t),1)';disp([m [n;t;flipud(n)] m])
This is something special for you
#####################################
#                                   #
# This is something special for you #
#                                   #
#####################################
>> t=[32 input('','s') 32];m='#####'.';n=repmat('# ',numel(t),1)';disp([m [n;t;flipud(n)] m])
Hello World
###############
#             #
# Hello World #
#             #
###############
>> t=[32 input('','s') 32];m='#####'.';n=repmat('# ',numel(t),1)';disp([m [n;t;flipud(n)] m])
I <3 Code Golf StackExchange!
#################################
#                               #
# I <3 Code Golf StackExchange! #
#                               #
#################################


# Ruby, 83 bytes

I guess it could be golfed further, but since there's no Ruby answer yet, here it is:

s=ARGV
n=s.size
r="#"*(n+4)
t="\n#"+" "*(n+2)+"#\n"
puts r+t+"\n# "+s+" #\n"+t+r

• Here's a golfed version to update if you're still active. I might edit yours since mine is just a direct golfing of your code down to 61 chars. Update: s=$* puts r=?#*n=s.size+4,t=?#+' '*(n-2)+?#,"# #{s} #",t,r Yours: s=ARGV n=s.size r="#"*(n+4) t="\n#"+" "*(n+2)+"#\n" puts r+t+"\n# "+s+" #\n"+t+r Oct 19, 2021 at 7:49 ## Pyke (noncompetitive), 6 bytes .X".X#  Try it here! Pyke was written after the challenge and is therefore noncompetitive. .X" - surround string in spaces .X# - surround string in hashes  .X takes a string and a string constant arg and surrounds a string with that group of characters. The constant arg can be up to 8 characters and have different effects on how the string is surrounded. # PHP, 95 93 bytes Not exactly brilliant or anything similar, but it was actually fun! $l=strlen($s="$argv ");printf("#%'#{$l}s# #%1${l}s# #s# #%1${$l}s#
#%1\$'#{$l}s#",'');


Not exactly pretty or anything, but it works brilliantly!

Thanks to @Titus for saving 2 bytes.

• You can save two bytes by using $argv instead of $_GET and -r. Oct 8, 2016 at 14:38

# MAWP, 117 107 bytes

%|0_!!!3M[1A75W;]25W;%75W;1M[1A84W;]75W;25W;75W;84W;~[;]84W;75W;25W;%75W;1M[1A84W;]75W;25W;%3M[1A75W;]25W;.


-10 bytes from Dion.

Try it!

• This is impressive, it's the kinda thing even i didnt expect from MAWP... Well done!
– Dion
Aug 10, 2020 at 11:09
• It seems that i have outgolfed you by quite a bit... :D
– Dion
Aug 11, 2020 at 6:11
• It seems that your submission with newlines being 25W instead of 25W3M is a byte shorter than mine! 8dion8.github.io/MAWP/…
– Dion
Aug 12, 2020 at 10:37
• What luck! Changed it. Aug 12, 2020 at 10:39

# J, 21 bytes

'#'g' '&g=.|.@|:@,^:4


Try it online!

# Python 3, 82 bytes

d,h,s,m=input(),'#',' ','\n'
l=len(d)+2
print(l*h+h,l*s,s+d+s,l*s,l*h+h,sep=h+m+h)


Try it online!

• Welcome to CGCC! Your code actually only has 82 bytes. I've added a Try It Online link as well, which just lets people test / verify your code a bit more easily. Good first answer! May 11, 2021 at 19:35

