Shortest infinite loop producing no output

Question

Your task is to create the shortest infinite loop!

The point of this challenge is to create an infinite loop producing no output, unlike its possible duplicate. The reason to this is because the code might be shorter if no output is given.

Rules

• Each submission must be a full program.
• You must create the shortest infinite loop.
• Even if your program runs out of memory eventually, it is still accepted as long as it is running the whole time from the start to when it runs out of memory. Also when it runs out of memory, it should still not print anything to STDERR.
• The program must take no input (however, reading from a file is allowed), and should not print anything to STDOUT. Output to a file is also forbidden.
• The program must not write anything to STDERR.
• Feel free to use a language (or language version) even if it's newer than this challenge. -Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language. :D
• Submissions are scored in bytes, in an appropriate (pre-existing) encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score - if in doubt, please ask on Meta.
• This is not about finding the language with the shortest infinite loop program. This is about finding the shortest infinite loop program in every language. Therefore, I will not accept an answer.
• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainf**k-derivatives like Alphuck), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.
  
• There should be a website such as Wikipedia, Esolangs, or GitHub for the language. For example, if the language is CJam, then one could link to the site in the header like #[CJam](http://sourceforge.net/p/cjam/wiki/Home/), X bytes.
• Standard loopholes are not allowed.

(I have taken some of these rules from Martin Büttner's "Hello World" challenge)

Please feel free to post in the comments to tell me how this challenge could be improved.

Catalogue

This is a Stack Snippet which generates both an alphabetical catalogue of the used languages, and an overall leaderboard. To make sure your answer shows up, please start it with this Markdown header:

# Language name, X bytes

Obviously replacing Language name and X bytes with the proper items. If you want to link to the languages' website, use this template, as posted above:

#[Language name](http://link.to/the/language), X bytes

Now, finally, here's the snippet: (Try pressing "Full page" for a better view.)

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• CJ Dennis: I've got to start posting programs with a negative byte count to beat all these empty files!
• slebetman: This challenge is interesting because it brings out lots of 0 byte languages (some of which are NOT esolangs). FWIW, most declarative languages have an implicit infinite loop because declarative languages don't have loops in their syntax (they assume they're running in an infinite loop). Ladder diagrams are perhaps among the oldest such languages. Then you have the Instruction Language (IL), a sort of assembly for PLCs that also assume an infinite loop. ILs, like assembly are different between manufacturers
• ThisSuitIsBlackNot: Are programs that read and execute their own source code allowed, or does file I/O break the "must take no input" rule?
• user41805: @ThisSuitIsBlackNot Yes, file input is allowed.
• AAM111: Can you print "", an empty string?

Answer 1

Jelly, 1 byte

ß

Recursively calls the main link. Thanks to tail call optimization, this results in an actual infinite loop.

Try it online!

• Erik the Outgolfer: 01¿, without recursion.

Answer 2

JavaScript, 8 bytes

for(;;);

Since the expression to evaluate is omitted, it will always evaluate to true (i.e. your classic while(true){} loop in "disguise").

• Grant Davis: You don't need the ; at then end.
• Conor O'Brien: @GrantDavis Yes you do. The for loop still expects a statement, and the semicolon tells that there is no statement.
• rydwolf: Although much longer, my personal favorite is eval(s="eval(s)")., because it consumes a LOT of memory

Answer 3

Binary Lambda Calculus, 3 bytes

Before I dive into the explanation, let's start with the program itself.

F†€

Trust me, it's pure luck that they're all printable, and I'll get to why it is 2 1/4 bytes instead of 3 after the explanation. I'll explain this by walking through the process I took to create this program.

To start, BLC programs are just lambda calculus programs encoded in a special way. With this in mind, let's begin with the lambda calculus program that enters an infinite loop, known as omega.

(λx.xx)(λx.xx)

This results in an infinite loop because, according to Wikipedia, it reduces to itself after a single beta-reduction. To convert this into BLC, we must first convert it to De Bruijn indices. It converts into the following:

.λ.11λ.11 (The dots after the λs are necessary for BLC but not part of De Bruijn indices)

Okay, now that it's in De Bruijin indices we can now convert it into BLC where λ translates to 00, function application or . translates to 01, and numbers are represented as 1^n0 where n is the number. Knowing this, it translates into the following binary:

01 00 01 10 10 00 01 10 10

This is why it's 2 1/4 bytes. As BLC instructions aren't full bytes (with the exception of 7), it is rare for programs to fit exactly into a certain byte count. To turn this into hex, we have to pad it in order to make it fit into 3 bytes. Doing this yields the following:

46 86 80

There we have the hex dump of our program! It runs in an infinite loop, doesn't run out of memory to my knowledge, doesn't output anything, and is a complete program that can be saved and run by piping the contents of the file to the official interpreter. You can also pipe the binary text to the interpreter and add the -b flag, to demonstrate that the non-padded version can be run.

