I am trying to define a function func->symbol that takes a function and returns its name as a symbol. For example:
(define (pythagoras a b)
(sqrt (+ (* a a) (* b b))))
;; #1
(func->symbol pythagoras) ; Returns: 'pythagoras
;; #2
(func->symbol (if #t pythagoras sqrt)) ; Returns: 'pythagoras
;; #3
(let ((f (if #t pythagoras sqrt)))
(func->symbol f)) ; Returns: 'pythagoras
;; #4
(let ((f (if #t pythagoras sqrt)))
(let ((g f))
(func->symbol g))) ; Returns: 'pythagoras
This is a follow-up question on How do I get a definition's name as a symbol? which only deals with case #1. For case #1, a simple macro def->symbol is sufficient:
(define-syntax def->symbol
(syntax-rules ()
((_ def) 'def)))
However, this macro definition does not pass cases #2, #3, #4. Is it possible to define func->symbol, or is Scheme not expressive enough for this?
In Racket, in many cases, you can get a function's name using object-name. But it is probably a bad idea to rely on this result for anything other than debugging.
Perhaps it's worth an answer which shows why this is not possible in any language with first-class functions.
I'll define what I mean by a language having first-class functions (there are varying definitions).
Functions can be passed as arguments to other functions, and returned as values from them.
Functions can be stored in variables and other data structures.
There are anonymous functions, or function literals.
Scheme clearly has first-class functions in this sense. Now consider this code:
(define a #f)
(define b #f)
(let ((f (lambda (x)
(+ x 1))))
(set! a f)
(set! b f))
Let's imagine there is a function-name function, which, given a function, returns its name. What should (function-name a) return?
Well, the answer is that there's simply no useful value it can return (in Racket, (object-name a) returns f, but that's clearly exposing implementation details which might be useful for debugging but would be very misleading as a return value for a function-name procedure.
This is why such a procedure can't exist in general in a language with first-class functions: the function which maps from names to values is many-to-one and thus has no inverse.
Here is an example of the sort of disgusting hack you could do to make this 'work' and also why it's horrible. The following is Racket-specific code:
(define-syntax define/naming
;; Define something in such a way that, if it's a procedure,
;; it gets the right name. This is a horrid hack.
(syntax-rules ()
[(_ (p arg ...) form ...)
(define (p arg ...) form ...)]
[(_ name val)
(define name (let ([p val])
(if (procedure? p)
(procedure-rename p 'name)
p)))]))
And now, given
(define/naming a
(let ([c 0])
(thunk
(begin0
c
(set! c (+ c 1))))))
(define/naming b a)
Then:
> (object-name a)
'a
> (object-name b)
'b
> (eqv? a b)
#f
> (a)
0
> (b)
1
> (a)
2
So a and b have the 'right' names, but because of that they are necessarily not the same object, which I think is semantically wrong: if I see (define a b) then I want (eqv? a b) to be true, I think. But a and b do capture the same lexical state, so that works, at least.
Related
This Common Lisp macro and test function
(defmacro test (body)
`(let ,(mapcar #'(lambda (s)
`(,s ,(char-code (char-downcase (char (symbol-name s) 0)))))
'(a b))
,body))
(test (+ a b))
expands into
(let ((a 97) (b 98))
(+ a b))
and gives 195 when evaluated
Trying to do that in Racket
(define-syntax (test stx)
(syntax-case stx ()
[(_ body)
#`(let #,(map (lambda (x)
(list x
(char->integer (car (string->list (symbol->string x))))))
'(a b))
body)]))
(test (+ a b))
When I run the macroexpander, the macro form expands to:
(let ((a 97) (b 98)) (+ a b))))
which is what I thought I wanted.
But it fails with:
a: unbound identifier in context..
Disabling macro hiding gives a form that ends with:
(#%app:35
call-with-values:35
(lambda:35 ()
(let-values:36 (((a:37) (quote 97)) ((b:37) (quote 98)))
(#%app:38 + (#%top . a) b)))
(print-values:35)))
I don't understand why my nice expansion (let ((a 97) (b 98)) (+ a b)) doesn't work, and I'm puzzled by (#%top .a)... I wonder if it's trying to find a function called "a"?
When I copy the expanded form into the REPL, it works...
I'm grateful for any help!
Racket has hygienic macro. Consider:
(define-syntax-rule (or a b)
(let ([a-val a])
(if a-val a-val b)))
Then:
(let ([a-val 1])
(or #f a-val))
will roughly expand to:
(let ([a-val 1])
(let ([a-val2 #f])
(if a-val2 a-val2 a-val)))
which evaluates to 1. If macro is not hygienic, then it would result in #f, which is considered incorrect.
