Reverse stdin and place on stdout

Question

Requirements:

• Take an input on stdin including new lines / carriage returns of unlimited length (only bounded by system memory; that is, there is no inherent limit in the program.)
• Output the reverse of the input on stdout.

Example:

Input:

Quick brown fox
He jumped over the lazy dog

Output:

god yzal eht revo depmuj eH
xof nworb kciuQ

Shortest wins.

Leaderboard:

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Answer 1

Knight (v2.0-alpha), 37 bytes

;=s=t@;W=gP=s+s+g"
";Ws;=t+[s t=s]sOt

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Nothing super exciting, except for the fact that we can use @ instead of "" for the initial value of s and t.

Answer 2

Vyxal, 1 byte

Ṙ

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Answer 3

Thunno W, 1 byte

(actually \$ 1 \log_{256}(96) \approx \$ 0.82 bytes, but that doesn't show up on the leaderboard)

r

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Also works in Thunno 2: Attempt This Online!

   # Implicit input.
   # The W flag reads the whole of STDIN as a multi-line string,
   # rather than taking each line separately.
r  # Reverse the string on the top of the stack.
   # Implicit output.

Answer 4

ForWhile, 24 22 19 16 14 bytes

Saved 2 bytes due to new language feature of looping without pushing loop counter ([…) instead of (…))

9:`['1+_:~).[#)

Try it at the online interpreter!

This took me so long to figure out!! But I made it there in the end. I don't think the program can get shorter than this (but you never know…)

Thanks bsoelch for the tip to get large numbers.

Explanation:

9:`  \ 9 ^ 9
[    \ repeat (9 ^ 9) times:
  '  \   swap two items on the stack
  1+ \   add one (to string length)
  _  \   read a byte from STDIN
  :~ \   duplicate, bit flip; -1 (EOF) becomes 0
)    \ pop top of stack; if nonzero (not EOF) jump back to [
.    \ pop final -1
[    \ repeat (string length) times:
  #  \   print byte without popping
)    \ pop top of stack; if nonzero, jump back to [

---

ForWhile, 70 bytes

This one works for input of infinite size; ForWhile doesn't really support infinite loops, but you can do source-modification to make it work. Thanks bsoelch for writing this one.

""(,#)6~$~7~$(!;$1-:){}9+:57(.:@;57-$1-:).7~@_:~![..7~@34;$?];$1+7~$57

Try it at the online interpreter!

---

For the record: for ForWhile, reversing the string is the easy part of this challenge. The hard part is reading all of STDIN as a string. If the string could be passed as an argument to a procedure, this could be 5 bytes:

{[#)}

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Answer 5

05AB1E, 3 1 byte

R

Try it online! Takes input surrounded in """triple double quotes""".

R  # full program
R  # reverse...
   # implicit input
   # implicit output

Answer 6

Hassium, 62 Bytes

func main(){s=input();for(x=s.length- 1;x>=0;print(s[x--]))0;}

Run and see expanded here

Answer 7

Seriously, 2 bytes (non-competitive)

R

There is a formfeed character (ASCII 0x0C) as the first byte. Hexdump (reversible with xxd -r):

00000000: 0c52                                     .R

Try it online! (note that the formfeed shows up as the dingbat ♀ on TIO - one of the nice features Dennis has added is the parsing of CP437 dingbats into ASCII control codes).

Answer 8

Hexagony, 28 bytes (Non-competitive)

Compressed:

.$@.>;<..\'"/$\$/}{,<..>.../

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Answer 9

Sesos, 3 bytes (non-competing)

x

Since this contains unprintables, here is a hexdump:

0000000: 788c19                                            x..

This is the code that was used to generate it:

fwd 1,jnz,rwd 1,jmp,put,rwd 1

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This works, provided the input does not contain null characters (\0).
UTF-8 locale as far as I'm aware.

Explanation:

      ;(implicit nop) Start the input loop.
fwd 1 ;Store the input character for subsequent use.
jnz   ;(implicit jne) Store an input character. If EOF is reached, terminate input.
rwd 1 ;Let the output loop start.
jmp   ;Start the output loop.
put   ;Output a character.
rwd 1 ;Go to the next character.
      ;(implicit jnz) If the character is null (terminator), exit.

