# Convert a program to a palindromic program

A palindrome is a string that is the same forwards and backwards, such as "racecar".

Write a program in some language L, that takes any program P1 in language L as input, and outputs a palindromic program P2 in language L that does the same thing as P1.

You do not need to worry about handling input programs with syntax errors.

This is code golf, so the solution with the fewest number of bytes wins.

• Can we define the language L? – Greg Hewgill Jun 10 '14 at 3:13
• @GregHewgill Yes. L is the language you choose to write your program in. – Justin Jun 10 '14 at 3:14
• In some languages, this is surprisingly tough. – Justin Jun 10 '14 at 4:33
• With a turing complete subset of Python, this is a valid entry: x=input();print(x+'#'+x[::-1]). The subset is the set of all programs that don't include newlines. – Justin Jun 10 '14 at 4:58

# Perl, 55 54 bytes

undef$/;$a=<>."\n__END__\n";print$a,scalar reverse$a;


Reads program source from stdin and writes to stdout.

Result of running on itself:

undef$/;$a=<>."\n__END__\n";print$a,scalar reverse$a;

__END__

__DNE__

;a$esrever ralacs,a$tnirp;"n\__DNE__n\".><=a$;/$fednu

• +1 for not using comments – user16402 Jun 10 '14 at 15:33
• I like that it marks the apparent gibberish at the bottom with "DNE" - a common shorthand for "Do Not Erase" used to mark things on chalkboards/whiteboards so that people don't mistake them for unimportant scribblings and wipe them off. – anaximander Jun 10 '14 at 15:55
• how does it work, I don't know perl, more specifically how does it quine(get the line that it's reversing)? – Cruncher Jun 10 '14 at 20:05
• 1+ Works in most cases except when program ends with __DATA__ that is read.. eg. print while(<DATA>);\n__DATA__ will change behaviour. – Sylwester Jun 11 '14 at 9:51
• @Sylwester: True. This works for the subset of Perl scripts that don't use __DATA__. :) – Greg Hewgill Jun 11 '14 at 17:50

# Java, 225 bytes

class c{public static void main(String[]a){String s="";java.util.Scanner r=new java.util.Scanner(System.in);while(r.hasNext())s+=r.nextLine()+"\n";s=s.replace("\n","//\n");System.out.print(s+new StringBuilder(s).reverse());}}


Output on itself (when prettified beforehand):

class c {//
public static void main(String[] a) {//
String s = "";//
java.util.Scanner r = new java.util.Scanner(System.in);//
while (r.hasNext()) s += r.nextLine() + "\n";//
s = s.replace("\n", "//\n");//
System.out.print(s + new StringBuilder(s).reverse());//
}//
}//

//}
//}
//;))(esrever.)s(redliuBgnirtS wen + s(tnirp.tuo.metsyS
//;)"n\//" ,"n\"(ecalper.s = s
//;"n\" + )(eniLtxen.r =+ s ))(txeNsah.r( elihw
//;)ni.metsyS(rennacS.litu.avaj wen = r rennacS.litu.avaj
//;"" = s gnirtS
//{ )a ][gnirtS(niam diov citats cilbup
//{ c ssalc

• Problem if comment ending in *. See comment – edc65 Jun 17 '14 at 22:08

# Python 2, 68 bytes

import sys
x=''.join(l[:-1]+'#\n'for l in sys.stdin)
print x+x[::-1]


Doesn't work if run from IDLE, because you need to generate an EOF character to stop the program from waiting on input.

Output when run on itself:

import sys#
x=''.join(l[:-1]+'#\n'for l in sys.stdin)#
print(x+x[::-1])#

#)]1-::[x+x(tnirp
#)nidts.sys ni l rof'n\#'+]1-:[l(nioj.''=x
#sys tropmi


Thanks to Greg Hewgill for helping to hammer out problems and to golf.

