Write a script that outputs A to stdout infinitely.
There shold be no newlines or seperators between the characters
Standard loopholes apply
This is code-golf. The shortest solution in each language wins.
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x86-16, IBM PC DOS,
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x86-16, IBM PC DOS, 7 6 bytes
00000000: b041 cd29 ebfc .A.)..
Unassembled listing:
B0 41 MOV AL, 'A' ; put 'A' into AL
PRINT:
CD 29 INT 29H ; DOS fast console output char in AL
EB FC JMP PRINT ; loop infinitely
As a bonus, if you run this on your IBM 5151 monitor for a few hours this will actually produce infinite output on that screen until the end of time.
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3\$\begingroup\$ I'd hate to be the idiot that didn't put a screensaver on after running this for even a single minute. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 1:05
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3
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\$\begingroup\$ @S.S.Anne funny to read today that int 29h is undocumented... when virtually every resource on DOS programming tells about it. \$\endgroup\$ – Ruslan Mar 4 '20 at 9:22
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3\$\begingroup\$ Actually, this doesn't print to STDOUT as required in the OP. It prints directly to the screen, so you won't be able to e.g. redirect the output to a file as
SCREAM > OUT.TXT. \$\endgroup\$ – Ruslan Mar 4 '20 at 9:53 -
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Turing Machine Code, 9 bytes
0 * A r 0
One of the very few times that Turing Machine Code can compete overall.
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1\$\begingroup\$ indeed! +1 for spotting this really nice use case :D \$\endgroup\$ – RGS Mar 1 '20 at 19:45
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Bash + core utilities,
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Bash + core utilities, 16 14 13 bytes
Saved 2 3 bytes thanks to Mitchell Spector!!!
yes|tr \\ny A
Also for 13 bytes (written by Mitchell Spector):
Bash, 13 bytes
printf A;./$0
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\$\begingroup\$ Grr... I was going to do this next. Good job, though. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 0:40
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1\$\begingroup\$ You could also do
printf A;./$0-- this avoids both the filename and PATH issues, but is 3 bytes longer. \$\endgroup\$ – Mitchell Spector Feb 29 '20 at 17:36 -
1\$\begingroup\$ @MitchellSpector That's pretty cool! :-) \$\endgroup\$ – Noodle9 Feb 29 '20 at 18:17
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1\$\begingroup\$ @JDL Because you have to get rid of the newlines. You've got to generate a continuous stream of
As, not one ofA\ns thatyes Awill generate. They're actually\nAs which is what this post uses. \$\endgroup\$ – Noodle9 Mar 2 '20 at 12:38
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\$\begingroup\$ Traditionally, you don't answer your own question for a couple of days. However, that's usually for more complex challenges. This one is quite trivial, so I suppose it's OK. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 0:22
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3
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1
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2\$\begingroup\$ @BruceWayne,
print('A')includes the newline \$\endgroup\$ – PenumbraBrah Mar 2 '20 at 1:45
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Apple II 6502 Assembly code, 7 bytes
L1: A9 C1 LDA #'A'
20 ED FD JSR COUT
50 F9 BVC L1
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Vim,
Note: You'll probably have to
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Vim, 14 7 10 bytes
qqiA^[@qq@q
Added 3 bytes to fx a bug kindly pointed out by David.
If you fire up vim (with no command line options) and type in these key strokes (^[ is the esc key) then the screen will fill up with A's.
Explanation:
qqiA^[@qq@q
qq Start recording macro-q
i Enter insert mode
A Insert A
^[ Exit insert mode
@q Call macro-q from within macro-q
q Stop recording macro-q
@q Call macro-q
Note: You'll probably have to kill that session of vim to stop it!
You can try to stopping the macro with ctrl-c, if that works you can exit with :q!<Enter>.
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\$\begingroup\$ @S.S.Anne You'll have to
killit from another terminal on Linux. Windows you can use the task manager. \$\endgroup\$ – Noodle9 Feb 29 '20 at 1:11 -
\$\begingroup\$ Or you can just click on the screen,
^cand type:q!. I knew there was some way to do it. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 1:13 -
\$\begingroup\$ You have to click on the Vim window first. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 1:54
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1\$\begingroup\$ The trick being that you call the macro while still creating it. Nice! \$\endgroup\$ – Jonta Mar 2 '20 at 15:24
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><>, 4 bytes
'A'o
How it works
The instruction pointer begins at left, and its initial direction is to the right.
