From: Brian Candler <b.candler@...>
Date: 2010-02-22T06:42:43+09:00
Subject: Re: Ruby conditionals subtlety?

Eric Christopherson wrote:
> 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.

Almost - perhaps I muddied things a bit. At parse time, in the above 
code, foo is not 'undefined'; it is known that it must be a method name, 
because it's not a local variable. But it's not known whether there will 
be a method called foo at the time this code is executed.

So, the result of defined?(foo) is decided at runtime. But if foo is a 
local variable at that point in the source code (which is decided at 
parse time), then you always know that the result of defined?(foo) will 
be "local-variable", since the parse tree contains "NODE_LVAR foo"; the 
parser had already chosen foo to be a local variable, and this is a fact 
which cannot be altered subsequently.

I don't think the MRI interpreter actually optimises away defined?(LVAR) 
to a constant at parse time as I might have implied, but in theory it 
could.

For methods, the result is not known until runtime. Example:

>> 2.times { puts defined?(foo); def foo; end }
nil
method
=> 2
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