Golf you a quine for great good!

Question

Using your language of choice, golf a quine.

A quine is a non-empty computer program which takes no input and produces a copy of its own source code as its only output.

No cheating -- that means that you can't just read the source file and print it. Also, in many languages, an empty file is also a quine: that isn't considered a legit quine either.

No error quines -- there is already a separate challenge for error quines.

Points for:

• Smallest code (in bytes)
• Most obfuscated/obscure solution
• Using esoteric/obscure languages
• Successfully using languages that are difficult to golf in

The following Stack Snippet can be used to get a quick view of the current score in each language, and thus to know which languages have existing answers and what sort of target you have to beat:

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Answer 1

Forte, 66 bytes

Updated for the new Interpreter

2PUT34:LET1=3
4PUT34:END
1PRINT
"2PUT34:LET1=3
4PUT34:END
1PRINT
"

Which, in order is:

1: Print the first half of the code.
2: Print a ", then set line 3 to be line 1.
3: Print the second half of the code again.
4: Print another ", then end the program.

Try it online!

Answer 2

JavaScript (ES6 REPL), 22 bytes

f=_=>"f="+f+";f()";f()

Idea stolen from Kendall Frey but in less bytes.

Since I cannot comment on his answer because I don't have rep I decided to make a new answer.

Answer 3

Befunge-93, 15 14 13 bytes

+9*5x:#,_:@#"

Works in this interpreter. x is an unrecognized command which reflects the instruction pointer.

Thanks to Jo King for saving 1 byte.

This 14 byte version works in FBBI:

+9*5<>:#,_:@#"

Try it online!

Answer 4

JavaScript REPL, 21 bytes

(_=$=>`(_=${_})()`)()

It technically doesn't read its own file.

… kind of seems like 0 is also a quine for JavaScript the way this is evaluated, though.

Answer 5

Minecraft Java Edition, 241 bytes

This can be run as a function in a datapack, or by running each of the commands:

data modify storage z x set value ['["data modify storage z x set value ",{"nbt":"x","storage":"z"},"\\ntellraw @a {\\"storage\\":\\"z\\",\\"nbt\\":\\"x[0]\\",\\"interpret\\":true}"]']
tellraw @a {"storage":"z","nbt":"x[0]","interpret":true}

Explanation

data modify storage z x set value ...   # set the variable x in the storage minecraft:z

['...']                                 # to a list containing a string (in JSON rich text format) of

"data modify storage z x set value "    # (a literal part of the source code),

{"nbt":"x","storage":"z"}               # this variable (this will only be evaluated 
                                        # during the tellraw command when it is interpreted), and

"\\ntellraw @a {\\"storage\\":\\"z\\",
  \\"nbt\\":\\"x[0]\\",                 # the rest of the source code
  \\"interpret\\":true}"                # (\ and " must be escaped because this is part of a string).

tellraw @a {"storage":"z","nbt":"x[0]", # then interpret the JSON string and print it to the chat
 "interpret":true}                      # (this will place the variable inside)

The list is necessary so that when it is printed, it will contain the quotes and double backslashes.

Answer 6

C, 78 chars

#define Q(S)char*q=#S;S
Q(main(){printf("#define Q(S)char*q=#S;S\nQ(%s)",q);})

This version is shorter than the familiar 79-character C quine and also doesn't assume ASCII. It does still assume that it's safe to not include stdio.h. (Adding an explicit declaration of printf() brings the length up to 103 chars.)

Answer 7

Shell echo-sed quine:

echo sed -eh -es/[\\\(\\\\\\\\\\\)\\\&\\\|]/\\\\\\\\\\\&/g -es/^/echo\\ / -es/$/\\\|/ -eG|
sed -eh -es/[\(\\\\\)\&\|]/\\\\\&/g -es/^/echo\ / -es/$/\|/ -eG

I wanted to write a sed quine, but sed can only work on its input stream, not generate output spontaneously, so this is an echo-sed quine. This 154-character quine uses command-line sed, which automatically makes it hard to read, and uses three different sed commands, as well as two sequences of eleven backslashes in a row. This quine works in bash, ksh, and sh, but not csh or tcsh.

EDIT:

A blatant, and amusing, cheat: echo $BASH_COMMAND

Another, unreasonably silly, cheat: export PROMPT_COMMAND='echo $BASH_COMMAND';$PROMPT_COMMAND

Answer 8

C, 77 chars

Maybe the easiest one in C.

main(){char*c="main(){char*c=%c%s%c;printf(c,34,c,34);}";printf(c,34,c,34);}

34 is the ASCII decimal for ".

Answer 9

MUMPS, 9 bytes

R w $T(R)

This may fall afoul of the "you can't just read the source file and print it" restriction. Let me explain why I say may.

The line of code you see above constitutes a complete MUMPS "routine" (named R), which is sort of like a single source file in a conventional C-like language... but not quite.

The way MUMPS stores its routines is peculiar among programming languages. Routines are not files living in a regular filesystem. Instead, they are data structures internal to the database itself. The line of code I've supplied above is actually stored as part of the MUMPS global named ^ROUTINE (globals are basically trees). The "R" subtree (in MUMPS parlance, "subscript") of that global would look something like this:

^ROUTINE("R",0)=1
^ROUTINE("R",1)="R w $T(R)"

The first entry is the number of lines of code in the routine. The subsequent entries are the lines of code in the routine itself.

Why do I bring this up? Well, this means that in MUMPS, the routines themselves are first-class entries in the database! One can edit routines by directly manipulating the contents of the ^ROUTINE global, just as one can edit any other global. (Indeed, at the most basic level, if your MUMPS environment doesn't come with an editor, you must invent one for yourself that will edit the ^ROUTINE global on your behalf.)

The ability to manipulate routines in MUMPS code is so important that the standard even defines a function whose explicit purpose is to tell you what code is found at a given line of a given routine. That function is named $T[EXT], and if you give it a pointer to a line of code, it will return the code present at that location.

And that's what we do here. We w[rite] the result of a call to $TEXT(R) - that is, the contents of the line at the first line of the routine R - to the output stream, and since R is only one line long, that makes the program a quine.

This program involves no file IO at all. The whole thing is internal to the MUMPS environment. I claim that this is interesting enough to count as a legitimate quine, despite the fact that this has a surface-level resemblance to a program that just reads and prints the source file.

Answer 10

QBasic, 76 (110) 54 (72)

Tested with QB64 on Windows 7, with auto-formatting turned off.

READ a$:?a$;:WRITE a$:DATA"READ a$:?a$;:WRITE a$:DATA"

: is a statement separator, and ? is a shortcut for PRINT. The main trick here is using DATA and READ so we don't have to split the string up to add the quotes. Edit: I learned this week about the WRITE command, which outputs strings wrapped in double-quotes--a significant byte-saver here!

