Write a script that outputs A to stdout infinitely.
There should be no newlines or separators between the characters.
Standard loopholes apply.
This is code-golf. The shortest solution in each language wins.
Write a script that outputs A to stdout infinitely.
There should be no newlines or separators between the characters.
Standard loopholes apply.
This is code-golf. The shortest solution in each language wins.
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.
SCREAM > OUT.TXT. \$\endgroup\$ 0 * A r 0 One of the very few times that Turing Machine Code can compete overall.
AHHHHhhHHHhHHHhHHHhHHHHhHhhHHHHHHhhhhhhh The right language for the job.
AHHH Start program HhhH Increment cell to 1 HHhH Double cell to 2 HHhH Double cell to 4 HHhH Double cell to 8 HHHh Square cell to 64 HhhH Increment cell to 65 HHHH Loop while cell is nonzero: Hhhh Output cell as ASCII character hhhh End loop Go to Post Office:w 1 l 1 r 1 l.[A]A is waiting at Writer's Depot.A is waiting at Writer's Depot.Go to Writer's Depot:w 1 r 1 l 2 l.Pickup a passenger going to Post Office.Pickup a passenger going to Post Office.Go to Zoom Zoom:n.Go to Post Office:w 3 l 2 r 1 l.Switch to plan A. This is an interesting challenge because it requires both refueling infinitely for gas and getting enough passengers to pay for the gas. If passengers paid enough to pay for the gas used to transport them, the following would be enough:
Go to the Post Office: west 1st left, 1st right, 1st left. [loop] "A" is waiting at the Writer's Depot. Go to Go More: west 1st right, 1st left, 1st left, 2nd right. Go to the Writer's Depot: west 1st right. Pickup a passenger going to the Post Office. Go to the Post Office: north 1st right, 2nd right, 1st left. Switch to plan "loop". However, because passengers do not pay enough this results in only 54 * A, crashing the program once we are out of fuel. We can also go to Zoom Zoom to buy fuel, it's way cheaper to buy here allowing us to get 79 * A. Luckily, picking up two passengers at a time solves this problem of not earning enough per iteration. The code can be golfed further by removing quotes for strings that don't contain whitespace and minimising the directions. Ungolfed it looks like this:
Go to the Post Office: west 1st left, 1st right, 1st left. [loop] "A" is waiting at the Writer's Depot. "A" is waiting at the Writer's Depot. Go to the Writer's Depot: west 1st right, 1st left, 2nd left. Pickup a passenger going to the Post Office. Pickup another passenger going to the Post Office. Go to Zoom Zoom: north. Go to the Post Office: west 3rd left, 2nd right, 1st left. Switch to plan "loop". Before I realised I could go to Zoom Zoom with two passengers I found the following solutions of 321 bytes which have some interesting tricks not used in the shorter solution…
Go to Post Office:w 1 l 1 r 1 l.[A]"A"is waiting at Writer's Depot."A"is waiting at Writer's Depot."A"is waiting at Writer's Depot.Go to Writer's Depot:w 1 r 1 l 2 l.[B]Switch to plan C i.Pickup a passenger going to Post Office.Switch to plan B.[C]Go to Go More:s 1 l.Go to Post Office:e 1 l 1 r 1 r 1 l.Switch to plan A. or
Go to Post Office:w 1 l 1 r 1 l.[A]"AAA"is waiting at Writer's Depot.Go to Writer's Depot:w 1 r 1 l 2 l.Pickup a passenger going to Chop Suey.Go to Zoom Zoom:n.Go to Chop Suey:w 1 l 3 r.[B]Switch to plan C i.Pickup a passenger going to Post Office.Switch to plan B.[C]Go to Post Office:s 1 r 1 l 2 r 1 l.Switch to plan A. Using Go More to get Fuel, even picking up the maximum of three passengers does not give enough money to loop infinitely, so I had to think about it smarter. If we pick up passengers sooner and take them on a longer route through the city they will pay more for the ride. Going to Writer's Depot before getting fuel results in 70 * A. If we pick up three passengers using this route, we can loop infinitely.
