You are to write a program that will output source code that is
This is code-golf, so shortest answer wins.
a=[show,print]
".|>$([a;a])
".|>a
prints (79 bytes):
".|>Function[show, print, show, print]\n".|>Function[show, print, show, print]
then, the program is doubled each time:
".|>Function[show, print, show, print]\n".|>Function[show, print, show, print]
".|>Function[show, print, show, print]\n".|>Function[show, print, show, print]
based on this quine:
".|>[show,print]".|>[show,print]
<?echo fgets(fopen(__FILE__,'r')).';';
It will add a semicolon at each run.
';'. Then you'd get longer outputs.
\$\endgroup\$
<?=fgets(fopen(__FILE__,'r')).';';?
\$\endgroup\$
Commented
Feb 24, 2014 at 23:54
;<?echo fgets(fopen(__FILE__,'r'));
\$\endgroup\$
Commented
Oct 15, 2014 at 8:59
Place this code in any .bat file and it will continue to execute (in an infinite loop) and the file will grow as well.
echo echo %0 ^>^> %0 >> %0
%0 is the command used to invoke the script, which may not have a .BAT extension. You can use %~nx0 to get the full filename of the batch file.
\$\endgroup\$
.bat to be executed. If the filename is execute.bat, you can enter either execute or execute.bat. Both will work.
\$\endgroup\$
Commented
Feb 24, 2014 at 1:59
.bat extension to execute but you can leave the extension off when you execute it (when you type a command with no extension, Windows tries .com, .exe, then .bat in that order). If the filename is hello.bat, then >> %0 will write a file named hello, which isn't the original hello.bat (and can't be executed).
\$\endgroup\$
set d=time
copy c:\>loop.bat /B loop%d%.bat
call c:\>loop.bat
Javascript - 46
(function $(){console.log('('+$+'());$()')}())
P(Q,Q)
Prints it's own source code, separated by newlines. The second generation would be
P(Q,Q)
P(Q,Q)
and so forth.
0
0
1
1
1
4
1
11
4
1
11
1
2
2
1
12
18
10
18
1
32
22
1
18
26
4
26
8
18
11
6
17
4
26
It first prints two lines with a zero in them, and then compares each line to line 1, if they're equal (which is the case when the file is exhausted), we print what's in cell 26. The resulting output will look the same, except with an added 33, which doesn't do anything. Next time, another line will be added, and so on.
H
Prints Hello, World!.
Other characters than H are valid (and safe) source code (that print other variants of Hello World).
Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´αaα´ＡＦα⁺´´ια
Explanation:
first, go check out this answer on the quine challenge and its explanation.
This program is that program, but with the letter a at the end of the string variable.
Ａ α Assign to a
´α´´´Ａ´Ｆ´α´⁺´´´´´ι´αa "α´ＡＦα⁺´´ιαa", but with ´ escape character with each
character, except a
these are the variable being assigned to, and the
rest of the program that is not the string.
´Ａ Print Ａ to the grid. current grid: "Ａ"
Ｆα⁺´´ι For each character in a, print ´ + character
this results in the escaped version of the string
which is the literal string that is assigned at the
start, plus a escaped. current grid state: "Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´α´a"
α Print a ("α´ＡＦα⁺´´ιαa"), which is the commands after
the string assignment. final grid state vvv:
"Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´α´aα´ＡＦα⁺´´ιαa"
[implicitly print the grid: "Ａ´α´´´Ａ´Ｆ´α´⁺´´´´´ι´α´aα´ＡＦα⁺´´ιαa", the new program]
This new program works in the same way, except for two things:
the "a" is escaped (which has no effect, as "a" escaped is "a")
there is an "a" at the end of the program. this "a" will write an extra "a" after the source is produced. thus the program grows slowly
i;main(a){printf(a="i;main(a){printf(a=%c%s%1$c,34,a,i+2);for(i=%d;i--;)",34,a,i+2);for(i=1;i--;)printf(a="printf(a=%c%s%1$c,34,a,34),",34,a);printf(a="0;printf(a=%c%s%1$c,34,a);}",34,a);}
main(a){ declares the main function with an untyped variable a.printf prints the existing source code and a for loop. An integer is printf'd into the for loop. This is how the multi-quine works - each time the quine program is called, it increments d, making the for loop run more times.printf is inside the for loop and prints itself, so each time the quine program is called, there are more printf's, therefore generating bigger versions of itself. Be warned that this growth is exponential, due to the for loop incrementation and the multiple printfs.printf prints the remaining source code.Requires POSIX.
