From: Josh Cheek <josh.cheek@...>
Date: 2010-06-19T03:43:35+09:00
Subject: Re: String comparison. Why does Ruby consider this true?

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On Fri, Jun 18, 2010 at 12:46 PM, Abder-rahman Ali <
abder.rahman.ali@gmail.com> wrote:

> When I try for example to compare the following strings in Ruby, I get
> "true".
>
> puts 'Xeo' < 'ball'
>
> When I make 'Xeo' start with a lowercase letter, i get 'false'
>
> puts 'xeo' < 'ball'
>
> The second statement is clear, but why when I capitalize 'Xeo' I get
> true?
>
> Thanks.
> --
> Posted via http://www.ruby-forum.com/.
>
>
Well, this used to be easy to show, but apparently since ascii has been
abandoned, and I don't know unicode, I have to resort to hacky things like
this to explain it.


$chars = (1..128).inject(Hash.new) { |chars,num| chars[num.chr] = num ;
chars }

def to_number_array(str)
  str.split(//).map { |char| $chars[char] }
end

to_number_array 'Xeo'   # => [88, 101, 111]
to_number_array 'xeo'   # => [120, 101, 111]
to_number_array 'ball'  # => [98, 97, 108, 108]
to_number_array 'ABC'   # => [65, 66, 67]
to_number_array 'abc'   # => [97, 98, 99]



In this case, $chars is a hash that will take a 1 character string, and
return its ascii value. So the method receives a String, and returns an
array where each index is the ascii value of the character.

Then to understand why one would be less than or greater than the other, go
through index by index, comparing the number in that index. If the two
strings (or in this case, their array representations that I made) have
different numbers, then whichever has the smaller number is considered less
than the other. If you run out of indexes on one of them, then that one
comes before the other. If you run out of indexes on them both
simultaneously, then they are equal.

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