Picking up three passengers going to the same destination can be done more byte-efficient than using the same code thrice. Using Switch to plan "name" if noone is waiting we can pick up a single pasenger until all three are picked up. The quotes around the plan name are optional and it any word after it get's interpreted as if noone is waiting. Using this we get our first 321 byte solution, the code below is the ungolfed version:
Go to Post Office: west 1st left, 1st right, 1st left. [loop] [Setup 3 * "A"] "A" is waiting at Writer's Depot. "A" is waiting at Writer's Depot. "A" is waiting at Writer's Depot. Go to Writer's Depot: west 1st right, 1st left, 2nd left. [Pickup 3 * "A"] [before pickup] Switch to plan "after pickup" if noone is waiting. Pickup a passenger going to Post Office. Switch to plan "before pickup". [after pickup] [Get Gas] Go to Go More: south 1st left. [Print 3 * "A"] Go to Post Office: east 1st left, 1st right, 1st right, 1st left. Switch to plan "loop". But there's more! Setting up three different passengers to pick up costs quite a few of bytes. If we instead split up the string "AAA" we can eliminate two lines setting up passengers, at the cost of going past Chop Suey. With this we don't need to go out of our way to visit the Go More gas station anymore, Fueler Up is on our route and Zoom Zoom is very easily accessible too. Looking into Fueler Up, it's the most expensive, in fact, it's so expensive that you don't get enough money even taking three passengers! We can only print 561 * A, 187 loops of three passengers, but not infinite as we are trying. Luckily, Zoom Zoom the cheapest gas station does not require much extra routing at all and taking this route does allows us to enter another infinite loop.
Go to Post Office: west 1st left, 1st right, 1st left. [loop] [Setup 3 * "A"] "AAA" is waiting at Writer's Depot. Go to Writer's Depot: west 1st right, 1st left, 2nd left. Pickup a passenger going to Chop Suey. [Get gas] Go to Zoom Zoom: north. Go to Chop Suey: west 1st left, 3th right. [Pickup 3 * "A"] [before pickup] Switch to plan "after pickup" if noone is waiting. Pickup a passenger going to Post Office. Switch to plan "before pickup". [after pickup] [Print 3 * "A"] Go to Post Office: south 1st right, 1st left, 2nd right, 1st left. Switch to plan "loop". It's sad that this does not safe any bytes, but it's the second time I got two answers with the same number of bytes which I think is cool! (See this DDoouubbllee ssppeeaakk challenge.)
I also realised when I was about to publish that all strings without whitespace can be written without quotes. That golved my actual solution a bit more and does not leave the 321 byte solutions the same length anymore. Even so I still publish this explanation too, because the repeat to pickup three passengers is useful and using Chop Suey was fun :)
Saved 2 3 bytes thanks to Mitchell Spector!!!
yes|tr \\ny A Also for 13 bytes (written by Mitchell Spector):
printf A;./$0 Yet another 13 byter, this one written by pxeger:
yes|tr -c A A printf A;./$0 -- this avoids both the filename and PATH issues, but is 3 bytes longer. \$\endgroup\$ As, not one of A\ns that yes A will generate. They're actually \nAs which is what this post uses. \$\endgroup\$ why cant i cease,i say i scream"A"out loud in agony i cry,as i shouted on i scream"A"out loud in agony is the mood of current times. \$\endgroup\$ print(end="A"). \$\endgroup\$ Print('A')? \$\endgroup\$ print('A') includes the newline \$\endgroup\$ L1: A9 C1 LDA #'A' 20 ED FD JSR COUT 50 F9 BVC L1 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 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>.
ttttlIauqasj In grid form:
ttttli auqasj Represents the following Piet program:
pointer command stack t<4> -> l Push 4 [4] l -> i Dup [4, 4] i -> j Dup [4, 4, 4] j -> s * [4, 16] s -> a * [64] a -> q Push 1 [64, 1] q -> u + [65] u -> a OutC [] Print 'A' a -> t % [] Stack underflow; ignored A(){A(putchar(65));} -4 bytes thanks to ceilingcat!
How could I have forgotten recursion...