Since actual QBasic doesn't let you turn off auto-formatting, here's the same thing with proper formatting in 72 bytes:

READ x$: PRINT x$;: WRITE x$: DATA "READ x$: PRINT x$;: WRITE x$: DATA "

---

Original versions (76 bytes golfed, 110 formatted):

READ a$:q$=CHR$(34):?a$+q$+a$+q$:DATA"READ a$:q$=CHR$(34):?a$+q$+a$+q$:DATA"

or

READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "

Answer 11

RProgN, 3 bytes

0
0

Try it online!

This exploits a potential flaw in our definition of proper quine:

It must be possible to identify a section of the program which encodes a different part of the program. ("Different" meaning that the two parts appear in different positions.)

Furthermore, a quine must not access its own source, directly or indirectly.

The stack of RProgN is printed backwards, so the first 0 encodes the second 0, and vice versa.

This can be verified empirically; the program

1
2

prints

2
1

Try it online!

Answer 12

Klein, 11 + 6 = 17 bytes

3 additional bytes for the topology argument 001 and another 3 for ASCII output -A.

:?/:2+@> "

Try it online!

Let's start with the topology. The 1 at the end indicates that the north and south edges of the code are mapped to each other in reverse. So if the IP leaves the code through the south edge in the leftmost column, it will re-enter through the north edge in the rightmost column. We use this to skip to the end of the program.

:             Duplicate the top of the stack (implicitly zero).
?             Skip the next command if that value is non-zero (which it isn't).
/             Reflect the IP north.
              The IP leaves through the north edge in the third column from
              the left, so it will re-enter from the south edge in the third
              column from the right.
>             Move east.
":?/:2+@> "   Push the code points of the program, except for the quote itself
              to the stack.
:             Duplicate the top of the stack, now a 32 (the space).
?             Skip the next command (the /).
:             Duplicate the top of the stack again.
2+            Add 2, to turn the space into a quote.
@             Terminate the program.

Answer 13

Reflections, 4222 bytes

Since wastl out-golfed me by about... 1.810³⁷¹ bytes through a vastly superior encoding system, I've decided to have another look at the problem. Since my program is still quite long, here's the main section (with SOHs replaced with spaces):

\0=0#_(4:(2(4(40\
      /# 0v\/(1v/
      \+#@~ > ~<
/#@#_#_#_1^1/
+
\#1)(2:2)4=/

Try It Online! (but have patience) (ASCII-only points out that unchecking the time between steps will make it go faster, but beware of the javascript freezing up your browser)

This uses the same encoding as wastl's answer, where each character with byte value n is represented by n newlines followed by

+
#

and the first character of the code is \ to change the pointer's direction down. Additionally, it also encodes the \ as well as the #,+ and newline in this process to save on doing them later

The main code is a more streamlined version of wastl's, where quite a few shortcuts have been made. I've also replaced all the spaces with SOHs (byte value 1) to save on bytes.

Detailed explanation

\0=0        Create a copy of the data in stack 0
    #_      Print the `\`
      (4:(2(4(4   Push the +, \n, # to stack 4, and a copy of the newline to stack 2
               0\ Switch back to the intact copy of the data

            /(1v/ Reverse the data
            > ~<
          ^1/


          v\
          ~   While the stack exists
          ^

          v\
              
           1  Move data to stack 1

         4=/  Copy #, \n, +
    (2:2)     Copy newline
\#1)          Get top of data

+
\#        Redefine origin and move up

/
+     Push -2

/#@   Print the newline the value of the top of data times
   #_#_#_   Print the +, \n, #
         1^ Switch back to the data and loop again


      /# 0v When the data stack is empty
      \+#@~
/#@#_#_#_1^

         0  Switch to the other copy of the data

      /#    Redefine the origin to push 1
      \+
        #@  Print the whole stack
          ~ >  And end

Answer 14

J (REPL) - 20 (16?) char

Seems we're missing a J entry. Trivially, any sentence that doesn't evaluate gets itself printed in the REPL, so 1 or + or +/ % # are all quines in that sense. A non-trivial quine would be one that produces specifically a string containing the source code.

',~@,~u:39',~@,~u:39

u:39 is the ASCII character 39, i.e. the single quote, and ',~@,~u:39' is a string. , is the append verb. The main verb ,~@,~ evaluates as follows:

x ,~@,~ y      
y ,~@, x       NB. x f~ y => y f x       "Passive"
,~ (y , x)     NB. x f@g y => f (x g y)  "At"
(y,x) , (y,x)  NB. f~ y => y f y         "Reflex"

So the result is 'string'string when x is string and y is the single quote, and thus this is a quine when x is ,~@,~u:39.

If we're allowed the J standard library as well, then we can write the 16 character

(,quote)'(,quote)'

which appends the quote of the string (,quote) to itself.

Answer 15

Unary, ~6.1*10^4391 bytes

000000000...(more 0s than fits in the observable universe)...000

For reasons that may be somewhat obvious, you can't try this one online.

What the hell is Unary?

Unary is a joke(?) esolang with a pretty basic premise: the only valid symbol is the number 0. To interpret Unary, the 0s are first counted, then that number is converted into binary, then that number is converted into brainfuck (000 corresponds to +, 001 corresponds to -, and so on), then the brainfuck is run.

It's worth noting that you can't distinguish between the numbers 000001 and 001 even though they represent different brainfuck, so in all cases, Unary requires us to include an extra 1 bit at the top of the binary representation: we instead write the binary numbers 1000001 or 1001, which are distinguishable.

Basic overview/comments

Whew, this was a huge pain. As a forewarning, I'm almost certain it's not optimal (there are quite a few structures that are pretty space-inefficient used in the brainfuck) but I'm also almost certain there's no way this is ever fitting on any real computer, so I don't feel too bad about it.

The basic idea for this is probably about what you expect: there's a big array that represents the rest of the code (with a cell set to negative one at the start for navigation), and there's a bunch of code which takes an array and prints out unary code representing the code initializing the array (and the negative one flag), then the code represented by the array.

Obviously it's impossible to test this thing, but I tested the actual quining part on the smallest possible array input ("->+>>" representing "-++"), and it correctly output exactly 19,407,936 0s (corresponding to "->+>>-++"), which is good enough for me.

Finally, I'd like to point out that this is technically a Unary/Lenguage polyglot quine (as all Unary code is also valid lenguage with the same function), though a proper Lenguage golf adapted from this one would probably be a couple hundred orders of magnitude smaller.

Detailed overview

The core idea of this code is to use our array as a base-8 "number", with each entry being another base-8 "digit". We can cascade this downwards (subtract 1 from an entry and add 8 to the one directly after it), and then print a 0 for every entry in the last array entry. Doing this process on the original array prints Unary code for whatever brainfuck the array represented (the representation is as you might expect: every entry is a different symbol, and every entry contains a number 0-7, one corresponding to each brainfuck command).

You'll note that technically, we're doing this "backwards": printing all the 0s for the code before we print the 0s for the array. I'm not certain that this is necessary, but it does enable us to use some clever constructions later on to save quite a bit of work.