Notice that a-val is renamed to a-val2 automatically to avoid the collision. That's what happens to your case too.
One way to fix the problem in your case is to give a correct context to the generated identifiers so that the macroexpander understands that they should refer to the same variable.
(define-syntax (test stx)
(syntax-case stx ()
[(_ body)
#`(let #,(map (lambda (x)
(list (datum->syntax stx x) ; <-- change here
(char->integer (car (string->list (symbol->string x))))))
'(a b))
body)]))
(test (+ a b))
As a counterpart Sorawee Porncharoenwase's answer (which is the right answer) I think it's worth while thinking a bit about why your test macro is problematic in CL and why macros which do similar things are outright buggy.
Given your test macro, imagine some user looking at this code:
(let ((a 1) (b 2))
(test (+ a b)))
Well, I don't know about you, but what I would expect to happen is that the a and b inside test are the a and b I've just bound. But that's not the case at all, of course.
Well, perhaps the documentation for test describes in great detail that it binds two variables, and that this is what I should expect. And, of course, there are macros which do just that, and where it's fine:
(defmacro awhen (test &body forms)
`(let ((it ,test))
(when ,it ,#forms)))
And now:
(awhen (find-exploder thing)
(explode it))
And this is all fine, because the documentation for awhen will say that it binds it to the result of the test in its body.
But now consider this or macro stolen from the other answer:
(defmacro vel (a b)
`(let ((a-val ,a))
(if a-val a-val ,b)))
This is a disaster. It 'works', except it doesn't work at all:
> (let ((a-val 3))
(vel nil a-val))
nil
Now that's not just surprising in the way your test macro is: it's wrong.
Instead, you have to write vel like this in CL:
(defmacro vel (a b)
(let ((a-val-name (make-symbol "A-VAL")))
`(let ((,a-val-name ,a))
(if ,a-val-name ,a-val-name ,b))))
(You can of course use gensym instead of make-symbol, and most people do I think.)
And now
> (let ((a-val 3))
(vel nil a-val))
3
as you would expect.
This is all because the CL macro system is unhygenic – it relies on you to ensure that things like names do not clash. In CL you have to go slightly out of your way to write macros which are correct in many cases. The Racket macro system, on the other hand, is hygenic: it will by default ensure that names (and other things) don't clash. In Racket (and Scheme) you have to go out of your way to write macros which are either incorrect or do something slightly unexpected like introducing bindings visible from code which makes use of the macros.
Note that I'm not expressing a preference for either approach to macros: I've spent most of my life writing CL, and I'm very happy with its macro system. More recently I've written more Racket and I'm happy with its macro system as well, although I find it harder to understand.
Finally here is a variant of your macro which is less surprising in use (almost all of the noise in this code is sanity checking which syntax-parse supports in the form of the two #:fail-when clauses):
(define-syntax (with-char-codes stx)
(syntax-parse stx
[(_ (v:id ...) form ...)
#:fail-when (check-duplicate-identifier (syntax->list #'(v ...)))
"duplicate name"
#:fail-when (for/or ([name (syntax->list #'(v ...))])
(and (> (string-length (symbol->string
(syntax->datum name)))
1)
name))
"name too long"
#'(let ([v (char->integer (string-ref (symbol->string 'v) 0))] ...)
form ...)]))
And now
> (with-char-codes (a b)
(+ a b))
195
I am having problems with this
e.g. i have
(define (mypow x) (* x x))
and I need to eval expressions from given list. (I am writing a simulator and I get a sequence of commands in a list as an argument)
I have already read that R5RS standard needs to include in function eval as second arg (scheme-report-environment 5), but still I am having issues with this.
This works (with standard function):
(eval '(sqrt 5) (scheme-report-environment 5))
but this does not:
(eval '(mypow 5) (scheme-report-environment 5))
It says:
../../../../../../usr/share/racket/collects/racket/private/kw.rkt:923:25: mypow: undefined;
cannot reference undefined identifier
Eventhough simply called mypow in prompt returns:
#<procedure:mypow>
Any advice, please? (btw I need to use R5RS)
(scheme-report-environment 5) returns all the bindings that are defined in the R5RS Scheme standard and not any of the user defined ones. This is by design. You will never be able to do what you want using this as the second parameter to eval.
The report mentions (interaction-environment) which is optional. Thus you have no guarantee the implementation has it, but it will have all the bindings from (scheme-report-environment 5)
For completeness there is (null-environment 5) which only has the bindings for the syntax. eg. (eval '(lambda (v) "constan) (null-environment 5)) works, but (eval '(lambda (v) (+ 5 v)) (null-environment 5)) won't since + is not in the resulting procedures closure.