Answer 4

Cubix, 0 bytes

Alternatively, any single character does exactly the same.

Test it online!

Cubix is a stack-based 2D language, created by me in March 2016. Cubix differs from ordinary 2D languages in that it's not strictly 2D: the code is wrapped around a cube. Any one-byte program wraps to this cube net:

 ?
>. . . .
 .

where ? represents the character, and the IP (instruction pointer) starts at the arrow. . is simply a no-op, and when the IP reaches the right side of the cube net, it simply wraps back around to the left. Thus, any 1-byte program does nothing forever, no matter what the character is (as long as it's not whitespace).

Thanks to @ErikGolferエリックゴルファー for reminding me that the empty program does the same. When Cubix was originally created, this didn't work, because the interpreter created a cube of size 0 and threw an error when it tried to run. This was fixed a while ago, and it is now impossible to cause an error in Cubix.

Answer 5

Tcl/Tk, 0 bytes

Execute any empty file with the wish interpreter instead of tclsh.

Answer 6

Java6 : 25 Bytes

In Java 6 and previous versions you can execute static initialization block without having main in your class file.

class A{static{for(;;);}}

• ETHproductions: Wouldn't for(;;); work just as well as for(int i=1;i>0;);, or is there some restriction in Java 6?
• user8397947: while(true) is shorter than and as valid as for(int i=1;i>0;).
• Oskar Skog: Why is System.exit(0); needed?
• Akash Thakare: @OskarSkog It's not needed, just added to terminate the loop. Let me correct.

Answer 7

Processing, 8 bytes

for(;;);

This is based on Geobit's answer in Java

Although the code below is not the shortest, it is one of Processing's specialties.

void draw(){}

This draw statement repeats itself over and over again. It is one of the differences between Processing and Java.

• flawr: You can also add JavaScript to your title, code stays the same=)

Answer 8

CJam, 4 bytes

1{}h

Put a 1 on the stack, and loop until 1 is no longer truthy. Using h means that the number is never popped.

• Martin Ender: Alternatively, {1}g.

Answer 9

Beam, 2 bytes

There is a few ways of doing this from the very basic

><

to the very basic

>?

and

>|

• ETHproductions: I just came over here to post this, but then I noticed it in the catalogue snippet. +1

Answer 10

Rust, 17 bytes

Didn't see one in rust so:

fn main(){loop{}}

real original and interesting and so different from all the other entries.

• MutantOctopus: When I try this on the playground, it prints to stderr.
• Liam: @BHustus What is it printing to stderr? I'm not getting anything.
• MutantOctopus: Besides the standard compilation details, running results in /root/entrypoint.sh: line 7: 6 Killed timeout --signal=KILL ${timeout} "$@" via this link: play.rust-lang.org/… That -might- just be the playground and not the actual compiled product though, I didn't have the chance to test a compiled version.
• Liam: @BHustus It's an infinite loop and so the playground kills it after a timeout because they don't want to waste time running my code endlessly. If you run this on a desktop that shouldn't happen.
• MutantOctopus: In that case, disregard what I said.

Answer 11

Half-Broken Car in Heavy Traffic, (削除) 5 (削除ここまで) 4 bytes

^ov#

also in 4 bytes:

o
 #

• Sp3000: I just took the interpreter for a spin, and somehow I get the feeling the looping is not in the actual code execution, but in the interpreter checking for infinite loops. Can't confirm until I manage to understand this code properly though (I don't think that affects the validity of the answer, but it's interesting...)
• user63571: @Sp3000 Took it for a spin. Nice pun!

Answer 12

Linux shebang, 5 bytes

#!./a

Must be named as a.

Will run out of memory, but this is an infinite loop.