This process, incidentally, will destroy whatever array it's run on. Thus, before we do that, we need to copy the array (actually quite an obnoxious task given that the arrray is variable-length).

Thus after we print the code itself, we have data looking like this:

-1 flag | [original array] | -2 flag | [empty array with the same length] | -3 flag

This means we only need to print the Unary for the array itself, the -1 flag at the start of the array, and the 1 bit at the top of the binary representation which Unary requires.

Fortunately, we can actually use similar code to do this! In order to encode further data, we need to print out 0s in chunks of some very large power of 8: ie, if our code so far was 50 characters long, in order to encode a "-" before the code, we would have to print 8^50 0s.

The power of 8 we have to use is exactly the length of that big empty array. This means that if we put a 1 in the top register of that array, it will increment the last symbol encoded by 1, and if we put an 8 in the top register it will encode a new symbol (specifically a minus).

Furthermore if we move the flag at the end of the array to the right by 1 it'll print out 8 times as many 0s, and thus start editing the symbol before the last symbol encoded!

Since a plus is just "000", we can just move the flag to the right by the value of the last position in the array. This adds as many "000"s to our binary output as there were plusses in the original array representation. When more stuff is added to our array and cascaded down, this will correctly encode the plusses.

We then move the flag at the start of the empty array to the left by 1: this means we start editing the symbol before the plusses we just added. Conveniently, the space it moves into will have been emptied by the operation that moved the end of the array just before now.

All this means that if we put a 2 in the top register of the empty array and then cascade it'll actually print a right carat, and the blank spaces left from moving b will interpret to an appropriate amount of plusses.

We can then repeat this for every single entry in the filled array, thus encoding all of the array representation.

Memory now looks like this

-1 flag|-2 flag| [huge empty array, 2 times as long as the original array]| -3 flag

We now need to add one last minus (to initialize the -1 flag we just got to). All we need to do this is change the minus 2 to a plus 1, then cascade. This both performs the appropriate multiplication by 8, and encodes a -.

Finally, we just need to encode the 1 at the top of the binary representation. To do this, we move b over by 1 one last time and add a 1 at the top, then cascade.

And there it is! We've now printed out 0s representing every part of our original code.

Thank you, and I'm sorry.

Source code

The brainfuck is a little bit long when fully commented (~200 lines), so I'm not certain if I should post it, but I figure y'all can at least have the un-pretty source:

->++>+>+>++>++>+>+>++++++>+>+++>>+++++++>+>++>++++++>>>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>>>++++++>+>+++>>+++++++>+>++>+++++++>>>>++++++>+>+>+>++>>>>+++++++>++>+>+>++++++>+>+++>>+++++++>+>++>+>+>+>+>++>>>++++++>+>+>+++>++>++++++>>>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>>>>>>++++++>+>+>+>+>+++>>>>>+++++++>+>+>+>+>++>+++++++>>>>++++++>+>+>+>++>>>>+++++++>++>>++++++>+>+>+>+>+++>>>>>+++++++>++>+>+>+>+>++>>>+++++++>+>+>+++>>>>>>>>++++++>+>+>+>++>>>>+++++++>+++>+>+>++++++>+>+++>>+++++++>+>++>++>+>+>++++++>+>+>+>++>>>>+++++++>+>+>+>++>++>+>+>++++++>+>+>+++>>>+++++++>+>+>++>++++++>>>>>++++++>+>+>+>+>+>+++>>>>>>+++++++>+>+>+>+>+>+++>>>>>>++++++>+>+>+>+>++>>>>>+++++++>+>+>+>+>+++>>>>++++++>+>+>+++>>>+++++++>+>+>++>+++++++>+>++++++>+>+>+>+>+>+++>>>>>>+++++++>++>+>++++++>+>+>+>+>++>>>>>+++++++>++>>>++++++>+>+>+>+>+>+>+++>>>>>>>+++++++>+>+>+>+>+>+>++>++>>>>++++++>+>+>+>+++>++++++>>>>>++++++>+>+>+>+>+>+++>>>>>>+++++++>+>+>+>+>+>+++>>>>>>++++++>+>+>+>+>++>>>>>+++++++>+>+>+>+>+++>>>>>>>>++++++>+>+>+>+>+>+>+++>>>>>>>+++++++>+>+>+>+>+>+>++>+++++++>>>>>>++++++>+>+>+>+>+>+++>>>>>>+++++++>++>+>++++++>+>+>+>+>++>>>>>+++++++>++>>>++++++>+>+>+>+>+>+>+++>>>>>>>+++++++>++>+>+>+>+>+>+>++>++>>>>+++++++>+>+>+>+++>+++>>>>>>>>>>>>>>++++++>+>+>+>+>+>+>+>+++>>>>>>>>+++++++>>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>++++++>>++++++>+>+>+++>>>+++++++>+>+>+++>>>>>>++++++>+>+>+>+>++>>>>>+++++++>+>+>+>+>+++>>>>>++++++>+>+>+>+++>>>>+++++++>+>+>+>+++>+++++++>++>++>++++++>>++++++>+>+>+++>>>+++++++>+>+>++>>>>>++++++>+>+>+>++>>>>+++++++>+>+>+>++>+++++++>>>++++++>+>+>+++>>>+++++++>+>+>++>++>++>+>+>+>++++++>+>+>+>++>>>>+++++++>+>+>+>++>+>+>+>+>+>++>++>>>>>++++++>+>+>+>+>+++>++++++>>>>>>++++++>+>+>+>+>+>+>+++>>>>>>>+++++++>+>+>+>+>+>+>+++>>>>>>>++++++>+>+>+>+>+>++>>>>>>+++++++>+>+>+>+>+>++>+++++++>>>>>>>++++++>+>+>+>+>+>+>+++>>>>>>>+++++++>++>+>++++++>+>+>+>+>+>++>>>>>>+++++++>++>+>+>+>+>+>++>++>>>>>+++++++>+>+>+>+>+++>+++>>>>>>>>>>>>>>>++++++>+>+>+>+>+>+>+>+>+>+++>>>>>>>>>>+++++++>>++++++>+>+>+>+>++>>>>>+++++++>+>+>+>+>+++>++++++>>>++++++>+>+>+>+++>>>>+++++++>+>+>+>+++>>>>>>++++++>+>+>+>+>++>>>>>+++++++>+>+>+>+>+++>+++++++>>>>++++++>+>+>+>+++>>>>+++++++>+>+>+>++>++>>>>>>>++++++>+>+++>>>>>>>>>++>+++++++>+++>++>+>+>+>+>+>+++>+++>+++>++++++>+>++>++>++++>+++>+++>+++++++>>>>++++++>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>+++>++++++>+>++>++>++>++++>++++>++++>++++>++++>++++>++++>++++>+++>+++>+++>+++++++>>++++++>+>++++++>++>+>++>+++++++>+++>>+++++++>+++>+>++>>>>>>>>>+++>>>>+++++++>+>+>++>+>+>+>+>++++++>+>+>+>+>++>>>>>+++++++>>>>>>++++++>+>+>+>+>+>+++>>>>>>+++++++>>>++++++>+>+>+++>>>+++++++>+>+>+++>>++++++>+>++++++>>>++++++>+>+>+>++>>>>+++++++>+>+>+>++>++++++>+>++>>+++>+++++++>+++>>>>++>+>++++++>+>+>+++>>>+++++++>+>+>+++>+++++++>++>>>+++>+>+>++>>>>>>++++++>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>+++>++++++>+>++>++>++>++++>++++>++++>++++>++++>++++>++++>++++>+++>+++>+++>+++++++>>++++++>+>++++++>++>+>++>+++++++>+++>>+++++++>+++>+>++>>>>>>>>>+++>>>>+++++++>+>+>++>+>+>+>+>+>+>++++++>+>+>+++>>>+++++++>+>+>+++>>+++++++>+>++>>>>>>++++++>>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>+++>++++++>+>++>++>++>++++>++++>++++>++++>++++>++++>++++>++++>+++>+++>+++>+++++++>>>++++++>+>+>++++++>++>+>+>++>+++++++>+++>>>+++++++>+++>+>++>>>>>>>>>+++>>>+++++++>+>++>+>+>+>+>+>++++++>+>+>+>++>>>>+++++++>+>+>+>++>++++++>+>++>>+++>+++++++>+++>>>>++>+>+>++++++>+>+++>>+++++++>+>++>>>>++++++>>++++++>+>+>+>++>>>>+++++++>+>+>+>+++>+++>++++++>+>++>++>++>++++>++++>++++>++++>++++>++++>++++>++++>+++>+++>+++>+++++++>>>++++++>+>+>++++++>++>+>+>++>+++++++>+++>>>+++++++>+++>+>++>>>>>>>>>+++>>>+++++++

