4 ms·

You're fired. IO.write(ARGV[1], ARGF.lines.sort.join)

• Aaronneyer 13y ago
  I believe you still need to join on newline, the default is space. However you can save a couple characters by using `$*` instead of ARGV. Also, trim out the space between the arguments and kill the parenthesis for the optimal golfing.
• danmaz74 13y ago
  Wow, this is really sweet and esoteric (didn't know about ARGF) - but, reading the docs for ARGF, wouldn't it also try to ready from ARGV[1]? Maybe this would fix it: IO.write(ARGV.pop, ARGF.lines.sort.join('\n'))
  • Freaky 13y ago
    Yes, good catch. join('\n') is wrong though - IO#lines preserves newlines, the default join is the empty string, and you meant "\n" :)
    • danmaz74 13y ago
      :)
• Aaronneyer 13y ago
  Here's the shortest I've gotten after looking at some of the comments: IO.write$*.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.