Answer 10

tcl, 30

puts [string rev [read stdin]]

To terminate input press Ctrl+D.

Available to run on: https://goo.gl/18Sqz0

Answer 11

Python 2, 39 bytes

import sys;print sys.stdin.read()[::-1]

Answer 12

Triangular, 21 bytes

\.~/.?.`.<....p.]p@(<

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This formats into the triangle:

     \
    . ~
   / . ?
  . ` . <
 . . . . p
. ] p @ ( <

This part:

   \
  . ~
 / . ?
. ` . <

creates a loop to continue reading input as long as it's not EOF. ? jumps over the directional command < that restarts the loop.

This part:

 . . . . p
. ] p @ ( <

pops the EOF from the stack, directs control flow to the left, opens a loop, prints the top of stack, pops it, and continues the loop if there are any values left.

Answer 13

8th, 7 bytes

Code

s:rev .

Example

ok> "Quick brown fox\nHe jumped over the lazy dog" s:rev .
god yzal eht revo depmuj eH
xof nworb kciuQ

Answer 14

Chip, 59 bytes

))))))))-v~.
ABCDEFGHS<8<
01234567 >9s
))))))))~\t
abcdefgh

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Requires either a null terminator, or the -z flag (which handles that for you). Since I use the flag only for the benefit of TIO, it is not included in the byte count.

ABCDEFGH            These are the eight bits of the input.

01234567            These are the eight bits of the stack head.

abcdefgh            These are the eight bits of the output.

))))))))-v~.        When at least one bit is on, enable writing to the stack
ABCDEFGHS<8<        (9) and suppress output (S). When all bits are zero,
         >9s        read from the stack (8) and suppress input (s) instead.

ABCDEFGH            When stack is in write mode, copy input bits directly
01234567            onto the stack.

01234567            When the stack is in read mode, send the values directly
))))))))~\t         to output. Once the stack is empty (all bits are zero),
abcdefgh            terminate the program (t).

Answer 15

K (oK), 12 bytes

Solution:

`0:||:'"\n"\

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Example:

`0:||:'"\n"\"Quick brown fox\nHe jumped over the lazy dog";
god yzal eht revo depmuj eH
xof nworb kciuQ

Explanation:

We can't easily take STDIN, so take a string, split on newline, reverse each line and then reverse the list of lines before printing to STDOUT:

`0:||:'"\n"\ / the solution
       "\n"\ / split (\) on newline "\n"
    |:'      / reverse (|:) each (')
   |         / reverse (|)
`0:          / print to STDOUT

Answer 16

Jelly, 6 bytes

⁸ƈ;$ÐL

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Answer 17

Matlab (56)

a=1;b=0;while(a)a=input('','s');b=[flipud(a) 10 b];end,b

Execution:

abc
def


b =


fed
cba

Answer 18

QBIC, 28 bytes

{_?~A=B|_X\Z=_fA|+chr$(13)+Z

Explanation

I can't just grab a multi-line string off of the cmd line parameters with the ; command, that doesn't work in QBasic. Also, when we ask the user for input with _? it terminates on enter. We work around that with a loop:

{          DO infinitely
_?         Ask the user for input, store it in A$
~A=B       IF A$ is empty (or equal to B$ which is unassigned and therefore == '')
|_X        THEN QUIT, and print Z$ on the way out
\Z=+Z      ELSE set Z to be
_fA|       The last line entered reversed
+chr$(13)  Plus a newline
+Z         plus whatever already was in Z$
           The IF and the DO-loop are closed implicitly

Answer 19

Implicit, 3 bytes

©®"

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©    consume all input
 ®   reverse the stack
  "  stringify entire stack
     implicit output

For an alternate and faster version, use ©"\, which reads all input, stringifies it, and reverses the string. It's faster than reversing an entire stack.

Version without builtins:

(~.);(¸@;)
(~.);       read all input with each character incremented by 1
     (¸@;)  decrement each character, print, pop loop

Answer 20

Forked, 32 bytes

v
>-v
| ~&-:-v
| |  | !
\-:-p^-<

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The first block is the same as in the 44-byte solution. The second changes a bit direction-wise:

   &-:-v
     | !
    p^-<

Still fairly self-explanatory once you read the 44-byte explanation.

