This is a collection of counter-intuitive or otherwise obscure trivia about the behavior of Scheme implementations and the interpretation of Scheme standards.
Different lambdas may return the same procedure
(eq? (lambda (x) x) (lambda (x) x)) can be either true of false.
define does not work
This procedure works as expected:
(define (foo) (let ((x 1)) x))
let outside the
define, the code no longer compiles:
(let ((x 1)) (define (foo) x))
The reason is that
foo is now defined in the scope inside
in the scope outside the
let as it was before. A
let body cannot
contain only definitions.
lambda which does not deal with definitions, so
let over a
lambda works fine:
(define foo (lambda () (let ((x 1)) x))) (define foo (let ((x 1)) (lambda () x)))
Common Lisp does not support inner definitions, so hoisting a
over a definition works there:
(defun foo () (let ((x 1)) x)) (let ((x 1)) (defun foo () x))
Applying a procedure to a circular list
The R7RS syntax
1=(+ 1 2 #1) means the same as
(+ 1 2 (+ 1 2 ( + 1 2 …))).
How to write
We show a syntax-rule macro-predicate that can be applied to any expression and can soundly and reliably determine if that expression is a symbol/literal (as opposed to a pair, a vector, a literal string, a literal number, etc). The mere existence of such a predicate is surprising and counter-intuitive. For completeness and reference, we also show two macro-identity predicates on identifiers: id-eq?? and id-eqv??.
It is well-known that syntax-rules are limited in expressive power. For example, we can easily write a low-level macro that can tell the literal type of its argument — whether the argument a symbol/identifier, a literal string, a literal character, or a literal number:
(define-macro (typeof x) (cond ((integer? x) "an integer") ((symbol? x) "a symbol/identifier") ((pair? x) "a pair") ((string? x) "a string"))) (typeof "str") ==> "a string" (typeof xxx) ==> "a symbol/identifier"
We can write a similar discriminator with syntax-case. The syntax-case macro system has "guards" specifically for that purpose:
(define-syntax typeof (lambda (x) (syntax-case x () ((typeof x) (integer? (syntax-object->datum #'x)) "an integer") ((typeof x) (identifier? #'x) "an identifier") ((typeof x) (pair? (syntax-object->datum #'x)) "a pair") ((typeof x) (string? (syntax-object->datum #'x)) "a string")))) (typeof "str") ==> "a string" (typeof 1) ==> "an integer" (typeof typeof) ==> "an identifier"
The latter example runs on Petite Chez Scheme.
Syntax-rules can determine the literal type of some arguments: a pair, a vector, an empty list, a boolean:
(define-syntax typeof (syntax-rules () ((typeof (x . y)) "a pair") ((typeof #(x ...)) "a vector") ((typeof #f) "a boolean") ((typeof #t) "a boolean") ))
We can also write a syntax-rule that tests if its argument is the specific number, the specific character, the specific string. A syntax-rule cannot determine if its argument is a string or an integer.
For a long time I used to think that we cannot write a syntax-rule that tests if its argument is a symbol. In particular, I wanted a macro-expand-time equivalent of the library function symbol?. We will call that macro symbol?? to avoid the confusion with the library function. That macro should take three arguments: any syntactic form plus two continuations, kt and kf. The macro symbol?? will expand to kt if and only if its first argument is a symbol. The macro would expand to kf if its argument is anything other than a symbol. I believed such a macro symbol?? is impossible with syntax-rules — until about a month ago I wrote it.
(define-syntax symbol?? (syntax-rules () ((symbol?? (x . y) kt kf) kf) ; It's a pair, not a symbol ((symbol?? #(x ...) kt kf) kf) ; It's a vector, not a symbol ((symbol?? maybe-symbol kt kf) (let-syntax ((test (syntax-rules () ((test maybe-symbol t f) t) ((test x t f) f)))) (test abracadabra kt kf)))))
The macro is based on the observation that if a form F is an identifier and a syntax-rule pattern P is anything but a literal identifier, then P can match F if and only if P is an identifier (symbol). The final form of this macro incorporates an improvement by Al Petrofsky.
As a bonus, the following two macros, id-eq?? and id-eqv??, test the equivalence of identifiers. Both macros take two identifiers, and two continuations, kt and kf. The macros expand into kt if the two identifiers are equivalent. The macros expand into kf otherwise.
(define-syntax id-eq?? (syntax-rules () ((id-eq?? id b kt kf) (let-syntax ((id (syntax-rules () ((id) kf))) (ok (syntax-rules () ((ok) kt)))) (let-syntax ((test (syntax-rules () ((_ b) (id))))) (test ok)))))) (define-syntax id-eqv?? (syntax-rules () ((id-eqv?? a b kt kf) (let-syntax ((test (syntax-rules (a) ((test a) kt) ((test x) kf)))) (test b)))))
For the macro id-eq??, two identifiers are equivalent if only if they have the same color, or to put it differently, are two occurrences of the same identifier. In other words, the two identifiers must be inter-changeable at macro-expand time. This is the strictest notion of equivalence. For a macro id-eqv??, the identifiers are equivalent if they refer to the same binding (or both identifiers are unbound and have the same spelling). Thus macro id-eqv?? can find two identifiers equivalent even if they have different colors. The last two test cases show the distinction.
(From Oleg. Source: http://okmij.org/ftp/Scheme/macro-symbol-p.txt)
Where does the type-op procedure naming convention come from?
John Cowan did a fair amount of research on where
procedure names were born. It is definitely earlier than R2RS; going
mailing list archive shows that
vector-ref is already in the
earliest draft. At that time, MIT Scheme was already in version 7, so
the evidence may be there.