>-->>--[-<+]->[++[--->+++]---<++[-<+]->]+++[--->+++]>--[-<+]->---->++[--<>[++[--->+++]---<+++++[----<++++]---->]+++[--->+++]>+[----<++++]>---->++]--<+++++++[--->+++]<--[-<+]->>--[--->+++]--->>--[--<++]-->[++++[-----<+++++]-----<+++++[---->++++]----<+++[--<++]-->]-[-----<+++++]>-[---->++++]>++[------<++++++]------>>+++[---<[++++[-----<+++++]-----<+++++[---->++++]----<+++++++[------<++++++]------>]+++++[-----<+++++]>-[---->++++]>++[------<++++++]>------>>+++]---<<+++++++++++++[-------<+++++++]+[--->+++]---<[+[--<++]--<+++++[---->++++]----<++++[---<+++]---<]>>[+[--<++]-->++++[--->+++]--->]++[--<++]-->>>---[--->+++]--->----->>++++[----<[+++++[------<++++++]------<++++++[----->+++++]----->]++++++[------<++++++]>-[----->+++++]>----->>++++]----<<++++++++++++++[---------<+++++++++]+[---->++++]----<[++[---<+++]---<+++++[---->++++]----<]+++[---<+++]--->>++++++[-<++++++++>]<>-----<<<[->>.<<]+++[[--->+++]---<<[->>>........<<<]+[-[>->]<+]<->++++++++<+++]-->----[---->++++]+++++[-----<+++++]++[--<++]--<+[-[++[--->+++]--->[->+<]<+++>-[--<++]--<]>++<-->+++++[[--->+++]---<<[->>>........<<<]+[-[>->]<+]<->++++++++<+++]-->------[--<++]--<+]->+++++[+[--->+++]---<<[->>>........<<<]++[--[>-->]<++]<->++++++++<++]->-----[--->+++]--->[->+<]<+++>--[-<+]->+++[+[--->+++]---<<[->>>........<<<]++[--[>-->]<++]<->++++++++<++]

The exact number of 0s used is

616143364761046940121393482224894339867345380201767130133313486919095927124457356445367231747521140266173370022946551757495986436536081196868045297628272330390384059772395971191779286883783813697712702083311380720802629542181528279259974270122382291334259076670078283456907144090063837123521728109038972289254725923131778417128787123839239037575809566518515144698909511013229597825122095407702198996899541417882553289474401282217717777845438435506256387112472797473268235689408744655098869728879606757978358505806156421963379547318142883884528468751816610374042589741195416986282360597637951748580316227437951083206547340002950237138353873683827224872420688269051148041914106540181545636126975204020515616888610209347943212212684032994981872040279134760921950441974664462626740824769006477972532358261625821022468550562561067773665916456352101740599621566669636512759138416019741339956533219082231457782020998061560188458949523569043543472825311694153186492127832152368263433627346401816742081184272285138868588737702544161051266511345401776302508848404989044851701234612868569930533447734786659774504348270829815069786996595352931052563253833910670459061338664835111006064026972821593395459376847379338727645638102959876509529598180508190944759740278032543135323831096032264282759700516599819159593380726609694016158768372040897232130064627016970719454234890810133557942103242082428193652525717784679766009840567745646644331050845934041815088097332504352438215802005681361724732417128846494405432135929762912067831306775322080038483165731774473085357566563624402414177907348890170209026776293825006497117681077216606902562225184543780956129658804093641090955334441702108200564658744439989789201224157261892503165643308457788492416371138829568712401818445624554600923862455172225972137155277042400428634635794267865563766239731245919228046855426376607293646700303578427596008362291239931429658436763436719678326692915408656227680419507941034924280704994802195387552837470839012206296262948274121554534825929221937317579576287409893717258523001476821273741591905466514991100795003027995820718624627560000685829593037893516218245771605918736777569490302646860116831211143429524438818972119953427472167111184521816913965170021208796837058611193110390517567196364486837447552746813630060986477162908564149642028449379763465797559696051374230746360712939114777230926431439196151657705180265020483638629061181115856223950249806419818375856529987900246040227082601855626646444080216637310658434683889414278770982063250184353729183793452944251724697854181617951614799763586644028479802980036647045105801552589931160634006431597294780772117688186110086846079813170103829768160563126247208942693587842950980471492950218296552846456932278774596999132699204842933464261728949352373762279520360130690593476460298502885309846199176184758132580237138986886575217114963071503275472309839043003811612480724199448586309349213643468070828586583244591263370384564485550086863984151080517206405235910197380819828141069270492593578986828345436437174028486982241355518138451897126139232242703752607826960024771549490116659322453183855123857138475614101348080310760135256588738232612996648332732281155595814319181966181574401320500067489163573212418944844484950259082074896272135947770797061836016279282123019467117684670513267319113631747092736545960755034809236690912731932588771491682676453979945982754468716932978590114642741630083944274379070387978397185461190324579966999262505612480431374579002240579057976399577489156724969889391725011201833796208889208642431523656036515892951544697280547449403879151875515288785339517008449261777065296448378328263083714549226346273500515673503831635754141766922846713000935292132391822069931132294405910519960461798776924488932623464859144582786720457864275598371518594949910770504649195142874620028228890448589489597208338562687621880437805894822197524541279873734273522403771848708628352939445662430217734644866430785751657273790572085485900108956813143322699032604055731135389248624818347278303261582515845621280853277496120155330289546717304073439125810011869192307295393119864362022104242206053603655047887527558361040028989920694801323274066616457054148538214643602629113440692125709885109559566032777838457552957583786148605526507117591555800832794105245862353696525934364943795116979478987279494334241959368528163214344004159915398916504737171079910682142420502023

