From: Eric Christopherson <echristopherson@...>
Date: 2010-02-21T13:09:55+09:00
Subject: Re: Ruby conditionals subtlety?

On Thu, Feb 18, 2010 at 5:18 PM, Ryan Davis <ryand-ruby@zenspider.com> wrote:
> Your latter code snippet treats "fooo" in defined? as a method call. This is because the assignment inside the conditional hasn't been parsed yet, and hasn't affected the lookup tables.
>
> The former doesn't have this problem because the body of the conditional is parsed first.

And:

On Sat, Feb 20, 2010 at 5:28 AM, Brian Candler <b.candler@pobox.com> wrote:
> Maybe it's clearer like this:
>
>  if false
>    foo = 123
>  end
>  puts foo      # nil
>
>  puts bar      # undefined local variable or method 'bar'
>
> That is, for a bare word expression like 'foo' ruby has to decide
> whether to parse it as a method call - as foo() or self.foo - or as a
> local variable reference.
>
> It makes the decision based on whether there has been a previous
> assignment of the form "foo = ..." parsed earlier in the code. This is
> regardless of whether the code is actually executed, because we haven't
> started executing any of it yet.

I think I understand the gist of this -- if the parser has seen foo
already, it is considered "defined" -- but how does the distinction of
method vs. local variable matter in this context? As far as I can
tell, defined? works the same on local variable names and method
names.