[plt-scheme] Undefined identifier? hey... it's defined!

On Wed, 2009-06-24 at 17:09 -0400, Ryan Culpepper wrote:
> Paulo J. Matos wrote:
> > On Wed, 2009-06-24 at 14:13 -0400, Matthias Felleisen wrote:
> >> The rhs of definitions are evaluated in sequence. In foo2, you are  
> >> evaluating (bar) in a context where foo1 is bound and evaluated and  
> >> bar isn't. Nothing new.
> >>
> >> Then again, in a module this kind of thing could go away.
> >>
> >>
> > 
> > You're right.. it does make sense... wierd I only stumbled at it now. 
> > I always thought that rhs of definitions were only evaluated in a
> > sequence outside of a module and inside it wouldn't make sense.
> 
> What would you like this to produce?
> 
>    #lang scheme
>    (define a (cons 1 b))
>    (define b (cons 2 a))
>    (write a)
> 
> (Perhaps the real lesson here is that procedure definitions are in no 
> way special. They're just like other value definitions.)
> 

You made your point... I retreat! :)

> Ryan
> 
> 
> 
> >> On Jun 24, 2009, at 2:02 PM, Paulo J. Matos wrote:
> >>
> >>> Hi all,
> >>>
> >>> When you think scheme hold no more strange surprises, you end up  
> >>> finding
> >>> something you can't explain. It's even worse when you understand the
> >>> error message but you have no idea why it's being thrown at you.
> >>>
> >>> So this is just small example showing the problem:
> >>> #lang scheme
> >>>
> >>> (define (foo1 x)
> >>>   (let ([y (bar x)])
> >>>     (+ y x)))
> >>>
> >>> (define foo2
> >>>   (let ([y (bar)])
> >>>     (lambda (x)
> >>>       (+ y x))))
> >>>
> >>> (define (bar (x 2))
> >>>   (* x x))
> >>>
> >>> I get in 4.2:
> >>> reference to an identifier before its definition: bar
> >>>
> >>> Why is bar undefined in foo2?
> >>>
> >>> Cheers,
> >>>
> >>> Paulo Matos
> >>>
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> > 
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>