Answer 16

Bubblegum, 105 Bytes

Hexdump:

00000000: 0000 00ff ff00 0000 ffff 0000 00ff ff00  ................
00000010: 1400 ebff 0000 00ff ff00 0000 ffff 0000  ................
00000020: 00ff ff00 1400 ebff 4288 21c4 0000 1400  ........B.!.....
00000030: ebff 4288 21c4 0000 1400 ebff 4288 21c4  ..B.!.......B.!.
00000040: 0000 1400 ebff 4288 21c4 0000 1400 ebff  ......B.!.......
00000050: 0000 00ff ff00 0000 ffff 0000 00ff ff03  ................
00000060: 1300 0000 0313 0000 00                   .........

Try it online!

(You might want to verify it offline - since the input is hexdump and the output is raw.)

---

This relies on the fact that Bubbleugum tries to DEFLATE decode its input first:

...
o = zlib.decompress(code, -zlib.MAX_WBITS)
...

So if we can find a fixpoint in DEFLATE compression, such that x = zlib.decompress(x, -zlib.MAX_WBITS), we are done. But how to do this?

Part I: Generic Compression Quine

Say we have a compression programming 'language' that has two operations:

• Pn: Print the following n tokens as literals, and skip interpreting them
• Rn: Print the last n tokens printed

Let's write some simple programs in this to understand how it works.

Input   | Output
P1 P0   | P0

Input   | Output
P1 P0   | P0
P1 P1   | P1

Input   | Output
P1 P0   | P0
R1      | P0

Input          | Output
P4 P0 P0 P0 P0 | P0 P0 P0 P0
R4             | P0 P0 P0 P0

Now the question is: Just with these two instructions, can we create a quine? The answer is yes, thanks to Russ Cox:

Input          | Output
P0             | 
P0             |
P0             |
P4 P0 P0 P0 P4 | P0 P0 P0 P4
R4             | P0 P0 P0 P4
P4 R4 P4 R4 P4 | R4 P4 R4 P4
R4             | R4 P4 R4 P4
P4 P0 P0 P0 P0 | P0 P0 P0 P0

(The tokens are not on the same line, but you can check they're the same).

This gives us hope we might be able to write a DEFLATE quine. But we're not close to done yet, since we have to deal with actual file formats and not made up tokens. Read on!

Part II: Zlib and DEFLATE

Zlib usually appends a 2 byte header and a 4 byte checksum to everything it compresses. The 4 byte checksum would make the creation of a quine much more difficult. But luckily, Bubblegum is designed using to utilize the -zlib.MAX_WBITS flag, which skips the header and the checksum! So we just have a raw DEFLATE stream. How does DEFLATE work? The full thing can be a bit complicated, but luckily we only need to pull out the bits that allow us to have our Pn and Rn building blocks.

Part III: The Pn building block

A deflate stream is made up of a series of blocks. Each block starts with the following:

• BFINAL: 1 bit, set to 1 if it's the last block.
• BTYPE: 2 bits. All we need to know is that it's 00 for 'no compression' (ie Pn) and 01 for 'fixed compression' (which turns out to map to Rn).

If we have a 'no compression' block, the rest of the bits in the current byte are set to zero and the next bytes look like:

+---+---+---+---+================================+
|  LEN  | NLEN  |... LEN bytes of literal data...|
+---+---+---+---+================================+

Where LEN is a 2-byte little endian unsigned number of bytes in the literal data, NLEN is the complement of LEN (also unsigned little endian) and we then have N literal bytes. Keeping in mind the first byte is packed from LSB to MSB, this means we can encode the following:

P0 = 00 00 00 ff ff
00000 00 0 | 00000000 | 00000000 | 11111111 | 11111111
^     ^  ^   ^^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^^^^^^^^
|     |  |   LEN = 0x0000          NLEN = ~LEN = 0xFFFF
|     |  |
|     |  \- BFINAL = 0 (not final block)
|     \---- BTYPE = 00 (no compression)
\---------- 5 bits padding in block

P4 = 00 14 00 eb ff
00000 00 0 | 00010100 | 00000000 | 11101011 | 11111111
^     ^  ^   ^^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^^^^^^^^
|     |  |   LEN = 0x0014          NLEN = ~LEN = 0xFFEB
|     |  |
|     |  \- BFINAL = 0 (not final block)
|     \---- BTYPE = 00 (no compression)
\---------- 5 bits padding in block

Why is P4 printing 0x14 = 20 bytes, you ask, instead of 4? Well, the previous token 'quine' had the units of 1 byte ~ 1 token, but we don't have that luxury. So instead, we have a fixed length of 5 bytes per token, since this is the minimum size of a print token. So 4 tokens is 20 bytes.

Part IV: The Rn building block

The BTYPE = 01 allows us to make queries of the form REPEAT(n, q):

Starting from q bytes away in the output, print n bytes.

It shouldn't be hard to see that REPEAT(n, n) gives us Rn. But there's a problem, since it turns out that R4 = REPEAT(20, 20) only takes up 3 bytes instead of 5! Since we are assuming all our tokens take up 5 bytes for our quine to work, this is no good. However, we can introduce some redundancy - it turns out if we define R4 = REPEAT(10, 20), REPEAT(10, 20), then we do the same thing but now the instruction takes up 5 bytes total!

The way these blocks are actually encoded as bytes is a little complex. I'll annotate the block, and to fill in the gaps read the RFC. For compressed blocks, the data is turned from bits into bytes LSB to MSB with a couple of exceptions.

P4 = 42 88 21 c4 00
01000 01 0 | 1 00010 00 | 001000 01 | 11 00010 0 | 0000000 0
^     ^  ^   ^  ^    ^    ^      ^    ^   ^    ^   ^       ^
|     |  |  [5] |    |    |      |    [8] |    |   padding [8]
[3]   |  \- [1] [4] [3]  [6]    [5]      [7]  [6]
      \---- [2]

[1]: BFINAL: 0 (not end block)
[2]: BTYPE: 01 (fixed compression) 
[3]: Literal code 264 (print 10 bytes...)
[4]: Distance code 8 (starting from 17 + ... )
[5]: Extra distance code bits ( ... 3 bytes back) (= 20 total)
[6]: Literal code 264 (print 10 bytes...)
[7]: Distance code 8 (starting from 17 + ... )
[8]: Extra distance code bits ( ... 3 bytes back) (= 20 total)

So we've got all our building blocks! P0, P4, R4 right? Are we done?

