So I'm playing around with a simple doc-string system as a warmup in scheme, the idea being you could do something like:
(def-with-doc (foo a b)
(desc "Takes two parameters and sums them")
(param 'a "First parameter")
(param 'b "Second parameter")
(return "Sum of arguments")
(+ a b)
Which would be turned into:
(begin
(begin
(desc 'foo "Takes two parameters and sums them")
(param 'foo 'a "First parameter")
(param 'foo 'b "Second parameter")
(return 'foo "Sum of arguments"))
(begin
(define (foo a b)
(+ a b))))
The macro I've written:
(define doc-symbol-list '(param desc return))
(define-macro (def-with-doc arg-list #!rest body)
;; Loop over body, splitting into doc calls and everything else
(let loop ((remaining body) (docs '()) (main '()))
(if (null? remaining)
; Reverse accumulation order of docs and main
; And build re-ordered begin tree
(let ((docs (cons 'begin (reverse docs)))
(main (cons 'begin (reverse main))))
(cons 'begin `(,docs ,`(define ,arg-list ,main))))
; Accumulate into docs list if expression is reserved
; Otherwise into the body list
(let ((sexp (car remaining)) (rest (cdr remaining)))
(if (member (car sexp) doc-symbol-list)
(loop rest (cons sexp docs) main)
(loop rest docs (cons sexp main)))))))
Takes the definition, moves the param/desc/return calls into the top level wrapped in begin statements and reconstructs the body of the function, that way the doc string calls are only executed once when the file is loaded rather than each time the function is called. I know I could manually put the doc-string stuff at the top level but I'm trying to emulate Python doc-strings.
Anyhow, the last think that I need to do is bind the function name (foo in above) into the doc-string calls, so that (param 'a "First parameter") becomes (param 'foo 'a "First parameter") so that the function each call is associated with is known. This is where I'm having trouble, every attempt I've made has failed to do what I want.
I would suggest using define-syntax as it is hygienic and its syntax-rules are pretty easy to understand. syntax-rules are in a pattern-to-result format; if you can understand cond, you can understand syntax-rules.
I think this does what you want, judging by the before and after snippets.
(define-syntax def-with-doc
(syntax-rules ()
;; this pattern
[(_ (func params ...)
(tag attributes ...)
...
code)
;; is converted into
(begin
(tag (quote func) attributes ...)
...
(define (func params ...)
code))]))
Forgive my terminology because I've never used doc-strings.
Basically, this matches against anything that follows that pattern of a function + params def, 0 or more tags with attributes, and a code statement.
Then, it just rearranges everything.
Related
I am trying to create a function that takes another function as a parameter and calls the functions in a loop.
The code below should get the function and the number of times the loop should execute:
(define (forLoop loopBody reps)
(let
(
(fun (car loopBody))
(str (cdr loopBody))
)
(cond
((eval (= reps 0) (interaction-environment)) "")
(else (cons str (forLoop '(fun str) (- reps 1))))
)
)
)
The code below is how i am calling the function
(define (printermsg)
(display msg)
)
(forLoop '(printer "text ") 4)
The expected output for the above code:
text text text text
There are several issues with your code:
The way you pass the loopBody parameter to your function is incorrect, a list of symbols will not work: you want to pass along a list with the actual function and its argument.
You are not obtaining the second element in the list, cdr won't work because it returns the rest of the list, you need to use cadr instead.
Avoid using eval, it's not necessary at all for what you want, and in general is a bad idea.
Why are you building a list? cons is not required here.
When calling the recursion, you are again passing a list of symbols instead of the proper argument.
You're not actually calling the function!
This should work:
(define (forLoop loopBody reps)
(let ((fun (car loopBody))
(str (cadr loopBody)))
(cond
((= reps 0) (void)) ; don't do anything when the loop is done
(else
(fun str) ; actually call the function!
(forLoop loopBody (- reps 1))))))
(define (printer msg)
(display msg))
(forLoop (list printer "text ") 4) ; see how to build the loop body
=> text text text text
For some code I was working I've needed to handle 'x inside macro. What is standard way of handling those values?
I have code like this:
(define (quoted-symbol? x)
(and (pair? x) (eq? (car x) 'quote) (symbol? (cadr x)) (null? (cddr x))))
(define-macro (test x)
(if (quoted-symbol? x)
`(begin
(display ',(cadr x))
(newline))))
(test 'hello) ;; 'hello will be expanded into list (quote hello)
Is this how this should be handled, or is just in macro you don't use quoted symbols?
NOTE: I'm not asking about hygienic macros (I'm asking about real lisp macros), so please no answers with hygienic macros.
EDIT:
My macro works correctly in Guile and BiwaScheme and in my own scheme like lisp in JavaScript. Here is better example:
(define-macro (test x)
(if (quoted-symbol? x)
`',(cadr x)))
(define (example arg)
(list arg (test 'world)))
(example 'hello)
the question was not about display, but about (cadr x).