#H}"'
This, I believe, is the smallest growing quine possible in Gol><>. It duplicates the #H} over and over...
($'#a_$93#"$_@("_$-.
Sends a dot leftward, printing the characters (@_$ with a literal and ' from the ASCII value 39. The dot then turns around and interprets the path as a string, printing the rest of the line.
(@_$'#a_$93#"$_@("_$-.
Prints (@_$' again, as well as a 0, thanks to the added @_ from the first run. The 0 ends up on the right side of the ' symbol, which means that all 0s printed in a previous run carry over. The dot safely ignores the additional 0s on its path.
(@_$'0#a_$93#"$_@("_$-.
(@_$'00#a_$93#"$_@("_$-.
(@_$'000#a_$93#"$_@("_$-.
`④⑩`④⑩
I've gone and done it now! No more ascii around here now folks. This will print:
`④⑩`④⑩④⑩
Which will print:
`④⑩`④⑩`④⑩`④⑩④⑩
Which prints:
`④⑩`④⑩`④⑩`④⑩`④⑩`④⑩④⑩④⑩
And so on.
`::.,`⑵④⑩
This uses the same approach as the 10 byte answer
Explained
`::.,`⑵④⑩
`::.,` #Push the string `::.,`
⑵ #Double the string
④⑩ #Print raw then nice without popping
`::.,`⑵:.,
This prints:
`::.,::.,`::.,::.,
Which prints:
`::.,::.,`::.,::.,`::.,::.,`::.,::.,
`::.,`6*:.,
This prints:
`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,
Which prints:
`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,`::.,::.,::.,::.,::.,::.,
And so on
`::.,`!⑨*:(.,
This prints:
`::.,::.,`::.,::.,
Which prints:
`::.,::.,`::.,::.,`::.,::.,`::.,::.,
And so on.
`::.,(:.,)`:.,
This will create pretty much the same chain as the previous 15-byter.
`::.,(::.,)`:.,
This prints:
`::.,(::.,)`::.,(::.,)
Which prints:
`::.,(::.,)`::.,(::.,)`::.,(::.,)`::.,(::.,)
And so on.
`:::.,(!2/|.,)`:.,
This will print
`:::.,(!2/|.,)`:::.,(!2/|.,)
Which in turn prints:
`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)
Which then prints:
`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)`:::.,(!2/|.,)
And so on.
a=_=>'a=_'+a
(a=_=>'(a=_'+a+')()')()
Every iteration adds an underscore to the parameter name
ToString[10#0][]&[]
Returns 10 (ToString[10 #0][] & )[], which, when run, returns 10*10 (ToString[2 #0][] & )[]. From there, the beginning number is multiplied by 10 at every step, which increases the length of the code by 1.
a
Add prompt at head every time, error is who cares
'rd3*9[r:r]>o<
First iteration duplicates the : in the middle: (it actually duplicates itself):
'rd3*9[r::r]>o<
Then since there are now 2 duplication signs it will duplicate itself twice:
'rd3*9[r::::r]>o<
etc. doubling each time.
':rd3*>o<
This simply duplicates the < at the end:
':rd3*>o<<
':rd3*>o<<<
etc. So the functionality of the program won't be effected at all. It's basically a comment.