Part V: The final tweak

Well, not so fast. Remember we had a bit saying which block was the end block? It turns out, for Python at least, that we need to include this on the last block, else it messes up our program. And unfortunately, if we let P*0 be a P0 end block token, the following is NOT a quine:

Input           | Output
P0              | 
P0              |
P0              | 
P4 P0 P0 P0 P4  | P0 P0 P0 P4
R4              | P0 P0 P0 P4
P4 R4 P4 R4 P4  | R4 P4 R4 P4
R4              | R4 P4 R4 P4 <-\
P*4 P0 P0 P0 P0 | P0 P0 P0 P0   |
^                               |
\--------------+----------------+
               |
          Not the same!

However, if we introduce an R*1, we can fix this quite easily:

Input            | Output
P0               | 
P0               |
P0               | 
P4 P0 P0 P0 P4   | P0 P0 P0 P4
R4               | P0 P0 P0 P4
P4 R4 P4 R4 P4   | R4 P4 R4 P4
R4               | R4 P4 R4 P4
P4 P0 P0 P0 R*1  | P0 P0 P0 R*1
R*1              | R*1

It turns out we can encode R*1 = 03 13 00 00 00, so we are done. Use the following Python program to assemble and verify our DEFLATE quine:

import zlib

P0 = b'\x00\x00\x00\xff\xff'
P4 = b'\x00\x14\x00\xeb\xff'
R4 = b'B\x88!\xc4\x00'
R1_F = b'\x03\x13\x00\x00\x00'

comp = b''
comp += P0
comp += P0
comp += P0
comp += P4 + P0 + P0 + P0 + P4
comp += R4
comp += P4 + R4 + P4 + R4 + P4
comp += R4
comp += P4 + P0 + P0 + P0 + R1_F
comp += R1_F

print(zlib.decompress(comp, -zlib.MAX_WBITS) == comp)

Well done! You are now a certified deflate quine expert™.

Answer 17

TECO, 20 bytes

<Tab>V27:^TJDV<Esc>V27:^TJDV

The <Esc> should be replaced with ASCII 0x1B, and the <Tab> with 0x09.

• <Tab>V27:^TJDV<Esc> inserts the text <Tab>V27:^TJDV. This is not because there is a text insertion mode which TECO starts in by default. Instead, <Tab> text <Esc> is a special insertion command which inserts a tab, and then the text. A string whose own initial delimiter is part of the text -- very handy.
• V prints the current line.
• 27:^T prints the character with ASCII code 27 without the usual conversion to a printable representation.
• J jumps to the beginning of the text.
• D deletes the first character (the tab).
• V prints the line again.

Answer 18

T-SQL 24

This statment reproduces itself in the EVENTINFO column of the output:

dbcc inputbuffer(@@spid)

Explanation:

• dbcc inputbuffer() - Displays the last statement sent from the client with the specified process id to the current instance of Microsoft SQL Server
• @@spid - Retrieves the current process id

tested with SQL Server 2008 R2 and 2012; probably working with other versions as well

Online demo: http://www.sqlfiddle.com/#!3/d41d8/2230

Answer 19

RETURN, 18 bytes

"34¤¤,,,,"34¤¤,,,,

Try it here.

First RETURN program on PPCG ever! RETURN is a language that tries to improve DUP by using nested stacks.

Explanation

"34¤¤,,,,"         Push this string to the stack
          34       Push charcode of " to the stack
            ¤¤     Duplicate top 2 items
              ,,,, Output all 4 stack items from top to bottom

Answer 20

Factor - 74 69 65 bytes

Works on the listener (REPL):

USE: formatting [ "USE: formatting %u dup call" printf ] dup call

This is my first ever quine, I'm sure there must be a shorter one! Already shorter. Now I'm no longer sure... _{(bad pun attempt)}

What it does is:

• USE: formatting import the formatting vocabulary to use printf
• [ "U... printf ] create a quotation (or lambda, or block) on the top of the stack
• dup call duplicate it, and call it

The quotation takes the top of the stack and embeds it into the string as a literal.

Thanks, cat! -> shaved 2 4 more bytes :D

Answer 21

F#, 90 bytes

let q="let q=%A
printf(Printf.TextWriterFormat<_>q)q"
printf(Printf.TextWriterFormat<_>q)q

F#’s smart printf comes back to byte us! We can’t write let q="...";;printf q q, as the first parameter to printf isn’t actually a string:

printf : TextWriterFormat<'T> -> 'T

F# uses some compiler magic under the hood to guarantee type-safe printf calls. For example, "yay %d wow!" is a valid TextWriterFormat<int -> unit> literal, but not a valid TextWriterFormat<double -> unit> literal. But if we define the format string separately, the compiler will see it as a regular old string and complain. Instead, we have to convert q ourselves in the first argument.

What about let q:TextWriterFormat<_>="..."? First of all, that’s two bytes longer. But second of all, the second argument to printf really needs to be a string, otherwise the typechecker will infer that we’re formatting a formatter, which in turn formats a formatter, which formats a…

error FS0001: Type mismatch. Expecting a
    'a    
but given a
    Printf.TextWriterFormat<('a -> unit)>    
The resulting type would be infinite when unifying ''a' and
    'Printf.TextWriterFormat<('a -> unit)>'

Yep, an infinite type. Oops.

Answer 22

S.I.L.O.S, 2642 2593 bytes

Credits to Rohan Jhunjhunwala for the algorithm.

A = 99
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 76
A + 1
set A 105
A + 1
set A 110
A + 1
set A 101
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 67
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 66
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 67
A + 1
set A 10
A + 1
set A 108
A + 1
set A 98
A + 1
set A 108
A + 1
set A 68
A + 1
set A 10
A + 1
set A 67
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 32
A + 1
set A 115
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 73
A + 1
set A 110
A + 1
set A 116
A + 1
set A 32
A + 1
set A 66
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 76
A + 1
set A 105
A + 1
set A 110
A + 1
set A 101
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 66
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 67
A + 1
set A 10
A + 1
set A 105
A + 1
set A 102
A + 1
set A 32
A + 1
set A 66
A + 1
set A 32
A + 1
set A 68
A + 1
set A 10
A + 1
set A 70
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 69
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 70
A + 1
set A 10
A + 1
set A 108
A + 1
set A 98
A + 1
set A 108
A + 1
set A 71
A + 1
set A 10
A + 1
set A 70
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 67
A + 1
set A 104
A + 1
set A 97
A + 1
set A 114
A + 1
set A 32
A + 1
set A 69
A + 1
set A 10
A + 1
set A 69
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 70
A + 1
set A 10
A + 1
set A 105
A + 1
set A 102
A + 1
set A 32
A + 1
set A 69
A + 1
set A 32
A + 1
set A 71
A + 1
printLine A = 99
C = 99
B = get C
lblD
C + 1
print set A 
printInt B
printLine A + 1
B = get C
if B D
F = 99
E = get F
lblG
F + 1
printChar E
E = get F
if E G

Try it online!