' starts string parsing mode. Everything until the next ' will be interpreted as individual characters, that will get pushed onto the stack. So A pushes that character, and then the second ' ends string parsing mode.
o pops the character from the stack and outputs it to STDOUT.
The instruction pointer has now reached the end of the code, so it wraps around to the initial position and keeps moving to the right, causing an infinite loop.
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Burlesque, 3 bytes
@'A
@ is an odd operator.
- For integers, it converts to double (
@1 => 1.0) - For two letters, it pushes both to the stack individually (
@az => 'a, 'z) - For characters it prints the character infinitely as a string (
@'a => "aaaaaaaaaa....") - For anything else, it does nothing and just remains on the stack (
@1.0 => @, 1.0).
@ # A symbol which does odd things. For a char, it repeats infinitely.
'A # Literal A
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\$\begingroup\$ Is it useful for the "anything else" scenario? \$\endgroup\$ – S.S. Anne Mar 5 '20 at 18:28
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\$\begingroup\$ @S.S.Anne Not really, I suspect you could push it into a block and eval it. But I've never found cause to. \$\endgroup\$ – DeathIncarnate Mar 6 '20 at 10:10
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C (gcc), 20 bytes
A(){A(putchar(65));}
-4 bytes thanks to ceilingcat!
How could I have forgotten recursion...
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Hexagony, 3 bytes
A.;
A .
; . .
. .
A sets the value of the current memory edge to A (ASCII 65).
. is a no-op that places the next command on a new row.
; prints the value of the current memory edge to stdout.
The no-op is required because the instruction pointer never returns to the top row after executing A.
It only loops over the second and third rows.
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2\$\begingroup\$
.A;also works. \$\endgroup\$ – user92069 Mar 2 '20 at 5:34
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Pyramid Scheme,
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Pyramid Scheme, 111 98 bytes
^
/ \
/do \
^-----^
/1\ / \
--- /out\
^-----
/ \
/chr\
^-----
/ \
/65 \
-----
Edit:
98 bytes thanks to @Jo King and height-0 pyramids. Also, chr 65 is truthy.
^
/ \
/do \
^-----^
-^ / \
-^ /out\
-^-----
/ \
/chr\
^-----
-^
-^
/ \
/65 \
-----
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\$\begingroup\$ O! Thanks! I must admit that I did not know about the height-0 pyramids. \$\endgroup\$ – MarcinKonowalczyk Jul 26 '20 at 8:33
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Malbolge,
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Malbolge, 2163 2069 1881 1787 bytes
b'a;$9"~}HG{iyxwuu?O=pL:]mHj5!3DCezRQ=+^:('&Y$#m!1So.QOO=v('98$65a!}^{@hyf<WV9sr%4#I20FEJVBfw)btOr@#!7~|4{y1xv.us+rp(om%lj"ig}fd"cx``uz]rwvYnslkTonPfOjiKgJeG]\EC_X]@[Z<R;VU7S6QP2N1LK-I,GF(D'BA#?>7~;:9y16w43s10)p-,l*#(i&%e#d!~``{tyxZpuXsrTTongOkdMhg`Hd]ba`_^W@[ZYXW9UNSRQPOHMLKJ-++FE''<A$?>=<;:387xw43s10/(-&m*)('&}${d!~}|^zyxwvutmVqpiRQlkjiKafedc\E`_^@\[ZYX;V9NMRQ42NGLK.IH*F?DCBA$#>7~;{{8xx5uu2rr/oo,ll)ii&f|e"!aw`{z\r[vXnmVTpongPkNihgJ_dcFa`B^]\UZ=RWV8TSLQ4ON0LE.IHA)E>'BA:?!7~5|38y6/v321q).-&m*)i'&%|{d!~}_{zs\wvutsUqTonPlOjiKgJedFbE`_A]@[Z<X;VU7S6QP22GL/JIB+FEDC%;@?