On Feb 11, 2014, at 10:06 AM, Matthew Johnson <mcoog...@gmail.com> wrote:
> So in terms of my example: > > We have the general form: ( (lambda (x) e) V) ~ e with all [free] x > replaced by V > > And for V <-- 2 :: ( (lambda (x) (* x y) 2) ~ (* 2 y) ? > > But for V <-- z^2 :: error and NOT ~ (* (* z z) y) ? Not error. It just doesn't reduce. But usually this doesn't happen, z will have been replaced with a value. > > Thanks > > On 11/02/2014, at 3:59 PM, Matthias Felleisen <matth...@ccs.neu.edu> wrote: > >> >> >> Ouch, I forgot to explain V. NO, you canNOT replace x with arbitrary >> expressions. You may use ONLY VALUES (V for Value). In this context a value >> is one of: #t/#f, number, or a lambda expression. -- Matthias >> >> >> >> >> >> On Feb 11, 2014, at 9:50 AM, Matthew Johnson <mcoog...@gmail.com> wrote: >> >>> The general form: ( (lambda (x) e) V) ~ e with all [free] x replaced by V >>> >>> In concrete terms means: ( (lambda (x) (* x y) 2) ~ (* 2 y) ? >>> >>> I figure best to be sure now. >>> >>> Thanks and best regards >>> >>> matt johnson >>> >>> >>> On 11/02/2014, at 3:35 PM, Matthias Felleisen <matth...@ccs.neu.edu> wrote: >>> >>>> >>>> On Feb 11, 2014, at 12:48 AM, Matthew Johnson <mcoog...@gmail.com> wrote: >>>> >>>>> I am still a little unclear. Is this a logical result, or a result of >>>>> the evaluator's rules? Or are they the same thing? >>>> >>>> >>>> They are the same thing. To calculate with the Racket that you see in the >>>> book/derivation, use this rule: >>>> >>>> ( (lambda (x) e) V ) ~ e with all [free] occurrences of x replaced by V >>>> >>>> [This is the rule you learn in pre-algebra, with the 'free' needed here >>>> because >>>> variables can occur at more places than in pre-algebra.] Using this rule, >>>> plus >>>> simple car/cdr/cons/+1 rules, you can confirm every step in the >>>> derivation. As it >>>> turns out, the compiler implements this rule totally faithfully BUT >>>> instead of >>>> copying actual code it "copies" references to code by moving things from >>>> registers >>>> to registers. Because the two are equivalent, a programmer can use the >>>> above >>>> rule to think about code and the compiler writer (one among 10,000s of >>>> language >>>> users) is the only one who has to keep references, registers, and >>>> replication >>>> in mind (kind of). >>>> >>>> -- Matthias >> ____________________ Racket Users list: http://lists.racket-lang.org/users
