Well, if you really want to golf it, you can just write:

  File.write ARGV[1], IO.readlines(ARGV[0]).sort.join('\n')

;)

You're fired.

    IO.write(ARGV[1], ARGF.lines.sort.join)

Here's the shortest I've gotten after looking at some of the comments:

  IO.write$*.pop,ARGF.lines.sort.join

And yes, it does work even without the space between `write` and `$*`.

Also after testing, I realized the ('\n') is not required for join. When you call 'lines', it still has the '\n' character in the string, and when you join, it defaults to join without a delimiter, so it's putting them back together with the newline still there.

Wow, this is really sweet and esoteric (didn't know about ARGF) - but, reading the docs for ARGF, wouldn't it also try to ready from ARGV[1]?

Maybe this would fix it:

  IO.write(ARGV.pop, ARGF.lines.sort.join('\n'))

Yes, good catch. join('\n') is wrong though - IO#lines preserves newlines, the default join is the empty string, and you meant "\n" :)

:)

I believe you still need to join on newline, the default is space. However you can save a couple characters by using `$*` instead of ARGV.

Also, trim out the space between the arguments and kill the parenthesis for the optimal golfing.

Is

> File.write ARGV[1], (IO.readlines ARGV[0]).sort.join('\n')

not valid Ruby?

  File.open(ARGV[1], 'w') { |file| file.write(File.read(ARGV[0]).lines.sort.join) }

http://stackoverflow.com/questions/8278680/ruby-undefined-me...