My first JAPH (a project to learn more perl)

As a project to teach myself more syntax, I decided to do a simple JAPH. So now I present, my first JAPH (please comment, and further learn me on the arts of JAPH:

perl()  ; sub perl {$h = "hacker" ; @j =
qw(Perl Just Rocks); @j=(@j, "another");
$j[2] = $h  ; @j = ($j[1], $j[3], $j[0],
$j[2]  )  ; print "\n@j[0 .. 3]\;\n" ; }
[download]

PS: It doesn't seem to return any errors for me...even with the -w flag.

Comment on My first JAPH (a project to learn more perl) Download Code

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RE: My first JAPH (a project to learn more perl) by chromatic (Archbishop) on Sep 28, 2000 at 08:05 UTC
I suppose trying to reduce this too far will end up with a single print statement. However, here's a more concise version: `perl(); sub perl { $h = "hacker"; @j = qw(Perl Just Rocks); $j[$#j+1]=("another"); print "\n@j[1, 3, 0] $h;\n"; }` [download] Well... concise with a little more punctuation.	[reply] [d/l]
RE: RE: My first JAPH (a project to learn more perl) by Coplan (Pilgrim) on Sep 28, 2000 at 16:34 UTC
Wouldn't the pound sign (#) in the second line make the rest of that line a remark?	[reply]
(ar0n: $# returns last el. in array) RE (3): My first JAPH (...) by ar0n (Priest) on Sep 28, 2000 at 16:42 UTC
No. `$#` returns the subscript of the last element in an array. So if you'd have an array called `@array` with six elements, `$#array` would return 5: `qw(one two three four five six); print $array[$#array]; # prints "six"` [download] `$j[$#j+1] = ("another");` [download] is actually just obfuscation for this: `push @j, "another";` [download] [~]	[reply] [d/l] [select]