# ><>, 145 bytes

i:0(?v
[1+2l57*o1
*o:&v>]~]ao57
2(?v>84*o1-:
aov>~~57*o
rv>84*:57*:@oo$o>l3(?v o57*ov>roo&:1[a :2(?v>84*o1- oaov>]~75* (?;>57*o1-:0 Probably very overcomplicated but it works and is kind of short. Just gotta love ><>. The code prints the output line by line. Every line has at least two lines responsible for it. One to print and the to "clean up" after the first line, deleting useless garbage in the stack, printing newlines, registering the register, etc. Try it here (This was the hardest golf I've done) # C++, 198 Bytes #include <iostream> #include <string> int i;int main(){std::string t,n,o;std::getline(std::cin,o);t="\n#";o="# "+o+" #";for(;i<o.size();i++){n+="#";if(i>1)t+=" ";}t+="#\n";std::cout<<n+t+o+t+n;}  My first stab at codegolf, and while I learned C++ is probably not the best language for golfing, I felt I did decently(?) for my first try. Ungolfed #include <iostream> #include <string> int i; //globals default to a value of 0 int main() { std::string t, n, o; std::getline(std::cin, o); t = "\n#"; // t needs hashes at start and end, places hash at start here o = "# " + o + " #"; // put hash at each end of input for(; i < o.size(); i++) { n += "#"; // fills n with hashes if(i > 1) { t += " "; // fill t with spaces between hashes, has two fewer spaces than n has hashes } } t += "#\n"; // puts final hash at end of t std::cout << n + t + o + t + n; // final output }  n, o and t represent the fully hashed lines, the input (with hashes at each end) and the lines between the input and the hashed lines respectively. ## ><>, 106 104 Bytes I get the feeling that ><> may not be the best language for this, but I've come too far to give up and not post this. The * at the end of line 4 is supposed to be a space. Don't you love how incredibly grotesque this code looks? Try it online. <v?(0:i v>~" ## "}}l::: >"#"o1-:?!v02.>~a"#"oo "-2ooa"#"~<.31v!?:-1o" 7v?=3loroo"#"a<.4 .>";^"e3pa2p093  Here's a version without anything but direction changers to give an idea of how the pointer moves (note that I've left out the "teleport" statements, i.e. .). Direction flow: <v v> > v > < v v < >  ## Explanation My visualization of the stack will be based off of the input input. ><> is a two dimensional language, so pay attention to where the pointer is moving between lines, as it executes code underneath it (in this code <>v^ are primarily used to change direction). I'll be starting my explanations from where the pointer starts. Note that there will be two lines repeated, as the pointer moves backwards after the fifth line. What I always find cool about ><> is its ability to modify its own source code, and I make use of it in this program. Lines 3 and 4 are reused to print the last two lines through a modification of a character in each. Line 1 : Input loop <v?(0:i < change direction to left (0:i checks if input is less than 0 (no input defaults to -1) v? change direction to down if so  Stack: [-1,t,u,p,n,i] Line 2: Generates third line of output v>~" ## "}}l::: >~" ## "}} remove -1 (default input value) from stack and pads with # and spaces l::: push 4 lengths of padded input  Stack: [9,9,9,9,#, ,t,u,p,n,i, ,#] Line 3: Prints first line of output >"#"o1-:?!v02.>~a"#"oo >"#"o print "#" 1- subtract 1 from length (it's at the top of the stack) :?!v move down if top of stack is 0  Stack: [0,9,9,9,#, ,t,u,p,n,i, ,#] Output: #########  Line 4: Prints second line of output "-2ooa"#"~<.31v!?:-1o"* -2ooa"#"~< pops 0, prints newline, "#", then decrements length by 2 " o"* prints space (* is supposed to be space char) -1 decrements top of stack .31v!?: changes direction to down if top of stack is 0, else jumps back to "  Stack: [0,9,9,#, ,t,u,p,n,i, ,#] Output (* represents space): ######### #*******  Line 5: Prints third line of output 7v?=3loroo"#"a<.4 oo"#"a< prints "#",newline r reverses stack 7v?=3lo .4 outputs until stack has 3 values, then changes direction to down  Stack: [9,9,0] Output: ######### # # # input #  Line 6: Sets itself up to print fourth and fifth lines of output .>";^"e3pa2p093 >";^" push ";",then "^" e3p place "^" as the fifteenth character on line 4 a2p place ";" as the eleventh character on line 3 0 push a value (value doesn't matter -- it will be deleted) . 93 jump to the tenth character on line 4  Stack: [0,9,9,0] Line 4: Print fourth line of output "-2ooa"#"~<.31^!?:-1o"* ooa"#"~< delete 0 (unnecessary value pushed), then print newline,"#" -2 subtract two from value on top of stack (length) " .31^!?:-1o"* print space until top of stack is 0, then change direction to up  Stack: [0,9,0] Output (* represents space): ######### # # # input # #*******  Line 3: Print last line of output "#"o1-:?!;02.>~a"#"oo >~a"#"oo pop top of stack, print "#", newline "#"o1-:?!;02. print "#" until top of stack is 0, then terminate  Stack: [0,0] Output: ######### # # # input # # # #########  # PHP, 93 91 bytes $b=str_pad("",$e=strlen($s=" $argv "));echo$h=str_pad("",2+$e,"#")," #$b#
#$s# #$b#
$h";  Takes input from command line argument; escape spaces or use single quotes. Run with -r. ## C# - 142 bytes (method body is 104) class P{static void Main(string[]a){for(int i=0;++i<6;)System.Console.Write("#{0}#\n",i==3?$" {a} ":new string(" #"[i%2],a.Length+2));}}