Other ways to get things done
Usually you could get away without using eval alltogether. eval should be avoided at almost all costs. The last 16 years I've used eval deliberately in production code twice.
By using a thunk instead of data:
(define todo
(lambda () ; a thunk is just a procedure that takes no arguments
(mypow 5))
(todo) ; ==> result from mypow
Now imagine you have a list of operations you'd like done instead:
(define ops
`((inc . ,(lambda (v) (+ v 1))) ; notice I'm unquoting.
(dec . ,(lambda (v) (- v 1))) ; eg. need the result of the
(square . ,(lambda (v) (* v v))))) ; evaluation
(define todo '(inc square dec))
(define with 5)
;; not standard, but often present as either fold or foldl
;; if not fetch from SRFI-1 https://srfi.schemers.org/srfi-1/srfi-1.html
(fold (lambda (e a)
((cdr (assq e ops)) a))
with
todo)
; ==> 35 ((5+1)^2)-1
In an attempt to emulate simple OOP in scheme (just for fun), I have caught myself repeating the following pattern over and over:
(define my-class ; constructor
(let ((let-for-name-encapsulation 'anything))
; object created from data is message passing interface
(define (this data)
(lambda (m)
(cond ((eq? m 'method1) (method1 data))
((eq? m 'method2) (method2 data))
(else (error "my-class: unknown operation error" m)))))
;
(define (method1 data)
(lambda (arg1 ...)
... )) ; code using internal 'data' of object
;
(define (method2 data)
(lambda (arg2 ...)
... ))
;
; returning three arguments constructor (say)
;
(lambda (x y z) (this (list 'data x y z)))))
I decided to wrap everything inside a let ((let-for-name-encapsulation ... so as to avoid leaking names within the global environment while still
being able to use the define construct for each internal function name, which enhances readability. I prefer this solution to the unsightly construct (let ((method1 (lambda (... but I am still not very happy because of the somewhat artificial let-for-name-encapsulation. Can anyone suggests something simple which would make the code look even nicer?. Do I need to learn macros to go beyond this?
I use that pattern often, but you don't actually need to define any variables:
(define binding
(let ()
(define local-binding1 expression)
...
procedure-expression)))
I've seen it in reference implementations of SRFIs so it's a common pattern. Basically it's a way to make letrec without the extra identation and lambdas. It can easily be made a macro to make it even flatter:
(define-syntax define/lexical
(syntax-rules ()
((_ binding body ...)
(define binding
(let ()
body ...)))))
;; test
(define/lexical my-class
(define (this data)
(lambda (m)
(cond ((eq? m 'method1) (method1 data))
((eq? m 'method2) (method2 data))
(else (error "my-class: unknown operation error" m)))))
(define (method1 data)
(lambda (arg1 ...)
... )) ; code using internal 'data' of object
(define (method2 data)
(lambda (arg2 ...)
... ))
;; returning three arguments constructor (say)
(lambda (x y z) (this (list 'data x y z))))
;; it works for procedures that return procedures as well
(define/lexical (count start end step)
(define ...)
(lambda ...))
Of course you could use macros to simplify your object system as well.
I am trying to write a function that can check whether or not some input is a quotation for a syntax checker.
I have the following code:
(define is-quote-expression?
(lambda (v)
(equal? (car v) 'quote)
(is-quotation? (cdr v)))))
(define is-quotation?
(lambda (v)
(or (number? v)
(boolean? v)
(char? v)
(string? v)
(symbol? v)
(null? v)
(and (pair? v)
(is-quotation? (car v))
(is-quotation? (cdr v)))))
When I try to evaluate, I get the following:
(is-quote-expression? '1)
#f
(is-quote-expression? (cons 'quote 1))
#t
I think my TA told me that the Scheme environment replaced all "'" with "'quote", however, this does not seem to be the case. We are running Petite Chez Scheme.
What am I not seeing?
Thanks in advance.
There are a couple of problems with your code, for starters the (lambda (pair? v) part in is-quote-expression? is almost certainly a mistake (you're defining a lambda with two parameters called pair? and v).
Also, I believe you intended to call is-quotation? from is-quote-expression? only if v isn't a pair, so it doesn't make sense to ask again if (pair? v) in is-quotation?. And who said that a pair is a quotation only if both its car and cdr are quotations?
Here, I believe this is what you intended:
(define is-quote-expression?
