I'm designing a program which has 2 parts. One on my computer and one on a Raspberry Pi.
I often send a message over TCP/IP, from a procedure on my computer to a procedure on the Raspberry Pi. So basically what it does is, sending the destination, the message and the list of arguments.
Then on the Raspberry Pi I read the destination (a certain object that is living on the raspberry pi) and send the message to that objet with the arguments.
In the objects on the raspberry pi I have many procedures which look like this :
(define (a-procedure argument1 argument2 ... argument-N)
...)
But the problem is that on the raspberry pi I receive a list of arguments.
I don't want to change all the procedures (on the raspberry pi) so they take an argument list instead of multiple arguments.
So intuitively I thought macros could do the job.
I never used macros and it was never teached to me so I'm very new to the concept. I searched a bit in the documentation and this is what I came up with :
(define-syntax rotate
(syntax-rules ()
[(a-procedure (list a)) (x a)]
[(a-procedure (list a ...)) (x a ...)]))
But the problem with this is that it always calls the procedure "x".
So now I would like that the procedure it calls could also be variable (not hardcoded, because I don't want to make such a macro for each single procedure on the raspberry pi...).
For example :
(callWithArguments (+ (list 1 2 3))) ; --> (+ 1 2 3)
(callWithArguments (* (list 1 2 3))) ; --> (* 1 2 3)
To do so I tried this :
(define-syntax callWithArguments
(syntax-rule ()
[(a-procedure (list a)) (a-procedure a)]
[(a-procedure (list a ...)) (a-procedure a ...)]))
But in this example "a-procedure" isn't variable.
If I try to call it like mentioned above I get :
a-procedure: undefined;
cannot reference an identifier before its definition
I tried to do it with "syntax-rules-id" but didn't succeed.
I also read the documentation about multiple values.
I saw a procedure "call-with-values".
So I tried this :
(call-with-values (lambda ()
(for-each (lambda (arg) arg) argumentList))
a-procedure)
Now the procedure is variable (not a hardcoded procedure) but the problem is that it only takes into account the last value returned by for-each.
I didn't found a way to turn a list into multiple values '(1 2 3 4) --> (values 1 2 3 4)
I think you're over designing this. If you want to call a procedure with multiple arguments that come in a list, just use apply. So this:
(callWithArguments (+ (list 1 2 3)))
Can be expressed as this:
(apply + (list 1 2 3))
But if you want to use a macro for evaluating the expression, try this:
(define-syntax callWithArguments
(syntax-rules ()
[(_ (a-procedure a-list))
(apply a-procedure a-list)]))
(callWithArguments (+ (list 1 2 3)))
=> 6
Or if you're interested in transforming the expression without actually evaluating it, use list splicing:
(define-syntax callWithArguments
(syntax-rules ()
[(_ (a-procedure a-list))
`(a-procedure ,#a-list)]))
(callWithArguments (+ (list 1 2 3)))
=> '(+ 1 2 3)
Related
Given a Chez Scheme record with many numeric fields that are contantly being mutated by small increments and decrements, usually by one, is there a way to write a macro that can mutate a field value by passing it the field? The way I accomplish this now is something like the following REPL transcript:
Chez Scheme Version 9.5.4
Copyright 1984-2020 Cisco Systems, Inc.
> (define-record-type r (fields (mutable x) (mutable y)
;; and so on...
))
> (define my-r (make-r 3 5
;; and so on...
))
> (r-x-set! my-r (+ (r-x my-r) 1))
> my-r
#[#{r gak6l6ll8wuv7yd61kiomgudo-2} 4 5]
It would be nice to have a simple macro, say inc!, that could do the mutating increment/decrement operations on the fields in the record. I started with something like a Scheme version of Lisps incf and decf,
(define-syntax inc!
(syntax-rules ()
((_ x) (begin (set! x (+ x 1)) x))))
(inc! (r-x my-r)) ;; Syntax error
Which works for "normal" variables (and makes it easy to implement dec!), but it doesn't use the mechanism to set mutable record fields, r-x-set! in this case.
