From: Ryan Davis <ryand@...>
Date: 2010-02-19T10:24:52+09:00
Subject: Re: Ruby conditionals subtlety?


On Feb 18, 2010, at 16:04 , Farhad Farzaneh wrote:

> Ryan Davis wrote:
>> On Feb 18, 2010, at 13:44 , Farhad Farzaneh wrote:
>> 
>>>>> irb(main):002:0> unless defined?(fooo) ; fooo = true ; end
>>> 
>>> Cool, any chance you could give a short description for those of us that 
>>> have never really thought about the parser or used parse_tree_show?
>> 
>> 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.
> 
> Thanks. To make sure I understand:  In the first case,
> 
>> foo = true unless defined?(foo)
> 
> the parser encounters 'foo', adds it to the symbol table, and then sees 
> the defined? call, so the net result is that foo is defined, but has nil 
> value.  As such, if we're ever going to use defined? as a conditional, 
> it should precede any other mention of the token (foo).

No, it isn't necessarily defined. AT PARSE TIME it sees the initial foo and decides it is a local variable so it treats the foo in defined?(foo) as a local variable. AT RUN TIME it evaluates the defined?(foo) and acts accordingly.