(lambda (v)
(if (pair? v)
(equal? (car v) 'quote)
(is-quotation? v))))
(define is-quotation?
(lambda (v)
(or (number? v)
(boolean? v)
(char? v)
(string? v)
(symbol? v)
(null? v))))
I agree with Óscar's post, though, is-quote-expression? accepts a list rather than a pair and returns whether it was a quotation.
(define is-quote-expression?
(lambda (v)
(and (proper-list-of-given-length? v 2)
(equal? (car v) 'quote)
(is-quotation? (cadr v)))))
Also, your original question shows some confusion as to what quote actually does. This is what really should happen:
> (is-quote-expression? '1)
#f
> (is-quote-expression? (cons 'quote 1))
#f
> (is-quote-expression? (quote 42))
#f
> (is-quote-expression? (quote (quote 42)))
#t
Note how the built-in quote procedure simply returns what it is passed. In the case of (quote 42) it simply returns 42. But (quote (quote 42)) returns (quote 42), which is what you wish to pass to is-quote-expression?.
The behavior of quote is specified in R6RS, appendix A.3. For '1, the relevant rule is 6sqv:
(store (sf1 ...) S1[ 'sqv1 ]) → (store (sf1 ...) S1[ sqv1 ])
Time to break this down.
The "S → T" notation defines one step in evaluating a term, where "S" evaluates to "T".
Since store and the sf1 non-terminals appear the same on both the left and right sides, you don't need to understand them to understand how '1 evaluates. If you need something, think of store as "storage environment" and the sfn as pairs of names and values; (store ((x 1) (y 2)) S) means the identifier x is associated with the value 1 and y with 2 when evaluating the term S.
If you're not familiar with the S[e] notation, it refers to a term with one "hole" (an term with a [] in it) filled with e. There are two related syntax elements: terms with holes (S[]) and terms with a hole filled by a value (S[e]). Holes are a little (but only a little) like unnamed variables. One important difference is that a term is allowed only one hole. Holes are perhaps best explained by example. All of the following are S[]:
(+ [] 1 2)
(list [] 'a "foo")
(cond ((= x 0) [])
((> x 0) 1)
(else -1))
Note that a hole can only appear where a sub-term is syntactically valid; [] 2) is not a term-with-hole. S[0] would be a term with 0 substituted into the hole:
(+ 0 1 2)
(list 0 'a "foo")
(cond ((= x 0) 0)
((> x 0) 1)
(else -1))
When a value is given for a hole, the term S[] is also called a "context". This comes from one of the primary uses for terms-with-holes: to match any term containing the given value. S[e] is any term that contains e as a valid sub-term, so S[] is the context that e appears in. In short, S1['sqv1] is a stand-in for any term that contains a quote.
(+ 'foo 1 2)
(list 'bar 'a "foo")
(cond ((= x 0) 'baz)
((> x 0) 1)
(else -1))
Note the second term provides two different contexts for quote-terms: (list [] 'a "foo"), (list 'bar [] "foo"). This suggests that you shouldn't think of holes too much as just being unnamed variables.
If you're wonder why context terms and holes are used, they're an alternative to recursive definitions. Without contexts, → would have to be defined recursively over the structure of terms (Scheme's grammar defines the structure). Substitution in lambda calculus is an example of structural recursion, as are any tree-processing functions you might define in Scheme (though Scheme syntax is quite different than the syntax used to define → and lambda calculus substitution).
(define (tree-depth tree)
(if (pair? tree)
(max (tree-depth (car tree))
(tree-depth (cdr tree)))
1
) )
Next, let's examine the meaning of sqv, which is short for "self-quoting values". This is a non-terminal from the Scheme grammar, given in appendix A.2.
sqv ::= n | #t | #f
n ::= [numbers]
sqv is simply a number or boolean.
All together, the 6sqv evaluation rule means that a quoted number or boolean evaluates to the number or boolean; the quote is simply discarded.
What this signifies for your homework is that you can't tell the difference between 1 and '1 with a normal function, since sub-terms are evaluated before the function is called. You'll need a macro.
Why go through all this just to say that "'1 evaluates to 1"? Firstly, answering "why" takes more than answering "what". Also, it will hopefully help you go beyond the question, learning a little bit how to read Scheme's formal semantics, give you a taste of computational theory concepts and lead to many more questions.
Does Scheme or do any dialects of scheme have a kind of "self" operator so that anonymous lambdas can recur on themselves without doing something like a Y-combinator or being named in a letrec etc.
