[plt-scheme] HTDP Section 22.1 some confusion...

It's all substitution of values for parameters and renaming.

1. ... (local ((define f e)) b) ...
    is equal to

    ;; at top level
    (define f-never-mentioned-before e{with f replaced by f-never- 
mentioned-before})
    ;; where local used to be
    ... b{with f replaced by f-never-mentioned-before} ...


2. ((lambda (x) e) 5)
    is equal to
    e{x is equal to 5}

So you have:

  (define add (lambda (x) (local ((define (x-adder y) (+ x y))) x- 
adder)))

  (define f (add 5))
  =  ;; replace name by value
  (define f ((lambda (x) (local ((define (x-adder y) (+ x y))) x- 
adder)) 5))
  =  ;; replace x by 5 (as in middle school)
  (define f (local ((define (x-adder y) (+ 5 y))) x-adder))
  =  ;; lift local as in 1
  (define x-adder-never-mentioned-before (lambda (y) (+ 5 y)))
  (define f x-adder-never-mentioned-before)
  =
  (define x-adder-never-mentioned-before (lambda (y) (+ 5 y)))
  (define f x-adder-never-mentioned-before (lambda (y) (+ 5 y))))





On Apr 29, 2009, at 3:58 AM, Dave wrote:

>>> Hi,
>>> I'm a bit confused by the evaluation example given for the  
>>> application of add to a number.
>>> please forgive my notation, I've started my lines with >>
>
>
> (define (add x)
> (local ((define (x-adder y) (+ x y))) x-adder))
>
>
>>> So we've got add which I think I understand fine, and now we're  
>>> defining f as an application of add to a number.
>
>
> (define f (add 5))
> = (define f (local ((define (x-adder y) (+ 5 y))) x-adder))
>
>
>>> so far so good...
>
>
> = (define f (local ((define (x-adder5 y) (+ 5 y))) x-adder5))
>>>                                            
>>> ^?                           ^?
>>> now what I don't understand is where the 5 in x-adder5 comes from.
>>> There's nothing actually called x-adder5, right? that line could  
>>> equally say
>>> = (define f (local ((define (add5please y) (+ 5 y))) add5please))
>>> It's not like you can call (x-adder5 20) is it? you can only  
>>> invoke x-adder5 with (f x)
>>> x-adder5 is just there to show that the newly defined function  
>>> has the formerly free variable x now fixed to 5
>>> please let me know If I've got this the right way round in my  
>>> mozgy gulliver.
>
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