Answer 23

S.I.L.O.S, 3057 bytes

A = 99
def S set 
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 72
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 56
A + 1
S A 51
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 100
A + 1
S A 101
A + 1
S A 102
A + 1
S A 32
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 72
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 32
A + 1
S A 83
A + 1
S A 32
A + 1
S A 10
A + 1
S A 67
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 66
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 67
A + 1
S A 10
A + 1
S A 108
A + 1
S A 98
A + 1
S A 108
A + 1
S A 68
A + 1
S A 10
A + 1
S A 67
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 72
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 73
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 66
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 66
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 67
A + 1
S A 10
A + 1
S A 105
A + 1
S A 102
A + 1
S A 32
A + 1
S A 66
A + 1
S A 32
A + 1
S A 68
A + 1
S A 10
A + 1
S A 70
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 69
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 70
A + 1
S A 10
A + 1
S A 108
A + 1
S A 98
A + 1
S A 108
A + 1
S A 71
A + 1
S A 10
A + 1
S A 70
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 69
A + 1
S A 10
A + 1
S A 69
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 70
A + 1
S A 10
A + 1
S A 105
A + 1
S A 102
A + 1
S A 32
A + 1
S A 69
A + 1
S A 32
A + 1
S A 71
A + 1
printLine A = 99
H = 83
print def 
printChar H
printLine  S 
C = 99
B = get C
lblD
C + 1
printChar H
print  A 
printInt B
printLine A + 1
B = get C
if B D
F = 99
E = get F
lblG
F + 1
printChar E
E = get F
if E G

Try it online!

I am ashamed to say this took me a while to write even though most of it was generated by another java program. Thanks to @MartinEnder for helping me out. This is the first quine I have ever written. Credits go to Leaky Nun for most of the code. I "borrowed his code" which was originally inspired by mine. My answer is similar to his, except it shows the "power" of the preprocessor. Hopefully this approach can be used to golf of bytes if done correctly. The goal was to prevent rewriting the word "set" 100's of times.
Please check out his much shorter answer!

Answer 24

Charcoal, 64 31 32 (because of newlines)

My first answer in charcoal ever!

Similar to /// and other languages, just straight up ascii would print itself. however that is not payload and also boring, so here is an actual quine.

taking a golfing tip from Ascii-only, and my realisation that the second looping is pointless, I have reduced by >50%

Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´αα´ＡＦα⁺´´ια

Try it online!

Explanation

(thanks to ascii-only for making most of this.)

Ａ                     α            Assign to a
 ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´α             "α´ＡＦα⁺´´ια", but with ´ escape character with each
                                    character
                                    these are the variable being assigned to, and the
                                    rest of the program that is not the string.

                        ´Ａ         Print Ａ to the grid. current grid: "Ａ"
                           Ｆα⁺´´ι  For each character in a, print ´ + character
                                    this results in the escaped version of the string
                                    which is the literal string that is assigned at the 
                                    start. current grid state: "Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´α"

                                  α Print a ("α´ＡＦα⁺´´ια"), which is the commands after
                                    the string assignment. final grid state vvv:
                                                 "Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´αα´ＡＦα⁺´´ια"

[implicitly print the grid: "Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´αα´ＡＦα⁺´´ια", the source, with a trailing newline]

Answer 25

///, 204 bytes

/<\>/<\\\\>\\\\\\//P1/<>/<<>\><>/<<>\<>\<>\<>\><>\<>\<>\<>\<>\<>\<>/<>/P<>1<>/P<>2<>/<>/P<<>\<>\><>\<>\<>2<>/P<>1<>/<>/<<>\><>/<<<>\<>\>><>\<>\<>/<>/<<>\<>\><>/<>/P<>1//P<\\>\\2/P1//<\>/<<\\>>\\//<\\>//P1

Try it online!

With some helpful whitespace inserted:

/<\>/<\\\\>\\\\\\/
/P1/
    <>/<<>\><>/<<>\<>\<>\<>\><>\<>\<>\<>\<>\<>\<>/<>/P<>1<>/P<>2<>/<>/P<<>\<>\><>\<>\<>2<>/P<>1<>/<>/<<>\><>/<<<>\<>\>><>\<>\<>/<>/<<>\<>\><>/<>/P<>1
/
/P<\\>\\2/P1/
/<\>/<<\\>>\\/
/<\\>//
P1

How it works

• The long third line is the quining data. It is made from the entire rest of the program, with a P2 in the spot where the data itself would fit, and then with the string <> inserted before each character from the set \/12.
  • It would be harmless to put <> before all characters in the data, but only these are necessary - \/ because they need escaping to be copied, and 12 because it's vital to have a break inside P1 and P2 to prevent infinite loops when substituting them.
• The first substitution changes all the <> prefixes into <\\>\\\. The \ in the source <\> is there to prevent its final printable form from being garbled by the other substitutions.
• The second substitution includes the quining data, copying them to the other P1s in the program. The <\\>\\\ prefixes now become <\>\ in both copies.
• The third substitution copies one of the quining data copies (in the substitution itself) into the middle of the other (at the end of the program), marked by the string P<\>\2. In the inner copy, the <\>\ prefix now becomes <> again.
• The fourth substitution changes the inner copy's <> prefixes into <<\>>\. The change is needed to introduce the final backspace, protecting any following \s and /s that are to be printed. The inner <\> is necessary to prevent this substitution from infinitely looping – just a backslash here wouldn't do, as it would be garbled by the fifth substitution.
• The fifth substitution removes all instances of the string <\>, both those remaining in the outer copy of the quining data, and those produced by the fourth substitution.
• Finally, we reach the constructed copy of the program, with suitable backslashes prepended to some characters, ready for printing.

Answer 26

Husk, 8 bytes

S+s"S+s"

Try it online!

Husk is a new golfing functional language created by me and Zgarb. It is based on Haskell, but has an intelligent inferencer that can "guess" the intended meaning of functions used in a program based on their possible types.

Explanation

This is a quite simple program, composed by just three functions:

S is the S combinator from SKI (typed) combinator calculus: it takes two functions and a third value as arguments and applies the first function to the value and to the second function applied to that value (in code: S f g x = f x (g x)).

This gives us +"S+s"(s"S+s"). s stands for show, the Haskell function to convert something to a string: if show is applied to a string, special characters in the string are escaped and the whole string is wrapped in quotes.

We get then +"S+s""\"S+s\"". Here, + is string concatenation; it could also be numeric addition, but types wouldn't match so the other meaning is chosen by the inferencer.

Our result is then "S+s\"S+s\"", which is a string that gets printed simply as S+s"S+s".