Something like:
(lambda (n)
(cond
((= n 0) 1)
(else (* n (self (- n 1)))))))
No. The trouble with the "current lambda" approach is that Scheme has many hidden lambdas. For example:
All the let forms (including let*, letrec, and named let)
do (which expands to a named let)
delay, lazy, receive, etc.
To require the programmer to know what the innermost lambda is would break encapsulation, in that you'd have to know where all the hidden lambdas are, and macro writers can no longer use lambdas as a way to create a new scope.
All-round lose, if you ask me.
There is a tradition of writing “anaphoric” macros that define special names in the lexical scope of their bodies. Using syntax-case, you can write such a macro on top of letrec and lambda. Note that the definition below is as hygienic as possible considering the specification (in particular, invisible uses of alambda will not shadow self).
;; Define a version of lambda that binds the
;; anaphoric variable “self” to the function
;; being defined.
;;
;; Note the use of datum->syntax to specify the
;; scope of the anaphoric identifier.
(define-syntax alambda
(lambda (stx)
(syntax-case stx ()
[(alambda lambda-list . body)
(with-syntax ([name (datum->syntax #'alambda 'self)])
#'(letrec ([name (lambda lambda-list . body)])
name))])))
;; We can define let in terms of alambda as usual.
(define-syntax let/alambda
(syntax-rules ()
[(_ ((var val) ...) . body)
((alambda (var ...) . body) val ...)]))
;; The let/alambda macro does not shadow the outer
;; alambda's anaphoric variable, which is lexical
;; with regard to the alambda form.
((alambda (n)
(if (zero? n)
1
(let/alambda ([n-1 (- n 1)])
(* (self n-1) n))))
10)
;=> 3628800
Most people avoid anaphoric operators since they make the structure of the code less recognizable. In addition, refactoring can introduce problems rather easily. (Consider what happens when you wrap the let/alambda form in the factorial function above in another alambda form. It's easy to overlook uses of self, especially if you're not reminded of it being relevant by having to type it explicitly.) It is therefore generally preferable to use explicit names. A “labeled” version of lambda that allows this can be defined using a simple syntax-rules macro:
;; Define a version of lambda that allows the
;; user to specifiy a name for the function
;; being defined.
(define-syntax llambda
(syntax-rules ()
[(_ name lambda-list . body)
(letrec ([name (lambda lambda-list . body)])
name)]))
;; The factorial function can be expressed
;; using llambda.
((llambda fac (n)
(if (zero? n)
1
(* (fac (- n 1)) n)))
10)
;=> 3628800
I have found a way using continuations to have anonymous lambdas call themselves and then using Racket macros to disguise the syntax so the anonymous lambda appears to have a "self" operator. I don't know if this solution is possible in other versions of Scheme since it depends on the Call-with-composable-continuation function of racket and the Macro to hide the syntax uses syntax parameters.
The basic idea is this, illustrated with the factorial function.
( (lambda (n)
(call-with-values
(lambda () (call-with-composable-continuation
(lambda (k) (values k n))))
(lambda (k n)
(cond
[(= 0 n) 1]
[else (* n (k k (- n 1)))])))) 5)
The continuation k is the call to the anonymous factorial function, which takes two arguments, the first being the continuation itself. So that when in the body we execute (k k N) that is equivalent to the anonymous function calling itself (in the same way that a recursive named lambda would do).
We then disguise the underlying form with a macro. Rackets syntax-parameters allow the transformation of (self ARGS ...) to (k k ARGS ... )
so we can have:
((lambda-with-self (n)
(cond
[(= 0 n) 0]
[(= 1 n) 1]
[else (* n (self (- n 1)))])) 5)
The complete Racket program to do this is:
#lang racket
(require racket/stxparam) ;required for syntax-parameters
( define-syntax-parameter self (λ (stx) (raise-syntax-error #f "not in `lambda-with-self'" stx)))
(define-syntax-rule
(lambda-with-self (ARG ... ) BODY ...)
(lambda (ARG ...)
(call-with-values
(lambda ()(call/comp (lambda (k) (values k ARG ...))))
(lambda (k ARG ...)
(syntax-parameterize ([self (syntax-rules ( )[(self ARG ...) (k k ARG ...)])])
BODY ...)))))
;Example using factorial function
((lambda-with-self (n)
(cond
[(= 0 n) 0]
[(= 1 n) 1]
[else (* n (self (- n 1)))])) 5)
This also answers my previous question about the differences between the different kinds of continuations.
Different kinds of continuations in Racket
This only works because unlike call-with-current-continuation, call-with-composable-continuation doesn't abort back to a continuation prompt but invokes the continuation at the place it was invoked.