• Nice job, beats my kinda lame Python attempt. – Greg Hewgill Jun 10 '14 at 4:00
• @GregHewgill I prefer a nice upvote to a nice comment ;-) – Justin Jun 10 '14 at 4:06
• Ok okay... I don't usually vote against myself. :) – Greg Hewgill Jun 10 '14 at 4:07
• @GregHewgill I vote "against" myself a lot. I upvote answers based on their merits, not based on whether or not I answered. – Justin Jun 10 '14 at 4:08

# GolfScript, 10 9 bytes

"
}"+.-1%


Quite similar to minitech's solution, but it works well with newlines. It relies on GolfScript's funny (and undocumented) behavior to ignore an unmatched (and uncommented) }, as well as everything that follows it.

It will fail if the input contains an unmatched {, but that would technically constitute a syntax error.

### How it works

"
}"   # Push the string "\n}".
+    # Concatenate it with the input string.
.    # Duplicate the modified string.
-1%  # Reverse the copy.


# Example

$echo -n '1{"race{car"} > {"foo\"bar"} > if#' | golfscript make-palindrome.gs 1{"race{car"} {"foo\"bar"} if# }} #fi }"rab"\oof"{ }"rac{ecar"{1$ echo '1{"race{car"}
> {"foo\"bar"}
> if#
> }}
> #fi
> }"rab"\oof"{
> }"rac{ecar"{1' | golfscript
race{car

• Try 1\n2# (\n would be an actual newline character) as your input. – Justin Jun 10 '14 at 5:57
• @Quincunx: Pesky comments... A newline before the curly bracket should fix that. – Dennis Jun 10 '14 at 6:00
• Before and after. Need to remain a palindrome. – Justin Jun 10 '14 at 6:06
• @Quincunx: Of course. It should work now. – Dennis Jun 10 '14 at 6:09

## x86 machine code on DOS (.com file) - 70 bytes

Dealing with .COM files, creating a palyndrome is easy - since the COM "loader" just puts the content of the file at address 100h and jumps there, the program must already hardcode its end somehow and ignore everything after it, so we can just append the reverse of the first N-1 bytes (only caveat: if the program somehow tries to do tricks with the length of file everything breaks).

Here is the hex dump of my .COM-palyndromizing .COM:

00000000  31 db 8a 1e 80 00 c6 87  81 00 00 ba 82 00 b8 00  |1...............|
00000010  3d cd 21 72 30 89 c6 bf  ff ff b9 01 00 ba fe 00  |=.!r0...........|
00000020  89 f3 b4 3f cd 21 3c 01  75 18 b4 40 bb 01 00 cd  |...?.!<.u..@....|
00000030  21 85 ff 75 e5 89 f3 f7  d9 88 ee b8 01 42 cd 21  |!..u.........B.!|
00000040  eb d8 47 74 f0 c3                                 |..Gt..|


It takes the input file on the command line, and writes the output on stdout; the expected usage is something like compalyn source.com > out.com.

Commented assembly:

    org 100h

section .text

start:
; NUL-terminate the command line
xor bx,bx
mov bl, byte[80h]
mov byte[81h+bx],0
; open the input file
mov dx,82h
mov ax,3d00h
int 21h
; in case of error (missing file, etc.) quit
jc end
; si: source file handle
mov si,ax
; di: iteration flag
; -1 => straight pass, 0 reverse pass
mov di,-1
loop:
; we read one byte at time at a bizarre memory
; location (so that dl is already at -2 later - we shave one byte)
mov cx,1
mov dx,0feh
mov bx,si
mov ah,3fh
int 21h
; if we didn't read 1 byte it means we either got to EOF
; or sought before the start of file
cmp al,1
jne out
; write the byte on stdout
mov ah,40h
mov bx,1
int 21h
; if we are at the first pass we go on normally
test di,di
jnz loop
back:
; otherwise, we have to seek back
mov bx,si
; one byte shorter than mov cx,-1
neg cx
; dl is already at -2, fix dh so cx:dx = -2
mov dh,ch
mov ax,4201h
int 21h
jmp loop
out:
; next iteration
inc di
; if it's not zero we already did the reverse pass
jz back
end:
ret


Tested on itself and the solutions to a previous question seems to work fine in DosBox, some more extensive testing on "canonical" DOS executables will follow.

## GolfScript, 8

.-1%'#'\


Doesn’t handle newlines, but nobody uses those in GolfScript.