Is there an obvious way to write such a macro? One where you can just pass a reference to the record field without having to write something different for each field?
You can construct a -set! mutator from the given accessor. This can be done by converting the symbol for the accessor to a string and appending "-set!" to it. Then eval can be used to get the actual mutator procedure. Here is a macro that increments a specified field by some amount n:
(define-syntax increment-n!
(syntax-rules ()
[(_ (acc rec) n)
(let* ((acc-name (symbol->string (quote acc)))
(mut-name (string-append acc-name "-set!"))
(mut! (eval (string->symbol mut-name))))
(mut! rec (+ (acc rec) n)))]))
This can be used to create an inc! macro:
(define-syntax inc!
(syntax-rules ()
[(_ (acc rec)) (increment-n! (acc rec) 1)]))
But, it would be nice to be able to increment multiple fields at the same time; here are inc! and dec! macros that do that:
(define-syntax inc!
(syntax-rules ()
[(_ (acc rec) ...) (begin (increment-n! (acc rec) 1) ...)]))
(define-syntax dec!
(syntax-rules ()
[(_ (acc rec) ...) (begin (increment-n! (acc rec) -1) ...)]))
Sample interaction:
> my-r
#[#{r n5an6pxs3wvid36v2gvn8z9zo-5} 3 5 7]
> (inc! (r-x my-r))
> my-r
#[#{r n5an6pxs3wvid36v2gvn8z9zo-5} 4 5 7]
> (dec! (r-z my-r))
> my-r
#[#{r n5an6pxs3wvid36v2gvn8z9zo-5} 4 5 6]
> (inc! (r-x my-r) (r-y my-r) (r-z my-r))
> my-r
#[#{r n5an6pxs3wvid36v2gvn8z9zo-5} 5 6 7]
A Note on the Use of eval
The increment-n! macro constructs a symbol which has already been bound to a mutator procedure. That symbol could then be bound to mut! directly, but then when the expression (mut! rec (+ (acc rec) n)) is evaluated an exception would be raised since mut! now evaluates to a symbol, e.g., r-x-set!. We want mut! to evaluate to a procedure in a procedure call. By calling eval on the constructed symbol first we get the mutator procedure which is bound to that symbol, binding it to mut! instead of the symbol.
Here is a REPL interaction that illustrates the problem, and will hopefully help to clarify:
> (define f (string->symbol "+"))
> f
+
> (f 1 2)
Exception: attempt to apply non-procedure +
Type (debug) to enter the debugger.
> (define f (eval (string->symbol "+")))
> f
#<procedure +>
> (f 1 2)
3
As an exercise in learning the Racket macro system, I've been implementing a unit testing framework, based on the C++ catch framework. One of the features of that framework is that if I write a check like this:
CHECK(x == y); // (check x y)
When the check is violated the error message will print out the values of x and y, even though the macro used is completely generic, unlike other test frameworks that require you to use macros like CHECK_EQUALS, CHECK_GREATER, etc. This is possible through some hackery involving expression templates and operator overloading.
It occurs to me that in Racket you should be able to do an even better job. In the C++ version the macro can't see inside subexpressions, so if you write something like:
CHECK(f(x, g(y)) == z); // (check (= (f x (g y)) z))
When the check is violated you only find out the values of the left and right hand side of the equal sign, and not the values of x, y, or g(y). In racket I expect it should be possible to recurse into subexpressions and print a tree showing each step of the evaluation.
Problem is I have no idea what the best way to do this is:
I've gotten fairly familiar with syntax-parse, but this seems beyond its abilities.
I read about customizing #%app which almost seems like what I want, but if for example f is a macro, I don't want to print out every evaluation of the expressions that are in the expansion, just the evaluations of the expressions that were visible when the user invoked the check macro. Also not sure if I can use it without defining a language.
I could use syntax-parameterize to hijack the meaning of the basic operators but that won't help with function calls like g(y).
I could use syntax->datum and manually walk the AST, calling eval on subexpressions myself. This seems tricky.