• Riker: I can't believe nobody's thought of this! +1
• manatwork: Which system? On Ubuntu exec is a Bash built-in, so I can't specify it as script interpreter.
• usernameak: @manatwork oh, sorry, i forgot that it is built-in. fixed
• manatwork: Not runs out of memory: "bash: ./a: ./: bad interpreter: Too many levels of symbolic links". (Though no actual symlink involved.) Well, infinite for 0.001 second.
• Emmanuel: Not working ...bad interpreter... better put ./a alone in the script. If you want to avoid out of memory use ./a& . That's fun if you ps after. That last script will put your computer in a race of creating processes and managing dead processes

Answer 13

Commodore Basic, 3 bytes

1R╭

PETSCII substitution: ╭ = SHIFT+U, ungolfs to 1 RUN.

Taking advantage of Commodore Basic's shortcut forms and the fact that any immediate-mode command can also be used in a program, this code simply runs itself, forever.

Alternatively, a more thoroughly infinite loop is the immediate-mode command

S|7

(PETSCII: | = SHIFT+Y, ungolfs to SYS 7).

This transfers execution to memory location 0x0007. The BASIC interpreter stores the current search character here; when running the above code, this character is the double-quotation mark with byte value 0x22. Opcode 0x22 is an undocumented HALT opcode, which works by putting the 6510's micro-operation interpreter into an infinite loop. The only way out of this loop is to reset the computer.

• Titus: How about a note on what R and Shift+U respectively Sand Shift+Y will expand to? And You could submit a 6510 assembler program.
• Shaun Bebbers: On the Commodore 64 and VIC-20 you can also use 0 GOTO - if you do not specify a number then GOTO assumes that you mean GOTO 0. As Commodore BASIC on the C64 and VIC are recycled from earlier CBM/PET machines, this may also work on those computers. It does not work on the C128 in native mode though.

Answer 14

TI-BASIC, 2 bytes

If recursion is allowed (is it a loop?)

prgmA

Runs a program named 'A', hence the program must be named the same. Some research revealed that prgm is 1 byte, plus 1 byte for A

Runs out of memory pretty quickly, but that doesn't seem to be addressed above.

• Stretch Maniac: : is the mandatory beginning of each line, and I believe that isn't counted. Also, the title shouldn't be counted. Do you count a .js file name into the byte count? I'll look into the prgm token, though it's a single command.
• Stretch Maniac: In this case, however, the actual title is irrelevant as long as it matches the command.
• The Fifth Marshal: Isn't it three bytes: 1 for the A in the program name
• 12Me21: Why do people always add : to the start of each line? It's just a part of the editor, not the actual code.
• The Fifth Marshal: @raddish0 "Even if your program runs out of memory eventually, it is still accepted as long as it is running the whole time from the start to when it runs out of memory"

Answer 15

Lost, 0 bytes

Try it online!

Another 2D language with a zero byte answer. Funnily enough, the minimum byte count to terminate consistently is 5 bytes:

%@
//

Try it online!

Answer 16

x86-16 Assembly, IBM PC DOS, 2 bytes

There's quite a few machine code answers, but thought I'd contribute a few different takes.

56 PUSH SI ; push 100H onto stack
C3 RET ; pop stack and jump to address (PUSH SI)

On DOS when a program is started the SI register contains the starting address of the program (generally 100H). Push that onto the stack and execute a RET which will pop the stack and jump to that address. Big ol' loop.

---

x86-16 Assembly, 1 byte

F4 HLT ; halt the processor and wait for signal

Okay, so this may depend a little on your definition of infinite loop. The HLT instruction (specifically in 8086 real mode) puts the processor in HALT state and awaits a signal in the form of BINIT#, INIT#, RESET# or interrupt (ref). While it's not technically executing our code, it is in a microcode loop of sorts waiting for one of those signals.

---

Motorola 6800, 2 bytes

9D DD HCF ; halt and catch fire

From Wikipedia:

When this instruction is run the only way to see what it is doing is with an oscilloscope. From the user's point of view the machine halts and defies most attempts to get it restarted. Those persons with indicator lamps on the address bus will see that the processor begins to read all of the memory, sequentially, very quickly. In effect, the address bus turns into a 16 bit counter. However, the processor takes no notice of what it is reading... it just reads.

