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# (r5rs.info.gz) Equivalence predicates Info Catalog (r5rs.info.gz) Standard procedures (r5rs.info.gz) Standard procedures (r5rs.info.gz) Numbers
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6.1 Equivalence predicates
==========================

A "predicate" is a procedure that always returns a boolean value (#t or
#f).  An "equivalence predicate" is the computational analogue of a
mathematical equivalence relation (it is symmetric, reflexive, and
transitive).  Of the equivalence predicates described in this section,
`eq?' is the finest or most discriminating, and `equal?' is the
coarsest.  `Eqv?' is slightly less discriminating than `eq?'.

-- procedure: eqv? obj1 obj2
The `eqv?' procedure defines a useful equivalence relation on
objects.  Briefly, it returns #t if OBJ1 and OBJ2 should normally
be regarded as the same object.  This relation is left slightly
open to interpretation, but the following partial specification of
`eqv?' holds for all implementations of Scheme.

The `eqv?' procedure returns #t if:

* OBJ1 and OBJ2 are both #t or both #f.

* OBJ1 and OBJ2 are both symbols and

(string=? (symbol->string obj1)
(symbol->string obj2))
==>  #t

_Note:_ This assumes that neither OBJ1 nor OBJ2 is an
"uninterned symbol" as alluded to in section
Symbols.  This report does not presume to specify the
behavior of `eqv?' on implementation-dependent
extensions.

* OBJ1 and OBJ2 are both numbers, are numerically equal (see
`=', section  Numbers), and are either both exact or
both inexact.

* OBJ1 and OBJ2 are both characters and are the same character
according to the `char=?' procedure (section
Characters).

* both OBJ1 and OBJ2 are the empty list.

* OBJ1 and OBJ2 are pairs, vectors, or strings that denote the
same locations in the store (section  Storage model).

* OBJ1 and OBJ2 are procedures whose location tags are equal
(section  Procedures).

The `eqv?' procedure returns #f if:

* OBJ1 and OBJ2 are of different types (section
Disjointness of types).

* one of OBJ1 and OBJ2 is #t but the other is #f.

* OBJ1 and OBJ2 are symbols but

(string=? (symbol->string OBJ1)
(symbol->string OBJ2))
==>  #f

* one of OBJ1 and OBJ2 is an exact number but the other is an
inexact number.

* OBJ1 and OBJ2 are numbers for which the `=' procedure returns
#f.

* OBJ1 and OBJ2 are characters for which the `char=?' procedure
returns #f.

* one of OBJ1 and OBJ2 is the empty list but the other is not.

* OBJ1 and OBJ2 are pairs, vectors, or strings that denote
distinct locations.

* OBJ1 and OBJ2 are procedures that would behave differently
(return different value(s) or have different side effects)
for some arguments.

(eqv? 'a 'a)                           ==>  #t
(eqv? 'a 'b)                           ==>  #f
(eqv? 2 2)                             ==>  #t
(eqv? '() '())                         ==>  #t
(eqv? 100000000 100000000)             ==>  #t
(eqv? (cons 1 2) (cons 1 2))           ==>  #f
(eqv? (lambda () 1)
(lambda () 2))                   ==>  #f
(eqv? #f 'nil)                         ==>  #f
(let ((p (lambda (x) x)))
(eqv? p p))                          ==>  #t

The following examples illustrate cases in which the above rules do
not fully specify the behavior of `eqv?'.  All that can be said
about such cases is that the value returned by `eqv?' must be a
boolean.

(eqv? "" "")                           ==>  _unspecified_
(eqv? '#() '#())                       ==>  _unspecified_
(eqv? (lambda (x) x)
(lambda (x) x))                  ==>  _unspecified_
(eqv? (lambda (x) x)
(lambda (y) y))                  ==>  _unspecified_

The next set of examples shows the use of `eqv?' with procedures
that have local state.  `Gen-counter' must return a distinct
procedure every time, since each procedure has its own internal
counter.  `Gen-loser', however, returns equivalent procedures each
time, since the local state does not affect the value or side
effects of the procedures.