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.
.A; also works. \$\endgroup\$ ⍞←⍣≠'A' -2 bytes from Bubbler.
-2 bytes from Adàm using forbidden hacky APL magic.
Older answer:
{⍞←'A'⋄∇⍵}0 {⍞←'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) ⍞←⍣≠'A' seems to work. Try it online! \$\endgroup\$ 'A'o 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.
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 -94 bytes thanks to @user100411.
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 REV JMP REV_JMP:REV JMP OUTPUT DUP JMP REV_JMP } JMP REV_JMP with REV JMP produced 1693 bytes of Malbolge (URL too long so not pastable) \$\endgroup\$ @'A @ is an odd operator.
@1 => 1.0)@az => 'a, 'z)@'a => "aaaaaaaaaa....")@1.0 => @, 1.0).@ # A symbol which does odd things. For a char, it repeats infinitely. 'A # Literal A '"""""'"""'"'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'""'"""
Explanation (A has codepoint 65):
'"""""'""" while(1≠0) '" print unicode character number '"" (×64) 1+1+1+… (64 times) '""" 1 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.
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 $0x21 jmp _start Another Clojure version saving 2 bytes over previous solution, posted as a separate answer because I don't have comment rights yet.
(while 1(pr 'A)) goto! Fun fact: solutions with while and recursion both use 25 bytes. \$\endgroup\$ ^ / \ /do \ ^-----^ /1\ / \ --- /out\ ^----- / \ /chr\ ^----- / \ /65 \ ----- 98 bytes thanks to @Jo King and height-0 pyramids. Also, chr 65 is truthy.
^ / \ /do \ ^-----^ -^ / \ -^ /out\ -^----- / \ /chr\ ^----- -^ -^ / \ /65 \ ----- (0)>0 (1)+ (2)Y[65,>1,>0] (3)p (4)<0 (2,1)<0 (0,1)<1 In essence, this just loops the ball on an add 1 loop until it reaches 65, then it bounces back and forth over p repeatedly printing "A"
(2)Y[65,<0,>0] instead, saving 1 dimension, 2 instructions and 16 bytes. \$\endgroup\$ ~1000~! >0010^ Output as a stream of bits. -9 thanks to DLosc. -2 thanks to JoKing.
01000001 not 00100001, and it doesn't appear to be trivial to fix. \$\endgroup\$ 00100001 on DLosc's thing, 01000001 on TIO. Oh well, it works on TIO. \$\endgroup\$ v->{for(;;)System.out.print("A");} Massive thanks to @Kevin for the lambda solution. I really need to learn how to do that.
class M{public static void main(String[]args){for(;;)System.out.print("A");}} It's a full program and can probably be golfed if I knew how lambdas work in java. Oh well.
args to a and class M{public static to interface M{static. Full program Java 5/6: enum A{A;{for(;;)System.out.print("A");}} (41 bytes); Lambda Java 8+: v->{for(;;)System.out.print("A");} (34 bytes) \$\endgroup\$ main-method (here the relevant tip). PS: Lyxal, I once created this program with explanation for someone else on how Java 8+ lambdas work, so maybe it's also useful for you. :) It mostly explains the different type of lambdas, like with/without parameter(s) and/or return. If you have any questions, feel free to ask. \$\endgroup\$ perl -e 'print A while 1' This is perl 5, version 16, subversion 3 (v5.16.3) built for x86_64-linux-thread-multi \$\endgroup\$ perl -e '$x = A; print $x while 1' \$\endgroup\$ A means: it's a file handle. Your program prints $_ to filehandle A; it doesn't print A to STDOUT. \$\endgroup\$ Derived from my answer to "Shortest code to produce infinite output".
"A"10ɑbuɒ This works because the ɒ instruction don't do anything with the loop index except to check if index < limit. If it is, it loops back to its associated ɑ. Otherwise it exits the loop (index manipulation is handled with e adn ø).
"A"10ɑbuɒ "A" (Push "A") 10 (Loop bounds: 0 to 1) ɑ (Start loop) b (Copy top) u (Print with no trailing) ɒ (End loop)