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Bash Version of C64 Code Art: 10 PRINT CHR$(205.5+RND(1)); : GOTO 10

Asked 2012-11-28T17:59:23.680
24

I picked up a copy of the book 10 PRINT CHR$(205.5+RND(1)); : GOTO 10

This book discusses the art produced by the single line of Commodore 64 BASIC:

10 PRINT CHR$(205.5+RND(1)); : GOTO 10

This just repeatedly prints randomly character 205 or 206 to the screen from the PETSCII set:

I'm not sure why the original uses the characters 205 and 206 instead of the identical 109 and 110. Also, I prefer to add a clear at the beginning. This is what I usually type into the C64:

1?CHR$(147)
2?CHR$(109.5+RND(1));:GOTO2
RUN

You can try this all for yourself in an emulator, such as this one using Flash or JavaScript:

I decided it would be amusing to write a bash line to do something similar.

I currently have:

clear; while :; do [ $(($RANDOM%2)) -eq 0 ] && (printf "\\") || (printf "/"); done;

Two questions:

  1. Any suggestions for making this more concise?
  2. Any suggestions for a better output character? The forward and backward slash are not nearly as beautiful since their points don't line up. Th
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2 Answers

6

After looking at some UTF stuff:

2571 BOX DRAWINGS LIGHT DIAGONAL UPPER RIGHT TO LOWER LEFT
2572 BOX DRAWINGS LIGHT DIAGONAL UPPER LEFT TO LOWER RIGHT

(╱‬ and ╲) seem best.

f="╱╲";while :;do print -n ${f[(RANDOM % 2) + 1]};done

also works in zsh (thanks Clint on OFTC for giving me bits of that)

answered 2012-11-28T18:31:43.493
3

Bash supports Unicode now, so we don't need to use UTF-8 character sequences such as $'\xe2\x95\xb1'.

This is my most-correct version: it loops, prints either / or \ based on a random number as others do.

for((;;x=RANDOM%2+2571)){ printf "\U$x";}
41

My previous best was:

while :;do printf "\U257"$((RANDOM%2+1));done
45

And this one 'cheats' using embedded Unicode (I think for obviousness, maintainability, and simplicity, this is my favourite).

Z=╱╲;for((;;)){ printf ${Z:RANDOM&1:1};}
40

My previous best was:

while Z=╱╲;do printf ${Z:RANDOM&1:1};done
41

And here are some more.

while :;do ((RANDOM&1))&&printf "\U2571"||printf "\U2572";done
while printf -v X "\\\U%d" $((2571+RANDOM%2));do printf $X;done
while :;do printf -v X "\\\U%d" $((2571+RANDOM%2));printf $X;done
while printf -v X '\\U%d' $((2571+RANDOM%2));do printf $X;done
c=('\U2571' '\U2572');while :;do printf ${c[RANDOM&1]};done
X="\U257";while :;do printf $X$((RANDOM%2+1));done

Now, this one runs until we get a stack overflow (not another one!) since bash does not seem to support tail-call elimination yet.

f(){ printf "\U257"$((RANDOM%2+1));f;};f
40

And this is my attempt to implement a crude form of tail-process elimination. But when you have had enough and press ctrl-c, your terminal will vanish.

f(){ printf "\U257"$((RANDOM%2+1));exec bash -c f;};export -f f;f

UPDATE:

And a few more.

X=(╱ ╲);echo -e "\b${X[RANDOM&1]"{1..1000}"}" 46
X=("\U2571" "\U2572");echo -e "\b${X[RANDOM&1]"{1..1000}"}" 60
X=(╱ ╲);while :;do echo -n ${X[RANDOM&1]};done 46
Z=╱╲;while :;do echo -n ${Z:RANDOM&1:1};done 44
answered 2013-05-02T14:07:49.337

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