EDIT2: You've asked so here you go, my macro:
(define-macro (--> expr . code)
"Helper macro that simplify calling methods on objects. It work with chaining
usage: (--> ($ \"body\")
(css \"color\" \"red\")
(on \"click\" (lambda () (print \"click\"))))
(--> document (querySelectorAll \"div\"))
(--> (fetch \"https://jcubic.pl\") (text) (match /<title>([^<]+)<\/title>/) 1)
(--> document (querySelectorAll \".cmd-prompt\") 0 \"innerText\")"
(let ((obj (gensym)))
`(let* ((,obj ,(if (and (symbol? expr) (not (null? (match /\./ (symbol->string expr)))))
`(.. ,expr)
`,expr)))
,#(map (lambda (code)
(let ((name (gensym))
(value (gensym)))
`(let* ((,name ,(cond ((quoted-symbol? code) (symbol->string (cadr code)))
((pair? code) (symbol->string (car code)))
(true code)))
(,value (. ,obj ,name)))
,(if (and (pair? code) (not (quoted-symbol? code)))
`(set! ,obj (,value ,#(cdr code)))
`(set! ,obj ,value)))))
code)
,obj)))
;; ---------------------------------------------------------------------------------------
(define (quoted-symbol? x)
"(quoted-symbol? code)
Helper function that test if value is quoted symbol. To be used in macros
that pass literal code that is transformed by parser.
usage:
(define-macro (test x)
(if (quoted-symbol? x)
`',(cadr x)))
(list 'hello (test 'world))"
(and (pair? x) (eq? (car x) 'quote) (symbol? (cadr x)) (null? (cddr x))))
the macro is used in my scheme like lisp in JavaScript, like the doc string suggest:
(--> document (querySelectorAll ".class") 0 "innerText")
I want to support:
(--> document (querySelectorAll ".class") 0 'innerText)
The code can be tested online at: https://jcubic.github.io/lips/ (You need to copy/paste the code since current version allow only method calls).
To get expansion you can use
(pprint (macroexpand (--> document (querySelector "x"))))
if it don't work (don't expand) it mean that macro is broken somehow.
dot is build in function that get property of an object and .. macro:
(define-macro (.. expr)
"(.. foo.bar.baz)
Macro that gets value from nested object where argument is comma separated symbol"
(if (not (symbol? expr))
expr
(let ((parts (split "." (symbol->string expr))))
(if (single parts)
expr
`(. ,(string->symbol (car parts)) ,#(cdr parts))))))
that can be use to get nested property like (.. document.body.innerHTML)
Scheme doesn't have "real lisp macros". Some implementations has something similar, but the forms have different names and uses. They are not portable at all.
The standard way of handling 'x is to handle it like an expression that gets evaluated in the expansion. Eg.
(define var 'x)
(test 'x)
(test var)
The two test forms should amount to the same even though the macro test gets (quote x) in the first and the symbol var in the second. At the time of the expansion var does not exist since the implementation can expand all the macros before starting.
You implementation of test will not work. Eg. the display might be run one or twice and then each time you call a procedure that uses it it will gfail since the expansion is the undefined value and it might not be fit for evaluation. eg.
(define (example arg)
(list arg (test 'w)))
When this is defined you get 'w or (quote w) printed with a newline and then the procedure it tries to store is:
(define (example arg)
(list arg #<undefined>))
Note that what constitutes the undefined value is chosen by the implementaion, but I know for sure that in many implementaions you cannot evaluate #<undefined>.
I'm reading The Little Schemer. And thanks to my broken English, I was confused by this paragraph:
(cond ... ) also has the property of not considering all of its
arguments. Because of this property, however, neither (and ... ) nor
(or ... ) can be defined as functions in terms of (cond ... ), though
both (and ... ) and (or ... ) can be expressed as abbreviations of
(cond ... )-expressions:
(and a b) = (cond (a b) (else #f)
and
(or a b) = (cond (a #t) (else (b))
If I understand it correctly, it says (and ...) and (or ...) can be replaced by a (cond ...) expression, but cannot be defined as a function that contains (cond ...). Why is it so? Does it have anything to do with the variant arguments? Thanks.
p.s. I did some searching but only found that (cond ...) ignores the expressions when one of its conditions evaluate to #f.
Imagine you wrote if as a function/procedure rather than a user defined macro/syntax:
;; makes if in terms of cond
(define (my-if predicate consequent alternative)
(cond (predicate consequent)
(else alternative)))
;; example that works
(define (atom? x)
(my-if (not (pair? x))
#t
#f))
;; example that won't work
;; peano arithemtic
(define (add a b)
(my-if (zero? a)
b
(add (- a 1) (+ b 1))))
The problem with my-if is that as a procedure every argument gets evaluated before the procedure body gets executed. thus in atom? the parts (not (pair? x)), #t and #f were evaluated before the body of my-if gets executed.
For the last example means (add (- a 1) (+ b 1)) gets evaluated regardless of what a is, even when a is zero, so the procedure will never end.
