From: Eric Christopherson <echristopherson@...>
Date: 2010-02-22T05:29:34+09:00
Subject: Re: Ruby conditionals subtlety?

On Sun, Feb 21, 2010 at 8:33 AM, Brian Candler <b.candler@pobox.com> wrote:
> Eric Christopherson wrote:
>> 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.
>
> irb(main):001:0> bar = 123
> => 123
> irb(main):002:0> defined?(bar)
> => "local-variable"
> irb(main):003:0> defined?(puts)
> => "method"
> irb(main):004:0> defined?(baz)
> => nil
>
> furthermore:
>
> irb(main):005:0> puts = "Hello"
> => "Hello"
> irb(main):006:0> defined?(puts)
> => "local-variable"
>
> So, defined?(xxx) will return non-nil if xxx is a local variable (a
> decision made at parse time), and if not then if the current object has
> a method called :xxx (which is only known at run time). Hence
>
> foo = 123
> bar if defined?(foo)
>
> is the same as
>
> foo = 123
> bar if "local-variable"
>
> because foo is known at parse-time to be a local variable.
>
> Regards,

Ah, thanks -- I wasn't grokking the parse-time/runtime distinction in
this case. So:

def foo; 456; end unless defined? foo
puts foo

will print 456, because at parse time foo is undefined, even though it
gets made into a method name at runtime.