From: Gary Wright <gwtmp01@...>
Date: 2008-01-25T02:11:10+09:00
Subject: Re: why must I initialize this variable?


On Jan 24, 2008, at 4:38 AM, 7stud -- wrote:
>
> So the parser marks hi as a method before the code hi="hello" is  
> parsed,
> and thereafter hi is marked as a variable?  Then execution evaluates
> hi="hello" and thereafter execution moves *backwards* to execute puts
> hi?

Execution isn't moving backwards.  You have to think of it as a two  
step process, parsing then execution.  The parser is going to prepare:

	puts "#{hi}" if (hi = "hello")
         puts hi

is such a way that the execution step sees:

        if (hi = "hello")
           puts "#{hi}"
        end
        puts hi

So during execution, nothing is going backwards.  But the parser has  
already tagged various parts of the code such that the 'hi' token in  
the conditional is an local variable assignment from a literal (thus  
the warning), the 'hi' token in the then clause as a method call, and  
the 'hi' token on the final puts as a local variable.

These three issues
1) 0-argument method calls and local variables
    are syntactically identical
2) local variables don't have to be declared.
3) Ruby's if/unless modifiers imply execution
    order that is backwards from textual order

when considered separately are quite nice, but when combined, have  
some awkward corner cases.  You could get rid of the problem by:

a) forcing all 0-arg method calls to be self.method or method()
b) requiring temporary variables to be declared
c) eliminating if/unless modifiers
d) implementing more elaborate multi-pass parser