You can make your own if with syntax:
(define-syntax my-if
(syntax-rules ()
((my-if predicate consequent alternative)
(cond (predicate consequent)
(else alternative)))))
Now, how you read this is the first part is a template where the predicate consequent and alternative represent unevaluated expressions. It's replaced with the other just reusing the expressions so that:
(my-if (check-something) (display 10) (display 20))
would be replaced with this:
(cond ((check-something) (display 10))
(else (display 20)))
With the procedure version of my-if both 10 and 20 would have been printed. This is how and and or is implemented as well.
You cannot define cond or and or or or if as functions because functions evaluate all their arguments. (You could define some of them as macros).
Read also the famous SICP and Lisp In Small Pieces (original in French).
Suppose I have something like this:
(define pair (cons 1 (lambda (x) (* x x))
If I want to return the front object of the pair I do this:
(car pair)
And it returns 1. However when the object is a procedure I don't get the exact description of it.
In other words:
(cdr pair)
returns #<procedure> and not (lambda (x) (*x x)).
How do I fix this?
Although there's no way to do this generally, you can rig up something to do it for procedures that you define.
Racket structs can define a prop:procedure that allows the struct to be applied (called) as a procedure. The same struct can hold a copy of your original syntax for the function definition. That's what the sourced struct is doing, below.
The write-sourced stuff is simply to make the output cleaner (show only the original sexpr, not the other struct fields).
The define-proc macro makes it simpler to initialize the struct -- you don't need to type the code twice and hope it matches. It does this for you.
#lang racket
(require (for-syntax racket/syntax))
;; Optional: Just for nicer output
(define (write-sourced x port mode)
(define f (case mode
[(#t) write]
[(#f) display]
[else pretty-print])) ;nicer than `print` for big sexprs
(f (sourced-sexpr x) port))
(struct sourced (proc sexpr)
#:property prop:procedure (struct-field-index proc)
;; Optional: Just to make cleaner output
#:methods gen:custom-write
[(define write-proc write-sourced)])
;; A macro to make it easier to use the `sourced` struct
(define-syntax (define-proc stx)
(syntax-case stx ()
[(_ (id arg ...) expr ...)
#'(define id (sourced (lambda (arg ...) expr ...)
'(lambda (arg ...) expr ...)))]))
;; Example
(define-proc (foo x)
(add1 x))
(foo 1) ; => 2
foo ; => '(lambda (x) (add1 x))
The procedure cons evaluates its arguments: 1 is self-evaluating to 1; (lambda ...) evaluates to an anonymous procedure. If you want to 'prevent' evaluation, you need to quote the argument, as such:
> (define pair (cons 1 '(lambda (x) (* x x))
> (cdr pair)
(lambda (x) (* x x))
I'm reading The Little Schemer. And thanks to my broken English, I was confused by this paragraph:
(cond ... ) also has the property of not considering all of its
arguments. Because of this property, however, neither (and ... ) nor
(or ... ) can be defined as functions in terms of (cond ... ), though
both (and ... ) and (or ... ) can be expressed as abbreviations of
(cond ... )-expressions:
(and a b) = (cond (a b) (else #f)
and
(or a b) = (cond (a #t) (else (b))
If I understand it correctly, it says (and ...) and (or ...) can be replaced by a (cond ...) expression, but cannot be defined as a function that contains (cond ...). Why is it so? Does it have anything to do with the variant arguments? Thanks.
p.s. I did some searching but only found that (cond ...) ignores the expressions when one of its conditions evaluate to #f.
Imagine you wrote if as a function/procedure rather than a user defined macro/syntax:
;; makes if in terms of cond
(define (my-if predicate consequent alternative)
(cond (predicate consequent)
(else alternative)))
;; example that works
(define (atom? x)
(my-if (not (pair? x))
#t
#f))
;; example that won't work
;; peano arithemtic
(define (add a b)
(my-if (zero? a)
b
(add (- a 1) (+ b 1))))
The problem with my-if is that as a procedure every argument gets evaluated before the procedure body gets executed. thus in atom? the parts (not (pair? x)), #t and #f were evaluated before the body of my-if gets executed.
For the last example means (add (- a 1) (+ b 1)) gets evaluated regardless of what a is, even when a is zero, so the procedure will never end.
You can make your own if with syntax:
(define-syntax my-if
(syntax-rules ()
((my-if predicate consequent alternative)
(cond (predicate consequent)
(else alternative)))))
Now, how you read this is the first part is a template where the predicate consequent and alternative represent unevaluated expressions. It's replaced with the other just reusing the expressions so that:
(my-if (check-something) (display 10) (display 20))
would be replaced with this:
(cond ((check-something) (display 10))
(else (display 20)))
With the procedure version of my-if both 10 and 20 would have been printed. This is how and and or is implemented as well.
You cannot define cond or and or or or if as functions because functions evaluate all their arguments. (You could define some of them as macros).
Read also the famous SICP and Lisp In Small Pieces (original in French).