r6rs-lib

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Chapter 14
Enumerations

This chapter describes the `(rnrs enums (6))`library for dealing with enumerated values and sets of enumerated values. Enumerated values are represented by ordinary symbols, while finite sets of enumerated values form a separate type, known as the enumeration sets. The enumeration sets are further partitioned into sets that share the same universe and enumeration type. These universes and enumeration types are created by the `make-enumeration` procedure. Each call to that procedure creates a new enumeration type.

This library interprets each enumeration set with respect to its specific universe of symbols and enumeration type. This facilitates efficient implementation of enumeration sets and enables the complement operation.

In the descriptions of the following procedures, enum-set ranges over the enumeration sets, which are defined as the subsets of the universes that can be defined using `make-enumeration`.

`(make-enumeration symbol-list)` procedure

Symbol-list must be a list of symbols. The `make-enumeration` procedure creates a new enumeration type whose universe consists of those symbols (in canonical order of their first appearance in the list) and returns that universe as an enumeration set whose universe is itself and whose enumeration type is the newly created enumeration type.

`(enum-set-universe enum-set)` procedure

Returns the set of all symbols that comprise the universe of its argument, as an enumeration set.

`(enum-set-indexer enum-set)` procedure

Returns a unary procedure that, given a symbol that is in the universe of enum-set, returns its 0-origin index within the canonical ordering of the symbols in the universe; given a value not in the universe, the unary procedure returns `#f`.

`(let* ((e (make-enumeration ’(red green blue)))
(i (enum-set-indexer e)))
(list (i ’red) (i ’green) (i ’blue) (i ’yellow)))
⇒ (0 1 2 `#f`)

`

The `enum-set-indexer` procedure could be defined as follows using the `memq` procedure from the `(rnrs lists (6))` library:

`(define (enum-set-indexer set)
(let* ((symbols (enum-set->list
(enum-set-universe set)))
(cardinality (length symbols)))
(lambda (x)
(cond
((memq x symbols)
=> (lambda (probe)
(- cardinality (length probe))))
(else `#f`)))))

`

`(enum-set-constructor enum-set)` procedure

Returns a unary procedure that, given a list of symbols that belong to the universe of enum-set, returns a subset of that universe that contains exactly the symbols in the list. The values in the list must all belong to the universe.

`(enum-set->list enum-set)` procedure

Returns a list of the symbols that belong to its argument, in the canonical order of the universe of enum-set.

`(let* ((e (make-enumeration ’(red green blue)))
(c (enum-set-constructor e)))
(enum-set->list (c ’(blue red))))
⇒ (red blue)

`

`(enum-set-member? symbol enum-set)` procedure
`(enum-set-subset? enum-set1 enum-set2)` procedure
`(enum-set=? enum-set1 enum-set2)` procedure

The `enum-set-member?` procedure returns `#t` if its first argument is an element of its second argument, `#f` otherwise.

The `enum-set-subset?` procedure returns `#t` if the universe of enum-set1 is a subset of the universe of enum-set2 (considered as sets of symbols) and every element of enum-set1 is a member of enum-set2. It returns `#f` otherwise.

The `enum-set=?` procedure returns `#t` if enum-set1 is a subset of enum-set2 and vice versa, as determined by the `enum-set-subset?` procedure. This implies that the universes of the two sets are equal as sets of symbols, but does not imply that they are equal as enumeration types. Otherwise, `#f` is returned.

`(let* ((e (make-enumeration ’(red green blue)))
(c (enum-set-constructor e)))
(list
(enum-set-member? ’blue (c ’(red blue)))
(enum-set-member? ’green (c ’(red blue)))
(enum-set-subset? (c ’(red blue)) e)
(enum-set-subset? (c ’(red blue)) (c ’(blue red)))
(enum-set-subset? (c ’(red blue)) (c ’(red)))
(enum-set=? (c ’(red blue)) (c ’(blue red)))))
⇒ (`#t` `#f` `#t` `#t` `#f` `#t`)

`

`(enum-set-union enum-set1 enum-set2)` procedure
`(enum-set-intersection enum-set1 enum-set2)` procedure
`(enum-set-difference enum-set1 enum-set2)` procedure

Enum-set1 and enum-set2 must be enumeration sets that have the same enumeration type.

The `enum-set-union` procedure returns the union of enum-set1 and enum-set2. The `enum-set-intersection` procedure returns the intersection of enum-set1 and enum-set2. The `enum-set-difference` procedure returns the difference of enum-set1 and enum-set2.

`(let* ((e (make-enumeration ’(red green blue)))
(c (enum-set-constructor e)))
(list (enum-set->list
(enum-set-union (c ’(blue)) (c ’(red))))
(enum-set->list
(enum-set-intersection (c ’(red green))
(c ’(red blue))))
(enum-set->list
(enum-set-difference (c ’(red green))
(c ’(red blue))))))
⇒ ((red blue) (red) (green))

`

`(enum-set-complement enum-set)` procedure

Returns enum-set’s complement with respect to its universe.

`(let* ((e (make-enumeration ’(red green blue)))
(c (enum-set-constructor e)))
(enum-set->list
(enum-set-complement (c ’(red)))))
⇒ (green blue)

`

`(enum-set-projection enum-set1 enum-set2)` procedure

Projects enum-set1 into the universe of enum-set2, dropping any elements of enum-set1 that do not belong to the universe of enum-set2. (If enum-set1 is a subset of the universe of its second, no elements are dropped, and the injection is returned.)

`(let ((e1 (make-enumeration
’(red green blue black)))
(e2 (make-enumeration
’(red black white))))
(enum-set->list
(enum-set-projection e1 e2))))
⇒ (red black)

`

`(define-enumeration <type-name>` syntax
`(<symbol> `...`)
<constructor-syntax>)`

The `define-enumeration` form defines an enumeration type and provides two macros for constructing its members and sets of its members.

A `define-enumeration` form is a definition and can appear anywhere any other <definition> can appear.

<Type-name> is an identifier that is bound as a syntactic keyword; <symbol> `...` are the symbols that comprise the universe of the enumeration (in order).

`(<type-name> <symbol>)` checks at macro-expansion time whether the name of <symbol> is in the universe associated with <type-name>. If it is, `(<type-name> <symbol>)` is equivalent to `<symbol>`. It is a syntax violation if it is not.

<Constructor-syntax> is an identifier that is bound to a macro that, given any finite sequence of the symbols in the universe, possibly with duplicates, expands into an expression that evaluates to the enumeration set of those symbols.

`(<constructor-syntax> <symbol> `...`)` checks at macro-expansion time whether every <symbol> `...` is in the universe associated with <type-name>. It is a syntax violation if one or more is not. Otherwise

`(<constructor-syntax> <symbol> `...`)

` is equivalent to `((enum-set-constructor (<constructor-syntax>))
’(<symbol> `...`)).

`

Example:

`(define-enumeration color
(black white purple maroon)
color-set)
(color black) ⇒ black
(color purpel) ⇒ `&syntax` exception
(enum-set->list (color-set)) ⇒ ()
(enum-set->list
(color-set maroon white)) ⇒ (white maroon)

`

> Note: In `(<type-name> <symbol>)` and `(<constructor-syntax> <symbol> `...`)` forms, only the names of the <symbol>s are significant.