>7~;:987w5v32r0)p-,+k)('~g$#"b~w|uz]xwvutsrqTinQlOjLhgfeH]bE`CB]\>ZSXWVUTSRQPON1LE.I,+*((&&$$""~~||zzxxv4u210/(-n+l)(i&g$ddy~}`u^]\ZZotsrTjShQOOMMKgfeG]F[DB^]?[T=R;9UTS5K4I200..,,*F)DC&A:#>=~;|9yyx/vutrrp.-,l$k"i~ge#"!aw`u^\\ZZXXVrqpRhQfOMMKKIeHcbECC^W\?>=;W:UT7R5PIN1L/.,,*FED&<%:#!!}}{987w/v-trrppnnllj(i&%ee"!xa|_^\x[vutWrqjSnQPNNLLJJHHFFDDB^A\[==XWVOT7R542N1LKJ-HGF?D'B%$""~<;:z2y0wu321q)p'nl*)(h~g|eccaa__]][[YuXsrTTonmleNiLgfeG]F[`C^]\?ZYXWP9T76442NML.D-B+)EDC%;$9"~<;:z2y0wuussqqoommk)j'&ff#"!~}v{^y\wvXtmVkpSnmlOjihgf_dGbEDBB@\?==R;PUTS5K4I200..,,**(DCB$:#8!}}{{yyw5v321r/.-,+*#j'h%$#cybw`^^s\ZvuWslUjSQQOOMMKgJedc\E`_B]@[==<QV9T76KPON0F/D-++))'CBA#9"7~||z87w5.u,sqqoommkki'h%$#d!xa`{^\\qZotsUqjShmPkjMhKfe^cFEDYB@@>>S<:VU7SL5J311//--++))'C&A@#>!<;49z76w4u2rr).-n%lkjhhffddb~}|^t]rwZXXmrUpoRmfONihgI_H]FD`_^@V?T=;;9977553311/K.IH+))>C&%@?>~6}49z76w4u,1rq.o,+l)j'~g$#d!b}__^yr[ZuXsrUSSnmfkjiLKfedFbaDY^A\[>Y<WVOTSRQ43H1FKJI+A*?(&BA@"8!6}{987w/v-trrppn,mkk"'&%e{dyb``^^\\ZZXXW22}Rn-O>Nvu(IeH6F[`~1A@hZSRuc9rrqK4\lMkK-CHAS(ubBN:L!J6}kXW1wfv3Prr`;o,%IH(4~}|d/@Q>v{;(\wZ$W4V1}/R-PxjvuKf_$G#nZ}B|z>-xwQc88qR^nO1GL|JVyGeEca&$$?8[6|GjWxg/AR2POq(o,JH6j4&C$0@@-a`^:y[q6H54rq0BR--N*chJ&_%cF!CY}Ai.-wwWV(s6%4o\lZkKDz,fdRQ
This was built with Prof. Masahiko Sakai's LAL toolchain from the following source code.
PROGRAM_START_TO ENTRY@Argh
ROUTINE Argh {
ENTRY:
ROT A
A: 0000021020t
JMP REV_JMP
REV_JMP:REV JMP
OUTPUT
DUP
JMP REV_JMP
}
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Unreadable, 208 bytes
'"""""'"""'"'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'"""
Explanation (A has codepoint 65):
'"""""'""" while(1≠0)
'" print unicode character number
'"" (×64) 1+1+1+… (64 times)
'""" 1
answered Jun 29 '20 at 9:13
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x86-16 machine code (DOS 1+), 8 bytes
Disassembled listing (objdump -D -bbinary -mi8086 scream.com):
0: b4 02 mov $0x2,%ah
2: b2 41 mov $0x41,%dl
4: cd 21 int $0x21
6: eb f8 jmp 0x0
Output (DOSBox 0.74, and a lot faster than it looks):
Here's my ldscript and command-line options for anyone interested:
OUTPUT_ARCH(i8086)
SECTIONS {
. = 0;
}
ENTRY(_start)
OUTPUT_FORMAT(binary)
commands:
as --32 scream.s -o scream.o
ld scream.o -Tldscript.lds -o scream.com
dosbox ./scream.com
and unmolested scream.s file:
_start:
mov $0x02, %ah
mov $0x41, %dl
int $0x21
jmp _start
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\$\begingroup\$ @640KB I guess I should give credit to Ralf Brown's Interrupt List while I'm at it. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 2:08
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\$\begingroup\$ @640KB I was into this a while back but I stopped. It was fun writing low-level code and learning about all the stuff you could and couldn't do. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 2:16
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1\$\begingroup\$ @640KB Roll 'em once and use 'em everywhere. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 2:33
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2
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Jelly, 4 bytes
A niladic link:
”AȮß
Try it online!, or check how it works below. If the "A" could be program input, we could get away with only two bytes: Ȯß
”A The character literal "A"
Ȯ Print it and return it,
ß and recursively call this same link.
