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

Java (JDK), 96 bytes

class M{public static void main(String[]a){System.out.print(new StringBuffer(a[0]).reverse());}}

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Answer 2

Zsh (builtins only), 22 bytes

<<<${(j::)${(Oas::)*}}

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Answer 3

BRASCA, 1 bytes

Another simple 1-byter.

,

Explanation

<implicit input>    - Push STDIN to the stack
,                   - Reverse the stack 
<implicit output>   - Output the stack, reversed.

Language Link

Github Repo

Answer 4

Burlesque, 3 bytes

<-Q

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Explanation:

      # Implicit input
<-    # Reverse
  Q   # Pretty format (removes wrapping "")
      # Output stack

Answer 5

Whispers v2, 79 bytes

> InputAll
> 0
>> #1
>> (3]
>> 2-L
>> Each 5 4
>> 1ⁿL
>> Each 7 6
>> Output 8

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Answer 6

Twue, 31 bytes

@!#␀::=
@!::>@!#.
#_::~_
::=
@!

Try it on the website! (Note, ␀ represents the byte 0x00, here in the snippet and in the rest of the explanation.)

This was a bit trickier than it might first seem, since we need to account for the fact that the input may contain any character, and hence, may bungle any particular replacement. To combat this, we use the two-character @! which is constantly replaced with @!#., where . is a byte of input, until EOF is hit, eventually looking something like this:

@!#␀#g#o#d# #y#z#a#l# #e#h#t# #r#e#v#o# #d#e#p#m#u#j# #e#H#
#x#o#f# #n#w#o#r#b# #k#c#i#u#Q

Our termination condition is @!#␀. Now, if we were to just use a single character, say @, to take input, the program would then continue to perform replacements on any such character in the input. By separating each character by #, we ensure that @! can never appear as a substring given any input.

The rest of the program is simple; #_ outputs each character, removing the # and that character, until no more replacements can be made. This will work even for # characters present in the input, as the first #_ in the string is matched every time, and we do not start processing these until the input is fully read.

Answer 7

Duocentehexaquinquagesimal, 4 bytes

;&Oo

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Answer 8

itflabtijtslwi, 233 204 bytes

/Z/\/\///Y/\\\\ZH/Y>ZF/P.ZE/C.ZD/.BZB/\/.ZA/Y.Y./B/<Y\HYYYZP1/GGE2GG./E1E3D\./F2.\D\DB/E1D\./F2.\./p.2.AAB/E3D\D/E1E2E3./F1./F3D/p.\2./F1D/<.HDB/F\3./F1D/<.H./<.A\.H.AAA/F2ZP\2/P1Z<HZ/P\3/P1Z<H/<Y\HYYY/P2

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Original

I am not familiar with the language well; I barely understand how the program works. This is a modification of cat program.

/./<\\\\>\\\\\\//P1/GGC.2GG./C.1C.3./.\./P.2.\./.\./././C.1./.\./P.2.\./p.2.\.\.\.\././C.3./.\././C.1C.2C.3./P.1./P.3././p.\2./P.1././<.\>./././P.\3./P.1././<.\>./<.\.\.\.\>.\.\.\.\.\.\./P.2//P\2/P1//<\>///P\3/P1//<\>/<\\\\>\\\\\\/P2

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Answer 9

Pxem, Filename: 15 bytes + Content: 0 bytes = 15 bytes.

• Filename (escaped): .w.i.c\001.+.a.s.p
• Content: empty

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Answer 10

Branch, 10 bytes

,[/,]^[.^]

Try it on the online Branch interpreter!

This works pretty much the same was as the BF answer does. Actually my language is proven TC by being a direct superset of BF, because if you only use the left branch, then you can use / and ^ in place of > and <, and { and } in place of - and + to translate any BF program into Branch (though you will have to do a bit of manipulation with the right branch if you want to depend on BF cells wrapping at 256).

Answer 11

Swift, 86 bytes

var o=""
while let s=readLine(){o="\n" + .init(s.reversed())+o}
print(o,terminator:"")

As far as I'm aware, Swift doesn't have any builtin facilities to get stdin as a sequence that I can just call reversed() on. There's something in the Foundation library that I could for try await over, but the extra setup makes it not worth it.