Answer 17

C (GNU-EFI), 0 bytes

Yes, a C GNU-EFI program won't exit. You need to return for it to exit. Will not say anything, just hang.
And yes, it works with a standart GNU-EFI Makefile.
I used this one:

ARCH = $(shell uname -m | sed s,i[3456789]86,ia32,)
OBJS = main.o
TARGET = main.efi
EFIINC = /usr/include/efi
EFIINCS = -I$(EFIINC) -I$(EFIINC)/$(ARCH) -I$(EFIINC)/protocol
LIB = /usr/lib
EFILIB = /usr/lib/
EFI_CRT_OBJS = $(EFILIB)/crt0-efi-$(ARCH).o
EFI_LDS = $(EFILIB)/elf_$(ARCH)_efi.lds
CFLAGS = $(EFIINCS) -fno-stack-protector -fpic \
 -fshort-wchar -mno-red-zone -Wall
ifeq ($(ARCH),x86_64)
 CFLAGS += -DEFI_FUNCTION_WRAPPER
endif
LDFLAGS = -nostdlib -znocombreloc -T $(EFI_LDS) -shared \
 -Bsymbolic -L $(EFILIB) -L $(LIB) $(EFI_CRT_OBJS)
all: $(TARGET)
main.so: $(OBJS)
 ld $(LDFLAGS) $(OBJS) -o $@ -lefi -lgnuefi
%.efi: %.so
 objcopy -j .text -j .sdata -j .data -j .dynamic \
 -j .dynsym -j .rel -j .rela -j .reloc \
 --target=efi-app-$(ARCH) $^ $@

Hope this counts. And yes I tested this on a virtual machine.

• rydwolf: Welcome to Code Golf! This is interesting, although I'm not sure if it counts as a loop.
• Mark: But does it loop?

Answer 18

Aussie++, 47 bytes

G'DAY MATE!
THE HARD YAKKA FOR f IS()<f();>f();

Tested in commit 0a5de7e. The official syntax for this is:

G'DAY MATE!
I RECKON I'LL HAVE A WALKABOUT UNTIL (YEAH, NAH) <
>
CHEERS C***!

However, that UNTIL loop is so verbose it's shorter to use a recursive function.

Answer 19

RISC-V machine code, 2 bytes

For CPUs implementing RV32I/E-C or RV64I-C:

01 a0 — c.j .

For RV32I/E-C CPUs only:

01 20 — c.jal .

RISC-V machine code (no extensions), 4 bytes

There are 4 sets of 32[1] instructions that work for this

6f 00 00 00 - j .
ef 00 00 00 - jal .
6f 01 00 00 - jal x2, .
...
ef 0f 00 00 - jal x31, .

The other 96 options are based on the conditional jump instructions beq, bge and bgeu with the same register as operands

[1]: 16 on RV32E

• Aaroneous Miller: Welcome to Code Golf, and nice first answer!

Answer 20

MUMPS, 1 byte

f

This is the equivalent of for ( ; ; ) ; in C-like languages. It runs from the prompt as is, though, and does not need to be wrapped in a function declaration or any such thing.

Answer 21

GNU dc, 6 bytes

[dx]dx

Tail recursion FTW, which GNU dc supports. Others might not.

• daniero: That's my length 6 dc snippet :) But on my machine it just crashes. Is that normal?
• Digital Trauma: Which machine do you have? (and which dc?). I'm running GNU dc on Linux, which supports infinite tail recursion. I think possibly other dcs don't support any kind of tail recursion, and end up crashing with a stack overflow.
• daniero: A mac, using the pre-installed dc. It says dc (GNU bc 1.06) 1.3. I see that it works on a Linux box with the slightly newer version (1.3.95)

Answer 22

MSM, 2 bytes

ee

Almost all 2 character strings will work, even two spaces, just don't use any of the following 6 commands :,/.?'.

Answer 23

Javascript (8 bytes)

for(;;);

Edit courtody of @KritixiLithos

Javascript ((削除) 10 bytes (削除ここまで))

while(1){}

• user41805: Try for(;;); It has 8 bytes.
• Daffy: @KritixiLithos I forgot that javascript for loops are basically infinite if used wrong. I'm way too used to languages like python which eventually stop for loops. Very clever :)
• The Fifth Marshal: Although, one can make an infinite for loop in python.

Answer 24

Scheme, 12 bytes

A tail-recursive infinite loop seems most appropriate for scheme :-)

(let l()(l))

even though (do(())()) (the CL variant of which is due to @JoshuaTaylor) would be 2 bytes shorter.

Answer 25

z80 Machine Code, 1 byte

c7 ; RST 00h

Or if assuming the code starts at 0000h is cheating, two bytes:

18 fe ; JR -2

These solutions make no assumptions about the rest of the environment's RAM. If it's filled with zero bytes, we are just spinning through NOPs forever, so we could have a 0-byte solution. (Thanks to Thomas Kwa for pointing this out.)