(define gen-counter
(lambda ()
(let ((n 0))
(lambda () (set! n (+ n 1)) n))))
(let ((g (gen-counter)))
(eqv? g g))                          ==>  #t
(eqv? (gen-counter) (gen-counter))
==>  #f
(define gen-loser
(lambda ()
(let ((n 0))
(lambda () (set! n (+ n 1)) 27))))
(let ((g (gen-loser)))
(eqv? g g))                          ==>  #t
(eqv? (gen-loser) (gen-loser))
==>  _unspecified_

(letrec ((f (lambda () (if (eqv? f g) 'both 'f)))
(g (lambda () (if (eqv? f g) 'both 'g))))
(eqv? f g))
==>  _unspecified_

(letrec ((f (lambda () (if (eqv? f g) 'f 'both)))
(g (lambda () (if (eqv? f g) 'g 'both))))
(eqv? f g))
==>  #f

Since it is an error to modify constant objects (those returned by
literal expressions), implementations are permitted, though not
required, to share structure between constants where appropriate.
Thus the value of `eqv?' on constants is sometimes
implementation-dependent.

(eqv? '(a) '(a))                       ==>  _unspecified_
(eqv? "a" "a")                         ==>  _unspecified_
(eqv? '(b) (cdr '(a b)))               ==>  _unspecified_
(let ((x '(a)))
(eqv? x x))                          ==>  #t

_Rationale:_ The above definition of `eqv?' allows
implementations latitude in their treatment of procedures and
literals:  implementations are free either to detect or to
fail to detect that two procedures or two literals are
equivalent to each other, and can decide whether or not to
merge representations of equivalent objects by using the same
pointer or bit pattern to represent both.

-- procedure: eq? obj1 obj2
`Eq?' is similar to `eqv?' except that in some cases it is capable
of discerning distinctions finer than those detectable by `eqv?'.

`Eq?' and `eqv?' are guaranteed to have the same behavior on
symbols, booleans, the empty list, pairs, procedures, and non-empty
strings and vectors.  `Eq?''s behavior on numbers and characters is
implementation-dependent, but it will always return either true or
false, and will return true only when `eqv?' would also return
true.  `Eq?' may also behave differently from `eqv?' on empty
vectors and empty strings.

(eq? 'a 'a)                            ==>  #t
(eq? '(a) '(a))                        ==>  _unspecified_
(eq? (list 'a) (list 'a))              ==>  #f
(eq? "a" "a")                          ==>  _unspecified_
(eq? "" "")                            ==>  _unspecified_
(eq? '() '())                          ==>  #t
(eq? 2 2)                              ==>  _unspecified_
(eq? #\A #\A)                          ==>  _unspecified_
(eq? car car)                          ==>  #t
(let ((n (+ 2 3)))
(eq? n n))                           ==>  _unspecified_
(let ((x '(a)))
(eq? x x))                           ==>  #t
(let ((x '#()))
(eq? x x))                           ==>  #t
(let ((p (lambda (x) x)))
(eq? p p))                           ==>  #t

_Rationale:_ It will usually be possible to implement `eq?'
much more efficiently than `eqv?', for example, as a simple
pointer comparison instead of as some more complicated
operation.  One reason is that it may not be possible to
compute `eqv?' of two numbers in constant time, whereas `eq?'
implemented as pointer comparison will always finish in
constant time.  `Eq?' may be used like `eqv?' in applications
using procedures to implement objects with state since it
obeys the same constraints as `eqv?'.

-- library procedure: equal? obj1 obj2
`Equal?' recursively compares the contents of pairs, vectors, and
strings, applying `eqv?' on other objects such as numbers and
symbols.  A rule of thumb is that objects are generally `equal?'
if they print the same.  `Equal?' may fail to terminate if its
arguments are circular data structures.

(equal? 'a 'a)                         ==>  #t
(equal? '(a) '(a))                     ==>  #t
(equal? '(a (b) c)
'(a (b) c))                    ==>  #t
(equal? "abc" "abc")                   ==>  #t
(equal? 2 2)                           ==>  #t
(equal? (make-vector 5 'a)
(make-vector 5 'a))            ==>  #t
(equal? (lambda (x) x)
(lambda (y) y))                ==>  _unspecified_

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