• Using newline in string literals can be used quite often ;-) – Howard Jun 10 '14 at 5:10

# Bash+coreutils, 39 bytes

f="cat
exit"
echo "$f" tac<<<"$f"|rev


Reads from STDIN and outputs to STDOUT:

$cat hello.sh #!/bin/bash echo 'Hello, World!'$ ./palin.sh < hello.sh
#!/bin/bash

echo 'Hello, World!'
exit
tixe
'!dlroW ,olleH' ohce

hsab/nib/!#
$ • @user23013 Seems to work fine. At least a simple test like ( echo 'Hello, World!' ). bash pretty much ignores everything after the exit. – Digital Trauma Jun 17 '14 at 20:05 # Javascript (ES6) Multi-line - 71 Kinda sorta stole Quincunx's comment method here: alert((x=prompt().replace(/\n/g,'//\n')+'/')+[...x].reverse().join(''))  # Single line - 49 alert((x=prompt()+'/')+[...x].reverse().join(''))  # C++, 214 209 bytes #include<cstdio> #include<stack> int main(){std::stack<char>s;int c;while((c=getc(stdin))>EOF){if(c=='\n')for(int i=2;i;i--)s.push(putchar('/'));s.push(putchar(c));}while(s.size()){putchar(s.top());s.pop();}}  Result of running on itself: #include<cstdio>// #include<stack>// int main(){std::stack<char>s;int c;while((c=getc(stdin))>EOF){if(c=='\n')for(int i=2;i;i--)s.push(putchar('/'));s.push(putchar(c));}while(s.size()){putchar(s.top());s.pop();}}// //}};)(pop.s;))(pot.s(rahctup{))(ezis.s(elihw};))c(rahctup(hsup.s;))'/'(rahctup(hsup.s)--i;i;2=i tni(rof)'n\'==c(fi{)FOE>))nidts(cteg=c((elihw;c tni;s>rahc<kcats::dts{)(niam tni //>kcats<edulcni# //>oidtsc<edulcni#  • Fail when continuation char '\' is used. Try [ideone.com/TCZHr9] – edc65 Jun 11 '14 at 7:42 • @edc65: Yeah, I thought about that later. The only obvious way I can think of to handle that would be to unfold the folded lines first. – Greg Hewgill Jun 11 '14 at 8:30 • can be done at small cost - se my C answer – edc65 Jun 11 '14 at 8:53 # Brainfuck, 749 without whitespace (not golfed) This produces brainfuck programs which mirrored palindromes, i.e. they are are mirror images of themselves. ++++++++++ [->++++>+++++++++<<]>+++.>+..<.>++. >>>>+[>,]<-[+<-] >[ [-<+<<+>>>] +<-------------------------------------------[-<+>>[-]<]>[-<<<.>>>] +<<-[->+>[-]<<]>>[-<<<.>>>] +<-[-<+>>[-]<]>[-<<<.>>>] +<<-[->+>[-]<<]>>[-<<<.>>>] +<--------------[-<+>>[-]<]>[-<<<.>>>] +<<--[->+>[-]<<]>>[-<<<.>>>] +<-----------------------------[-<+>>[-]<]>[-<<<.>>>] +<<--[->+>[-]<<]>>[-<<<.>>>] <[-]>> ] <<<<[<] <--.<.>++..--..<.>++. >>[>] <[ [->+>>+<<<] +>-------------------------------------------[->+<<[-]>]<[->>>.<<<] +>>-[-<+<[-]>>]<<[->>>.<<<] +>-[->+<<[-]>]<[->>>.<<<] +>>-[-<+<[-]>>]<<[->>>.<<<] +>--------------[->+<<[-]>]<[->>>++.--<<<] +>>--[-<+<[-]>>]<<[->>>--.++<<<] +>-----------------------------[->+<<[-]>]<[->>>++.--<<<] +>>--[-<+<[-]>>]<<[->>>--.++<<<] >[-]<< ] <--.<.>++..<.  Given a program it outputs +[[+]PROGRAM[+]][[+]MIRROR[+]]+  with PROGRAM and MIRROR replaced by the program (without non-brainfuck characters) and its mirror image. # C 168 175 Correctly handles escaped newline inside source code Edit 1 fixed bug when last newline missing Edit 2 fixed bug when line inside comment ends with *: add a tab char before the // comment (and golfed more) b[999999];main(c,z){char*p,for(p=b;(*p=c=getchar())>=0;z=c,p++)c-10||(z-92?