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\$\begingroup\$ How long does it take for the stack to overflow? \$\endgroup\$ – TheOnlyMrCat Mar 2 '20 at 20:49
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\$\begingroup\$ I didn't run it locally, only on TIO... And it seems TIO stops the execution because of the output size. So maybe the implementation of the recursive call is actually a loop? Because in Python the usual recursion depth is 1000 or 1024 and I manage to output way more than 1024 "A"s..? Just check the TIO link. \$\endgroup\$ – RGS Mar 2 '20 at 22:01
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1\$\begingroup\$ Here. The interpreter sets the recursion limit to
1 << 30. \$\endgroup\$ – user92069 Mar 3 '20 at 10:56
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Piet,
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Piet, 30 26 Codels
It's a .png 13x2 codels, a few are technically not used (4 white unused, 1 white as transition and 1 black to change direction). But since I don't know how to compress it further, I still count them.
Original file (codel size 1):
With codel size 10:
For some reason, the codel size 10 looks disproportional. Maybe it's an issue with the IDE
Pseudo code (incl. stack):
push 2 | Stack: 2
push 4 | Stack: 2, 4
push 2 | Stack: 2, 4, 2
push 4 | Stack: 2, 4, 2, 4
* | Stack: 2, 4, 8
* | Stack: 2, 32
* | Stack: 64
push 1 | Stack: 64, 1
+ | Stack: 65
dup | Stack: 65, 65
out(char) | Stack: 65 | Output: A
dup and out(char) are repeated indefinitely.
This is my first submission here and my first "real" program in Piet. I'm sure it still has some room for improvement, but I just wanted to share (what I think of as) a lovely language :D
Edit: Compressed down from 15x2 to 13x2.
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1\$\begingroup\$ I do think this is strictly suppossed to be an uppercase "A". Doesn't change your byte count though. \$\endgroup\$ – ouflak Mar 3 '20 at 8:24
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\$\begingroup\$ Dang it, completely forgot about
goto! Fun fact: solutions withwhileand recursion both use 25 bytes. \$\endgroup\$ – val says Reinstate Monica Mar 4 '20 at 10:57
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Labyrinth, 5 bytes
65
".
Simple square loop. Push the number 65, pop and print as charcode, and loop through a no-op.
Labyrinth, 5 bytes
<>.56
I think I found a way to loop through single line of program! (Except that the commands are necessarily backwards, and each of <> pops one value from the stack and uses it as an offset, so the loop should begin with <_>_ instead if the stack is non-empty at the boundary.)
<>.56 At start, IP runs "<" which cyclically shifts the row along with the IP
>.56< Now IP is at the end of the strip, which forces it to run backwards
6 Run commands in this order, printing an 'A'
5
.
> Cyclically shift the row to the right
<>.56 Continue running to the left, now stepping on "<" again
which causes IP to wrap through the edge and run in a loop
Labyrinth, 6 bytes
19
`
.
Uses -191 % 256 == 65. Runs back and forth along the linear path .`19, starting at 1 facing right.
Labyrinth, 5 bytes
~9
.1
Another -191.
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1\$\begingroup\$ I mean, why not just
_65.? Or*65.to avoid memory filling up \$\endgroup\$ – Jo King♦ Sep 30 '20 at 7:31
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APL (Dyalog Extended),
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APL (Dyalog Extended), 11 9 7 bytes
⍞←⍣≠'A'
-2 bytes from Bubbler.
-2 bytes from Adàm using forbidden hacky APL magic.
Older answer:
{⍞←'A'⋄∇⍵}0
Explanation
{⍞←'A'⋄∇⍵}0
⋄ separator (arguments will be evaluated left to right)
⍞←'A' Print 'A' without newline
∇⍵ Call function again with the same right argument
{ }0 Call first time with 0(any number works)
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1
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\$\begingroup\$ Totally a hack, and completely unsupported, but
⍞←⍣≠'A'seems to work. Try it online! \$\endgroup\$ – Adám Nov 10 '20 at 17:43 -
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Octave / MATLAB, 21 bytes
while fprintf('A')end
Explanation
fprintf('A') prints 'A' to STDOUT and returns the number of printed characters, that is, 1. This number is used as condition for the while...end loop, which thus becomes infinite.
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sed. I had to use the-zflag just to even remove one. \$\endgroup\$ – S.S. Anne Feb 29 '20 at 15:36