Answer 27

JavaScript (Firefox), 44 40 bytes

eval(e="alert('eval(e='+uneval(e)+')')")

Not sure how I haven't thought of this before; it's basically exactly the same as the standard function quine (f=_=>alert('f='+f+';f()'))(), but with a string. Funnily enough, I only thought of this while attempting to demonstrate how similar string-based quines are to function-based quines...

A cross-browser version (avoiding uneval) is 72 bytes:

Q='"';q="'";eval(e="alert('Q='+q+Q+q+';q='+Q+q+Q+';eval(e='+Q+e+Q+')')")

Or ES6, 50 bytes:

Q='"';eval(e="alert(`Q='${Q}';eval(e=${Q+e+Q})`)")

Previous answer, 74 bytes

".replace(/.+/,x=>alert(uneval(x)+x))".replace(/.+/,x=>alert(uneval(x)+x))

Simply takes the whole string and prepends its unevaluated form. Note: uneval may not work in all browsers. Here's a cross-browser version at 113 bytes:

".replace(/.+/,x=>alert(q+x+q+x.replace(/\\d/g,q)),q='1')".replace(/.+/,x=>alert(q+x+q+x.replace(/\d/g,q)),q='"')

Original answer, 118 bytes

Now, this certainly isn't a winner, but AFAIK, this is the first ever non-source-reading quine in JS! :D

alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))

How does it work, you ask? Well, if you look closely, you will see that it's really the same thing repeated twice:

alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))

The logic here is to A) place a copy of the real code in a string, and B) orient this string so the program can be split into two identical halves. But how could we get those quotes in there? Well, we could either navigate an insanely difficult path of inserting backslashes before a quote, or use the (painfully long) workaround String.fromCharCode(34) to retrieve one. The latter method is what I chose.

So, this code puts three copies of the string

,A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=

in an array, then joins them with quotes (using the mentioned workaround):

,A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=

and finally, slices off the unnecessary characters from the beginning and end:

,A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=
alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))alert([A=",A,A].join(String.fromCharCode(34)).slice(49,-9))

This leaves us with the text of the original program, which is alerted to the user.

If the alert is unnecessary, here's a 104-byte alternative:

[A=",A,A].join(String.fromCharCode(34)).slice(48,-3)[A=",A,A].join(String.fromCharCode(34)).slice(48,-3)

Answer 28

Bash, 48 bytes

Q=\';q='echo "Q=\\$Q;q=$Q$q$Q;eval \$q"';eval $q

Try it online!

Answer 29

Reflections, 1.81x10³⁷⁵ bytes

Or to be more accurate, 1807915590203341844429305353197790696509566500122529684898152779329215808774024592945687846574319976372141486620602238832625691964826524660034959965005782214063519831844201877682465421716887160572269094496883424760144353885803319534697097696032244637060648462957246689017512125938853808231760363803562240582599050626092031434403199296384297989898483105306069435021718135129945 bytes.

The relevant section of code is:

+#::(1   \/  \    /: 5;;\
          >v\>:\/:4#+     +\
     /+#   /   2 /4):_    ~/
     \ _   2:#_/ \  _(5#\ v#_\
         *(2 \;1^    ;;4) :54/
         \/ \    1^X    \_/

Where each line is preceeded by 451978897550835461107326338299447674127391625030632421224538194832303952193506148236421961643579994093035371655150559708156422991206631165008739991251445553515879957961050469420616355429221790143067273624220856190036088471450829883674274424008061159265162115739311672254378031484713452057940090950890560145649762656523007858600799824096074497474620776326517358755429533782443 spaces. The amount of spaces is a base 128 encoded version of the second part, with 0 printing all the spaces again.

Edit: H.PWiz points out that the interpreter probably doesn't support this large an integer, so this is all theoretical

How It Works:

+#::(1  Pushes the addition of the x,y coordinates (this is the extremely large number)
        Dupe the number a couple of times and push one of the copies to stack 1
            \
             >
                   Pushes a space to stack 2
            *(2
            \/


             /  \
             >v >:\
        /+#   /   2   Print space number times
        \ _   2:#_/


                #      Pop the extra 0
                \;1^   Switch to stack 1 and start the loop


                   /:4#+      +\
                    /4):_     ~/    Divide the current number by 128
                    \  _(5#\ v      Mod a copy by 128
                   ^      4) :
                           \_/


                             v#_\  If the number is not 0:
                   ^    ;;4) :54/  Print the number and re-enter the loop

                     /: 5;;\
             v\      4             If the number is 0:
                     4             Pop the excess 0
              :            \       And terminate if the divided number is 0
                                   Otherwise return to the space printing loop
              \     1^X

Conclusion: Can be golfed pretty easily, but maybe looking for a better encoding algorithm would be best. Unfortunately, there's basically no way to push an arbitrarily large number without going to that coordinate.

Answer 30

Brain-Flak, 1805 bytes

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Try it online!

-188 bytes by avoiding code duplication

Like Wheat Wizard's answer, I encode every closing bracket as 1. The assignment of numbers to the four opening brackets is chosen to minimize the total length of the quine:

2: ( - 63 instances
3: { - 41 instances
4: < - 24 instances
5: [ -  5 instances

The other major improvement over the old version is a shorter way to create the code points for the various bracket types.

The decoder builds the entire quine on the second stack, from the middle outward. Closing brackets that have yet to be used are stored below a 0 on the second stack. Here is a full explanation of an earlier version of the decoder:

# For each number n from the encoder:
{

 # Push () on second stack (with the opening bracket on top)
 <>(((((()()()()()){}){}){}())[()])<>

 # Store -n for later
 (([{}])

  # n times
  {<({}())

    # Replace ( with (()
    <>((({}())[()]))<>

  >}{}

  # Add 1 to -n
  ())

  # If n was not 1:
  ((){[()]<

     # Add 1 to 1-n
     (({}())<

       # Using existing 40, push 0, 91, 60, 123, and 40 in that order on first stack
       <>(({})<(([(({})())]((()()()()()){})({}{}({})(<>)))({})()()())>)

     # Push 2-n again
     >)

     # Pop n-2 entries from stack
     {({}()<{}>)}{}

     # Get opening bracket and clear remaining generated brackets
     (({}<{{}}>{})

      (<

        # Add 1 if n was 2; add 2 otherwise
        # This gives us the closing bracket
        ({}(){()(<{}>)}

         # Move second stack (down to the 0) to first stack temporarily and remove the zero
         <<>{({}<>)<>}{}>

        # Push closing bracket
        )

      # Push 0
      >)

     # Push opening bracket
     )

     # Move values back to second stack
     <>{({}<>)<>}

   # Else (i.e., if n = 1):
   >}{})

 {

  # Create temporary zero on first stack
  (<{}>)

  # Move second stack over
  <>{({}<>)<>}

  # Move 0 down one spot
  # If this would put 0 at the very bottom, just remove it
  {}({}{(<()>)})

  # Move second stack values back
  <>{({}<>)<>}}{}

}

# Move to second stack for output
<>