Sketchy LISP: An Introduction to Functional Programming in by Nils M Holm

By Nils M Holm

Sketchy LISP is a step by step creation to practical programming in Scheme. via a number of examples of various complexity, it takes the reader on an exciting and informative travel in the course of the language.

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By Nils M Holm

Sketchy LISP is a step by step creation to practical programming in Scheme. via a number of examples of various complexity, it takes the reader on an exciting and informative travel in the course of the language.

Show description

Read Online or Download Sketchy LISP: An Introduction to Functional Programming in Scheme, 3rd Edition PDF

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Extra resources for Sketchy LISP: An Introduction to Functional Programming in Scheme, 3rd Edition

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D) The result of this application is called an improper list. It is called ‘‘improper’’just because it does not end with (), which every proper list must do. Because the last member of an improper list is not the empty list, the external representation of such a list includes a dot marking its last cons. You can create improper lists not only by appending an atom to a list, but also by just typing them in: ’(a b c . d) => (a b c . d) However, most procedures of Scheme expect proper lists and reduce to bottom when an improper list is passed to them: (reverse ’(a b .

This is why null? can be used to detect the end of a list. ) Above illustration also makes clear why taking the cdr part of a list results in a list. Dotted pair notation also can be used to show why consing a list to a list does not append the lists: (cons ’(a b) ’(c d)) => ((a b) . (c d)) = ((a b) c d) Above definition for the list implies that the last element of each list must be (). Is it possible, though, to append an atom to a list? Let us see: (append ’(a b c) ’d) => (a b c . d) The result of this application is called an improper list.

It is even possible to create procedures accepting zero or more arguments using lambda. A pair obviously cannot be used to achieve this, because a pair with no car part would not be a pair. So how does it work? The signature of a procedure is a list containing the name and the variables of a procedure, as in (define (f2 a b . : ; Definition (define (f x) y) (define (f x . y) y) (define (f . x) x) Signature (f x) (f x . y) (f . x) Arguments (x) (x . y) x So the argument list of a variadic procedure with zero or more arguments is not a list at all, but a single variable.

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