Ungolfed:

class P
{
static void Main(string[] a)
{
for (int i = 0; ++i < 6;)
System.Console.Write("#{0}#\n", i == 3 ? $" {a} " : new string(" #"[i%2], a.Length + 2)); } }  # Haskell, 70 bytes x a=(\s->s++a++reverse s)$concat["##",'#'<$a,"##\n# ",' '<$a," #\n# "]


Try it online!

x a=id<>pure a<>reverse$concat["##",'#'<$a,"##\n# ",' '<$a," #\n# "]  Try it online! (Doesn't work since TIO uses an old GHC where <> isn't imported by default.) # Haskell, 63 bytes x a|s<-['#'<$a,"####\n# ",' '<$a," #\n# "]>>=id=s++a++reverse s  Try it online! • You should use a pattern guard rather than a lambda for your binding. Aug 7, 2020 at 20:06 • Also >>=id is usually shorter than concat. Aug 7, 2020 at 20:07 • Good ideas! Also combined two of the string literals. Aug 7, 2020 at 20:12 • I spent a little longer with this and came up with a 57 byte variation that abuses do notation to get a slight advantage. Aug 7, 2020 at 20:24 # Perl 5-n, 51 bytes map{say"#$_#"}($_="$_ ",y// /cr,y//#/cr)[2,1,0..2]


Try it online!

# MAWP, 108 bytes

%|_4M[75W;1A]%52W;75W;84W;_1M[84W;1A]%75W;25W;75W;84W;0__~[;]%84W;75W;52W;75W;1M![84W;1A]%75W;52W;2M[75W;1A]


Try it!

Now an actually valid solution

Explanation:

%              Remove the 1 already on stack
|              Push input as ascii
_4M            Push length of stack + 4
[75W;1A]%      Print that many #
52W;           Print a newline
75W;           Print #
84W;           Print a space
_1M            Push length of stack + 1
[84W;1A]%      Print that many spaces
75W;25W;       Print # with a newline
75W;84W;       Print # with a space
0              Push a 0
__             Push length of stack twice
~[;]           Reverse stack and print until 0
%              Remove top of stack
84W;75W;52W;   Print a space, # and newline
75W;           Print #
1M!            Add 1 to top and duplicate it
[84W;1A]%      Print that many spaces
75W;52W;       Print # with a newline
2M[75W;1A]     Add 2 and print that many #

• The second and fourth lines should have two hashes separated by length+2 spaces. Aug 11, 2020 at 6:32
• @Razetime welp... i knew something was wrong with it. facepalm
– Dion
Aug 11, 2020 at 6:53
• @Razetime now should be valid. Also, i noticed you were using ascii(13) for newlines instead of 10. Am i missing out on something? Aren't newlines ascii(10) be valid as well?
– Dion
Aug 11, 2020 at 7:29
• I just started golfing a week ago, so I still have a lot to learn, haha. I don't think there's any rule about that. Aug 11, 2020 at 7:40
• @Razetime you're doing great for the first week, way better than i did! Also, thanks for all the help with the interpreter, as this is my first actual project in javscript/html/css :)
– Dion
Aug 11, 2020 at 7:48

# R, 106 95 bytes

function(t,l=nchar(t),n="#
#",s=" ",&=strrep)cat(a<-"#"&l+3,b<-c(n,s&l+2,n),s,t,s,b,a,sep="")


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-5 bytes thanks to @Dominic, which led to further golfs.

• 101 bytes... May 11, 2021 at 9:57

# 05AB1E (legacy), 20 bytes

g4+'#×©,¹' .ø'#.ø,®,


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# Vyxalj, 29 bytes

L4+\#*:\#?L½›꘍m:„W2\#ð+Ḃ?\$++Ṁ


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For some reason Vertical mirror and center doesn't work