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Answer 12

K (ngn/k), 2 bytes

|:

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Trivial questions got trivial answers. If it wasn't for the Right operator (:) this could be a one-byter though.

K (ngn/k), 5 bytes

`0:|:

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This one outputs back to stdout, which is what the challenge specifies.

Answer 13

Z80Golf, 12 bytes

00000000: cd03 8038 03f5 e9ff f130 fc76            ...8.....0.v

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start:
    ; while (A = getchar() != EOF)
    ;     push A
.read:
    call   0x8003         ; CD 03 80 -> call getchar
    jr     c, .next       ; 38 03    -> jump to .next if carry (EOF)
    push   af             ; F5       -> Push input to stack
    jp     (hl)           ; E9       -> Jump to HL which == 0 (jr .read)
    ; while (A = pop() != EOF)
    ;     putchar(A)
.write:
    rst    0x38           ; FF       -> Print A
.next:
    pop    af             ; F1       -> Pop input to print
    jr     nc, .write     ; 30 FC    -> print and loop if no carry
    halt                  ; 76       -> Exit

The loops do have a bit of a tricky structure because of how the entry points are and how I am cheeky with jp (hl) instead of jr .read since hl == 0x0000. For all intents and purposes, these are two C style while loops.

Detecting when there is EOF in z80golf is easy, as the getchar syscall will set the carry flag. When the carry is not set, I push the input to the stack and loop again.

However, as for detecting when the input starts, that is a whole new problem. And to answer that, we first need to talk about parallel universes. Or rather...flags.

push af and pop af also push and pop the FLAGS register. It has the following structure

|  7  |  6  |  5  |  4  |  3  |  2  |  1  |  0  |
|  S  |  Z  | n/a |  H  | n/a | P/V |  N  |  C  |

Initially, the FLAGS are 00000000, so the stack looks like this if I type "foo":

      A  F    FLAGS
FFFA: 6F 00 (--------)
FFFC: 6F 00 (--------)
FFFE: 66 00 (--------)

However, when the stack is empty, it wraps around to 0x0000 which is where the code is, and notably, call 0x8003 will set the C flag, which I can test with a jr nc.

      A  F    FLAGS
0000: CD 03 (-----PNC)
0002: 80 38 (---H----)
...

Answer 14

GolfScript, 3 bytes

-1%

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Unless you can somehow represent -1 in one byte, I'm fairly certain 3 bytes is the lowest possible score here.

Answer 15

Go, 103 bytes

import(o "os";."io";."fmt")
func f(){s,_:=ReadAll(o.Stdin)
for i:=len(s)-1;i>=0;i--{Printf("%c",s[i])}}

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Answer 16

Python, 20 Bytes

print(input()[::-1])

[::-1] is the actual thing that reverses the string. Since I haven’t looked at all the answers and this is a trivial programme, I highly suspect this is a duplicate.

Answer 17

Lua, 31 bytes

io.write(io.read"*a":reverse())

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Answer 18

StackCell, 11 bytes

1[@:]`:[;:]

Assumes input does not contain any null bytes, as they would be treated as EOF

Explanation:

• 1: Skip the next instruction ([)
• ([): The start of a loop. Never executed
• @: Input a character from stdin
• :]: Duplicate the top value of the stack, and if it is non-zero jump back to the matching [, else continue
• `: Discard the top value of the stack (0)
• :[: Duplicate the top value of the stack, and if it is zero jump to after the matching ], else continue
• ;: Print the top value of the stack, as a character
• :]: Duplicate the top value of the stack, and if it is non-zero jump back to the matching [, else continue (and halt the program)

Answer 19

(,) 124 110 72 51 Chars or \$51\log_{256}(3)\approx\$10.1 Bytes

(,((()),(()(),(()())((())))),,((())),((()),(()))())

Into the explanation:
The code works by backindexing. Basically setting input to a negative number will next give that many characters from the back. It uses that to make input next give -1 by setting input to 0 (in the loop condition). Then, it sets input to (setting variable 2 to variable 2 + input). Then, it outputs input

Answer 20

Lua, 27 bytes

print(io.read'a':reverse())

This is Lua 5.4 and 5.3 compatible. If you want Lua 5.2, 5.1 and LuaJIT compatible then you need one more byte:

print(io.read'*a':reverse())