---

Forked, 44 bytes

v     &<
>-v    |
| ~  >-:
| |  | |
^-:-p^!<

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I love how self-explanatory this language is. However, I'll still explain it, as the conditional structure is a bit complex.

The first block thingy reads all input to the stack:

>-v
| ~
| |
^-:-

The fork at the bottom : directs the instruction pointer West if the inputted character is > 0, causing it to go back into the loop. When EOF is entered, it directs it East, causing it to enter the second block thingy:

      &<
       |
     >-:
     | |
    p^!<

First, it pops the EOF character. Then it goes North, then West, then hits the fork. While the top of stack is nonzero (i.e. it exists), the fork directs the IP South, then the < immediately directs it West, hitting ! (print as character and pop), and it goes back into the loop. The fork directs the IP North if it's zero (i.e. the stack is empty), where it hits < and then & (exit, in this scenario). (note that the redirect is entirely unnecessary but the bytecount is the same, so... whatever.)

Answer 21

Add++, 32 bytes

x:1
y:''
Wx,]getchar,`y,x+,`x
oy

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Fairly basic, although STDIN support was recently added to Add++, so I decided to show it off.

First, we set the two variables we need:

x:1
y:''

x to the integer 1 and y to the empty string. Next, we loop over each character in STDIN:

Wx,]getchar,`y,x+,`x

This is a while loop, with the condition simply being x. Each , denotes the separation of a new command, so the code is expanded into

Wx,
  ]getchar
  `y
  x+
  `x

]getchar is an additional, Add++ jargon for an extended command that is prefixed with a ]. Here, it simply reads a character from STDIN and assigns that to x. If the end of STDIN is reached, an empty string is returned.

Next, with

`y
x+
`x

We prepend this character to y, effectively building the string in reverse.

Once all input has been read, the x variable contains the empty string, and the while loop is terminated. Then we reach the final command

oy

This uses prefix notation: o is the command, and y indicates the variable to operate here. Here, o means output, without a trailing newline. y contains the input reversed, so this outputs our final result.

Answer 22

Pepe, 16 bytes

REEeREEEEeEeeReee

Try it online! (Compiler makes whitespaces when doing links between e and r)

Explanation:

REEeREEEEeEeeReee - full program

REEe              - insert input as string
    REEEEeEee     - reverse whole stack
             Reee - output

Answer 23

Ahead,  9  13 bytes

~ilj~#
 >dko@

This one will function properly when NULs are on the input.

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Answer 24

PowerShell, 38 bytes

Thanks to Joey for the $($input) expression.

$($input)-join'
'|% t*y|%{$s=$_+$s}
$s

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---

Alternative, 38 bytes

@($input)-join'
'|% t*y|%{$s=$_+$s}
$s

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Answer 25

Pip, 3 bytes

RVq

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Answer 26

Brain-Flak, 12 bytes

{({}<>)<>}<>

Not going to be winning with this, but it works

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Answer 27

Rust, 57 bytes

fn q(s:&str)->String{s.chars().rev().collect::<String>()}

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Dart, 34 bytes

f(s)=>s.split('').reversed.join();

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Pretty convoluted, you have to get a String List then reverse it and join it back for it to work.

Answer 28

Triangular, 14 bytes

(\~(#vp]<./)?<

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I am almost 100% certain that ,~#n^`>p/ (9 bytes) would work if IP switches behaved in accordance with their specification, but in the meantime, 14 ain't too shabby.

Ungolfed:

     ( 
    \ ~ 
   ( # v 
  p ] < .
 / ) ? <
----------------------------------------------
(                Set a point for the IP to jump to
 ~v<             Read a character from input, change directions twice
    ?)/          ) returns to the previously set point. ? will skip the jump back if ToS < 0
       p(        Pop the top value of the stack (the null-input), then set a new jump point
         /#<[    Pop the top value of the stack and print that value, then jump back if ToS > 0

Answer 29

Elixir, 42 bytes

IO.puts String.reverse IO.read :stdio,:all

Answer 30

MAWP, 16 bytes

[25W|]%%~%%0~[;]

Try it!