From: Logan Capaldo <logancapaldo@...>
Date: 2006-01-14T05:09:22+09:00
Subject: Re: [QUIZ.SUMMARY] Dice Roller (#61)


On Jan 13, 2006, at 2:57 PM, J. Ryan Sobol wrote:

> On Jan 12, 2006, at 10:22 AM, Christian Neukirchen wrote:
>
>> Matthew Moss <matthew.moss.coder@gmail.com> writes:
>>
>>> @dice = eval "lambda { #@src }"
>>>
>>> The morphed express is saved away in a lambda, which allows  
>>> Christian to
>>> reevaluate the expression repeatedly without performing those  
>>> substitutions
>>> every time roll is called.
>>
>> I could have evaluated @src each time without substituting anew,  
>> but a
>> lambda only makes it need to *parse* once.
>
> Hey Christian,
>
> I didn't have time to write my own solution to quiz #61, so I'm  
> going over yours, which is similar to what I was kicking around in  
> my head.  Unfortunately, I confused by your eval statement, which  
> is wrapped over a new Proc object.  Instead, why not wrap the  
> lambda around the eval like so?
>
> @dice = lambda { eval @src }
>
> The both _seem_ to produce the same output, yet conceptually, my  
> approach makes more sense to me.  Perhaps I'm missing something  
> crucial in my statement like...
>
> Does eval parse @src for each call() message?  Or does the parsing  
> happen on initialization of the Proc (like I would expect)?  I  
> suppose I could trace the execution using the debugger, but I'm  
> unaware how to step inside a Kernel method like eval or lambda with  
> it.
>
> ~ ryan ~
>

@dice.call has eval parse @src everytime (@src may have changed). By  
doing something like eval "lambda { #{@src} }" you parse @src once  
and get back a lambda.