Good.
On Sep 15, 2014, at 9:20 AM, Daniel Bastos <dbas...@toledo.com> wrote: > On Fri, Sep 12, 2014 at 6:43 PM, > Matthias Felleisen <matth...@ccs.neu.edu> wrote: > > > On Sep 12, 2014, at 4:40 PM, Daniel Bastos <dbas...@toledo.com> wrote: > > > Again, we start with (2) and apply a series of substitutions. > > > > (f f) > > = (<var> <var>) ;; since f is a <var> > > = (<exp> <exp>) ;; since <var> is a subset of <exp> > > > > However, (<exp> <exp>) is not by definition a value. It is an > > expression which may or may not /have/ a value. We know the > > definition of f, however, which tells us that it behaves like > > the identity function. So the evaluation of that (<exp> <exp>) > > yields a function, which is a value. We end with > > > > = <val> > > > > Therefore (2) is a value > > Is there a difference between the statements > > "Daniel is a hat." > > and > > "Daniel has a hat." > > or do they really mean the same thing? > > There's a difference. > > In this spirit, is there a difference between the statement > > "(f f) is a value." > > and > > "(f f) has a value." > > or do they really mean the same thing? > > There's a difference. The analogy is good. (I do remember having read this > analogy in the book, though I can't seem to find it right now.) > > Here's my final rewrite. Feel free to scrutinize it. (BTW, thanks for your > attention.) > > Exercise 20.1.1. Assume the Definitions window in DrScheme contains > (define (f x) x). Identify the values among the following expressions: > > (1) (cons f empty) > (2) (f f) > (3) (cons f (cons 10 (cons (f 10) empty))) > > Explain why they are values and why the remaining expressions are not > values. > > Solution. A value is anything of the form > > <val> = <boo> | <sym> | <num> | empty | <lst> > | <var> (names of defined functions) > | <prm> > > where <lst> is of the form > > <lst> = empty | (cons <val> <lst>). > > Now we start with (1) and apply a series of substitutions based on the > definitions we have. > > (cons f empty) > = (cons <var> <lst>) ;; since f is a <var> > = (cons <var> <lst>) ;; since empty is a <lst> > = (cons <val> <lst>) ;; since <var> is a subset of <val> > = <lst> ;; definition of <lst> > = <val> ;; definition of <val> > > Therefore (1) is a value. > > Next we consider (2). Here's a relevant part of the grammar for (2). > > <exp> = <var> > | <prm> > | (<exp> <exp> ...<exp>) > > Start with (2)... > > (f f) > = (<var> <var>) ;; since f is a <var> > = (<exp> <exp>) ;; since <var> is a subset of <exp> > > Now (<exp> <exp>) is not by definition a value. Expressions may have > values, but they're not values. (Daniel may have a hat, but Daniel is > not a hat.) > > Therefore (2) is not a value. > > Now (3). > > (cons f (cons 10 (cons (f 10) empty))) > = (cons <var> (cons <num> (cons (<var> <num>) <lst>))) > > The whole expression depends on the application (<exp> <exp>), > which may or may not have a value. Since we know f, we know it has a value, > but it /is not/ a value. > > Therefore (3) is not a value. ____________________ Racket Users list: http://lists.racket-lang.org/users