• lynn: Good point; I've written a note about it.
• Philip Kendall: For special fun, fill RAM with 0xDD (or 0xFD). Or maybe this isn't an infinite loop, because it's just one infinitely long opcode.
• mazzy: 1Byte = 76 ; Halt codegolf.stackexchange.com/a/166100/80745

Answer 26

Fortran, 11 bytes

It is not very creative, but here is the shortest I found:

1goto 1
end

I am pretty sure this is not standards compliant, but it compiles with ifort.

Answer 27

3var, 2 bytes

{}

To quote the docs:

{ Starts a loop which repeats forever
} Ends a loop which repeats forever

Answer 28

R, 8 bytes

repeat{}

I believe this is the shortest way to make a loop that won't stop. We would need an infinite length list to use a for loop and while(T) is longer than repeat.

• user5957401: I thought I had, but it sure as hell doesn't work -- dunno what I did. Sorry about that.
• digEmAll: +1 : This or equivalently repeat 1 (or repeat x with x being any valid single character expression)
• JayCe: @digEmAll doesn’t repeat 1 output something?
• digEmAll: @JayCe: I also thought so, but actually it doesn't...

Answer 29

NASM/YASM x86 assembly, 4 bytes

ja $

$ is the address of the current instruction, and ja jumps there if the carry and zero flags are both unset. (i.e. the Above condition is true.) This is the case in x86-64 Linux at process startup.

ja$ just defines a symbol, instead of being an instruction, so the space is not optional. I did test that this works without a trailing newline, so it really is 4 bytes.

Assemble/link with

$ yasm -felf64 foo.asm && ld -o foo foo.o
ld: warning: cannot find entry symbol _start; defaulting to 0000000000400080

The x86-64 ABI used by Linux doesn't guarantee anything about the state of registers at process startup, but Linux's actual implementation zeroes all the registers other than RSP for a newly-execed process. EFLAGS=0x202 [ IF ], so ja (jump if Above) does jump, because the carry and zero flags aren't set. jg (ZF=0 and SF=0) would also work. Other OSes that initialize flags differently might be able to use one of the other one-letter conditions that require a flag bit to be set: jz, jl, jc, jp, js.

Using ja instead of jmp (unconditional jump) makes the source one byte shorter, but the binary is the same size (2 bytes: one opcode byte, one rel8 displacement, for a total size of 344 bytes for a stripped ELF64 binary. See casey's answer for a 45 byte ELF executable if you're interested in small binary size rather than small source size.)

Answer 30

AT&T (PDP-11) Syntax Assembly: 4 bytes

br .

PDP-11 UNIX A.OUT binary output: (削除) 24 (削除ここまで) 18 bytes

0000000 000407 000002 000000 000000 000000 000000 000000 000000
0000020 000777 000000 000000 000004
0000030

This is the output produced by the assembler. As the sizes in the header show, the last three words are not necessary, it can be cut down to the first 18 bytes.

Some modern assemblers do not support the br instruction, so it would be five bytes for jmp .. And executable headers are generally much bigger these days.

Linux x86-64 binary output, after strip: 336 bytes

Now, OSX's assembler is much more strict. You must have a symbol (by default start, but here I use f) for the entry point, which balloons the size of the source. It also requires a newline at the end of the file.

Mac OS X x86-64 Assembly: 17 bytes

.globl f
f:jmp f

Mac OS Mach-O Binary Output: 4200 bytes

• casey: You can get smaller binary output if you golf the ELF header in Linux: see codegolf.stackexchange.com/a/59479/19694
• Peter Cordes: What machine is br . for? x86? Which assembler? GNU as rejects it for x86-64. In any case, I posted a 4-byte x86 YASM source, which assembles and links into a complete working infinite loop program on Linux: ja $. GNU ld defaults to 0x0400080 if it can't find a _start symbol, so this really is a complete program.
• Random832: @PeterCordes PDP-11 - and I assume also VAX, though I haven't had a chance to test it. It's common for older assemblers to use separate mnemonics for short jump instructions. The PDP-11 instruction contains the offset within the instruction word, with a range of even values from -64 to +62.
• Peter Cordes: @Random832: could you update your answer, then? "AT&T syntax" doesn't specify a machine. For example, there is an i386/amd64 flavour of "AT&T syntax", used by the GNU tools by default. add 16,ドル %rsp, for example.
• Random832: My answer goes into detail on the source required to make a usable x86-64 program on GNU/Linux and Mac OSX. By "AT&T" I meant the actual original assembler produced by AT&T, whose syntax is imitated by these other assemblers.