95&34_"
Non-competing as the language postdates the challenge. This outputs:
95 95 34
__"
which in turn outputs:
95 95 34
95 95 34
__"
and so on. After ~5 runs it looks like this:
95 95 34 95 95 34 95 95 34 95 95 34
95 95 34 95 95 34 95 95 34 95 95 34
__"__"__"__"
95 95 34 95 95 34 95 95 34 95 95 34
95 95 34 95 95 34 95 95 34 95 95 34
__"__"__"__"
f:{{(2*#x)#x}"f:",x,";`0:f@$f;"};`0:f@$f;
This program prints a copy of itself, and then doubles it. Thus it grows.
`0:{,//$("`0:";o;"[";x;"1];")}[1]
This program prints itself, but with an ever-growing number as the argument. This works in the closed-source interpreter.
" <@o&h:2d+2..
Subsequent versions will have an increasing number of spaces after the ".
" <@o&h:2d+2.." Push the code points of the source code, except for the
quotation marks.
< Send the IP back west.
"..2+d2:h&o@< " Push the code points of the source code, except for the
quotation marks, in reverse.
.. Make two copies of the space. One will be the additional
space, one will be turned into the quotation mark.
2+ Add two, to turn 32 (space) into 34 (quote).
d Push the stack depth. This is about twice the number of characters
we want to print, because there's still the first copy of the source
on the stack. But it's two less than twice that number, because the
additional space and the quote aren't duplicated.
2: Halve the value.
h Increment to account for the space and quote.
& Repeat the next command that many times.
o Print that many characters from the top of the stack.
@ Terminate the program.
i;ARa="i;ARa=%c%s%c,34,a,34,i+2);Oi=%d;i--;)",34,a,34,i+2);Oi=2;i--;)Ra="Ra=%c%s%c,34,a,34",34,a,34
v->{String s="v->{String s=%c%s%1$c+1;return s.format(s,34,s);}//";return s.format(s,34,s);}
Outputs (4 bytes larger):
v->{String s="v->{String s=%c%s%1$c+1;return s.format(s,34,s);}//"+1;return s.format(s,34,s);}//
Which outputs (2 bytes larger):
v->{String s="v->{String s=%c%s%1$c+1;return s.format(s,34,s);}//1"+1;return s.format(s,34,s);}//1
Which outputs (2 bytes larger):
v->{String s="v->{String s=%c%s%1$c+1;return s.format(s,34,s);}//11"+1;return s.format(s,34,s);}//11
etc. (2 bytes larger every new program)
Explanation:
quine-part:
String s contains the unformatted source code.%s is used to input this String into itself with the s.format(...).%c, %1$c and the 34 are used to format the double-quotes.s.format(s,34,s) puts it all togetherChallenge part:
+1 and // to make the output program larger and compilable.+1 adds a 1 to both the unformatted and formatted program.
Python 3, 54 chars
print((lambda s:s%s)('print((lambda s:s%%s)(%r)*2);'))
I simply modified this quine.
You can try it simply doing python this.py | python | python ... | python many times.
Edit: found this problem on the homepage and discovered too late that it is old.
It's not often you get to recycle answers from other questions.
x;*s="x;*s=%c%s%c;main(i){for(i=__LINE__;i--;puts(&x));printf(s,34,s,34);}";main(i){for(i=__LINE__;i--;puts(&x));printf(s,34,s,34);}
g=putStrLn("n=1"++show n++('-'<$[1..n]));main=(\s->putStr s>>print s>>g)$"g=putStrLn(\"n=1\"++show n++('-'<$[1..n]));main=(\\s->putStr s>>print s>>g)$"
n=2
A proper quine, except it grows very oddly.
pp34CS+"pp34CS+
Iterations (You can see that the output gets between and after the source code... ):
pp34CS+pp34CS+"pp34CS+
pp34CS+pp34CS+pp34CS+pp34CS+"pp34CS+pp34CS+
#define a(b)*s=#b;b
a(main(){printf("#define a(b)*s=#b;b\na(%s;)",s);})
Every iteration adds a semicolon after the closing brace of the main function.
f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))
Attempt This Online! Repeats the formatted string 9 times. This is the result of iteration 1:
f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))f="f=%qprint(f:format(f):rep(9))"print(f:format(f):rep(9))
:^)\$\endgroup\$