[ruby-talk:00123] Re: finished simple script

In message "[ruby-talk:00122] Re: finished simple script"
    on 98/12/18, Julian Fondren <[e-mail]> writes:

>> Humm... I can't find such situation for the above script :-(
>> Would you tell me what input shows that ``.'' is translated to 32, 
>> or explain me the reason why you think so? 
>
>I don't know. Yesterday 1 * . gave 32.. but then yesterday this script
>also allowed me to press enter 2000 times, 1, enter 2000 times, 2 and
>get the answer of 2. Strange.

To consider the problem, I tried as follows. 
I put ``p [ans1, ans2]'' at the end of your script, say `get2nums.rb', 
then the script was as follows: 
(`%' for the shell prompt)

  % cat get2nums.rb
  #!/usr/bin/env/ruby
  print "Input two numbers to multiply together: \n"
  ans1, ans2 = STDIN.gets.chomp.scan(/\d+/).collect { |x| x.to_i }

  while !ans1
	  if !ans1
		  ans1, ans2 = STDIN.gets.to_i, STDIN.gets.to_i
	  end
  end
  while !ans2
	  if !ans2
		  ans2 = STDIN.gets.to_i
	  end
  end
  p [ans1,ans2]
  %

With the above, I got: 

  % ruby -v
  ruby 1.1c9(98/11/26) [i386-freebsd2.1.0]
  % echo "1 * ." | ruby get2nums.rb
  Input two numbers to multiply together: 
  [1, 0]
  % ruby -e 'print ("\n"*2000,1,"\n"*2000,2)' | ruby get2nums.rb
  Input two numbers to multiply together: 
  [0, 0]
  %

... Maybe, I can't get as same as what you get :-(
Would you try the aboves and show me the results?

However, I see because of what I got. The pieces of the fact are

  "1 * .\n".chomp           #=> "1 * ."
  "1 * .".scan(/\d+/)       #=> ["1"]
  ["1"].collect{|x| x.to_i} #=> [1]
  ans1, ans2 = [1]          #=> ans1 == 1, ans2 == nil

  nil.to_i                  #=> 0
  "\n".to_i                 #=> 0

  etc...


By the way, `p' is a built-in function which is mimicked by

  def p(*args)
    STDOUT.print *args.collect{|elm| elm.inspect + "\n"}
    STDOUT.flush
    nil
  end

This function is often useful for quick debugging. 
I hope that you know `p' already. 

-- gotoken