From: Brian Candler <b.candler@...>
Date: 2012-02-23T05:08:54+09:00
Subject: Re: [Basic question] why a number does not match regular expression which represents a number?

Bdimych B. wrote in post #1048291:
> Hi
>
> C:\Users\bdimych>
> C:\Users\bdimych>
> C:\Users\bdimych>irb
> irb(main):001:0> RUBY_VERSION
> => "1.9.2"
> irb(main):002:0> x = 111
> => 111
> irb(main):003:0> x =~ /\d/
> => nil
> irb(main):004:0>
>
> It was unexpected for me, I thought since Ruby is dynamically typed it
> should handle such thing implicitly. Is this described in
> documentation?

Being dynamically typed is not the same as automatically converting one 
type to another. In Ruby, "0" and 0 are two completely different 
objects.

However the case in point here also highlights something else: operators 
in ruby are just syntactic sugar for method calls. a =~ b is really 
a.=~(b)

Now note:

>> /\d/ =~ 111
TypeError: can't convert Fixnum into String
        from (irb):2
>> 111 =~ /\d/
=> false

The first is calling Regexp#=~ which requires a string as its argument, 
which is sensible. So really the question is, why does Fixnum#=~ exist 
at all?

>> Object.new =~ /\d/
=> false

So it appears it has inherited this from Object#=~. I'm not really sure 
why Object has such a method, and it seems to be more a source of 
confusion than anything, but it *is* documented:

http://www.ruby-doc.org/core-1.9.3/Object.html#method-i-3D-7E

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