*p++=9,*p++=47,*p++=47,*p=c:(p-=2));*p=47;for(p=b;*p;)putchar(*p++);for(;p>b;)putchar(*--p);}  C99 Standard, valid code, many warnings Ungolfed b[999999]; // working buffer ~ 4M on 32 bit machine, max source size // c is current char, z is previous char, main(c,z) // z start as argv pointer, will be out of char range { char *p; for(p = b; (*p=c=getchar()) >= 0; // while testing EOF copy char to buffer set c variable z=c, p++) // at end loop increment p and set previous = current { c-'\n' || // if newline (z - '\\' // check if escaped ? *p++='\t',*p++='/',*p++='/', *p=c // if not escaped, add tab,/,/ and newline : (p-=2) // if escaped, drop both escape and newline ); } *p='/'; // if last newline missing, will add a comment anyway for(p=b;*p;) putchar(*p++); // ouput buffer for(;--p>=b;) putchar(*p); // outbut buffer reversed }  • there is a little bug in it. try /* *<NL> */int main(){} – jimmy23013 Jun 17 '14 at 19:38 # C# - 174 using System;using System.Linq;class c{public static void Main(){var a="";var b="";while((a=Console.ReadLine())!="")b+=a+"//\n";Console.Write(b+string.Concat(b.Reverse()));}}  Test Input: using System; using System.Linq; class c { public static void Main() { var a = ""; var b = ""; while ((a = Console.ReadLine()) != "") b += a + "//\n"; Console.Write(b+string.Concat(b.Reverse())); } }  Test Output: using System; using System.Linq; class c { public static void Main() { var a = ""; var b = ""; while ((a = Console.ReadLine()) != "") b += a + "//\n"; Console.Write(b+string.Concat(b.Reverse())); } } // } // } // ;)))(esreveR.b(tacnoC.gnirts+b(etirW.elosnoC // ;"n\//" + a =+ b // )"" =! ))(eniLdaeR.elosnoC = a(( elihw // ;"" = b rav // ;"" = a rav // { // )(niaM diov citats cilbup // { // c ssalc // ;qniL.metsyS gnisu // ;metsyS gnisu  • I think you may have misunderstood one of the instructions. Your program should be able to take any program as input, and write out a palindromic program that does the same thing as the original program. – Greg Hewgill Jun 11 '14 at 2:20 • It can.. If I input the C++ code from your answer, it returns exactly what you have. – mnsr Jun 11 '14 at 2:24 • All your program does is reverse its input. The output of your program is not a complete palindromic program. – Greg Hewgill Jun 11 '14 at 2:30 • Oh yeah, i get ya. Updated - better now? – mnsr Jun 11 '14 at 2:36 • Yup, that's it. Your test output should have the // at the end of each line now though. – Greg Hewgill Jun 11 '14 at 2:42 ## PHP, 96 bytes function a($b){
echo $c = "a('$b')" . strrev("a)'" . $b . "'(");$d = substr($c, 0, strlen($b) + 5);
eval("\$d;");
}


Sample Usage:

a('apple'); // echoes a('apple')('elppa')a until your bytes get exhausted


This is nothing clever. It's just a simple piece of code that does the job... I was in a mood for playing. I do know that this code is rife with bad programming practices!

Finally, I will gladly accept any criticism and edits to this code!

• Welcome to Code Golf. This is a function, not a program. See the other answers, they provide good examples. – A.L Jun 10 '14 at 9:21

# Cobra - 134

class P
def main
print '/#\n[i.reversed.join("\n")]\n#/#\n[i.join("\n")]\n#/'


# Racket 133

(require srfi/13)(let((r read-line)(w display))(let l((i(r)))(when
(not(eq? eof i))(w i)(w";\n")(l(r))(w"\n;")(w(string-reverse i)))))


Ungolfed (but still very imperative):

(require srfi/13)
(when (not (eof-object? instr))
(display instr)
(display ";\n")
(display "\n;")
(display (string-reverse instr))))


Output when given the ungolfed version as input:

(require srfi/13);
(when (not(eof-object? instr));
(display instr);
(display ";\n");