The trace library almost looks like what it does what I want, but you have to give it a list of functions upfront, and it doesn't appear to give you any control over where the output goes (I only want to print anything if the check fails, not if it succeeds, so I need to save the intermediate values to the side as execution proceeds).
What would be the best or at least idiomatic way to implement this?
Here is something to get you started.
#lang racket
(require (for-syntax syntax/parse racket/list))
(begin-for-syntax
(define (expression->subexpressions stx)
(define expansion (local-expand stx 'expression '()))
(syntax-parse expansion
#:datum-literals (#%app quote)
[x:id (list #'x)]
[b:boolean (list #'b)]
[n:number (list #'n)]
; insert other atoms here
[(quote literal) (list #'literal)]
[(#%app e ...)
(cons stx
(append-map expression->subexpressions (syntax->list #'(e ...))))]
; other forms in fully expanded syntax goes here
[else
(raise-syntax-error 'expression->subexpressions
"implement this construct"
stx)])))
(define-syntax (echo-and-eval stx)
(syntax-parse stx
[(_ expr)
#'(begin
(display "] ") (displayln (syntax->datum #'expr))
(displayln expr))]))
(define-syntax (echo-and-eval-subexpressions stx)
(syntax-parse stx
[(_ expr)
(define subs (expression->subexpressions #'expr))
(with-syntax ([(sub ...) subs])
#'(begin
; sub expressions
(echo-and-eval sub)
...
; original expression
(echo-and-eval expr)))]))
(echo-and-eval-subexpressions (+ 1 2 (* 4 5)))
The output:
] (+ 1 2 (* 4 5))
23
] +
#<procedure:+>
] 1
1
] 2
2
] (#%app * '4 '5)
20
] *
#<procedure:*>
] 4
4
] 5
5
] (+ 1 2 (* 4 5))
23
An alternative to printing everything is to add a marker for stuff that should be shown. Here's a rough simple sketch:
#lang racket
(require racket/stxparam)
(define-syntax-parameter ?
(λ(stx) (raise-syntax-error '? "can only be used in a `test' context")))
(define-syntax-rule (test expr)
(let ([log '()])
(define (log! stuff) (set! log (cons stuff log)))
(syntax-parameterize ([? (syntax-rules ()
[(_ E) (let ([r E]) (log! `(E => ,r)) r)])])
(unless expr
(printf "Test failure: ~s\n" 'expr)
(for ([l (in-list (reverse log))])
(for-each display
`(" " ,#(add-between (map ~s l) " ") "\n")))))))
(define x 11)
(define y 22)
(test (equal? (? (* (? x) 2)) (? y)))
(test (equal? (? (* (? x) 3)) (? y)))
which results in this output:
Test failure: (equal? (? (* (? x) 3)) (? y))
x => 11
(* (? x) 3) => 33
y => 22
I am struggling to find the right approach to solve the following function
(FOO #'– '(1 2 3 4 5))
=> ((–1 2 3 4 5) (1 –2 3 4 5) (1 2 –3 4 5) (1 2 3 –4 5) (1 2 3 4 –5))
The first Parameter to the foo function is supposed to be a function "-" that has to be applied to each element returning a list of list as shown above. I am not sure as to what approach I can take to create this function. I thought of recursion but not sure how I will preserve the list in each call and what kind of base criteria would I have. Any help would be appreciated. I cannot use loops as this is functional programming.
It's a pity you cannot use loop because this could be elegantly solved like so:
(defun foo (fctn lst)
(loop
for n from 0 below (length lst) ; outer
collect (loop
for elt in lst ; inner
for i from 0
collect (if (= i n) (funcall fctn elt) elt))))
So we've got an outer loop that increments n from 0 to (length lst) excluded, and an inner loop that will copy verbatim the list except for element n where fctn is applied:
CL-USER> (foo #'- '(1 2 3 4 5))
((-1 2 3 4 5) (1 -2 3 4 5) (1 2 -3 4 5) (1 2 3 -4 5) (1 2 3 4 -5))
Replacing loop by recursion means creating local functions by using labels that replace the inner and the outer loop, for example:
(defun foo (fctn lst)
(let ((len (length lst)))
(labels
((inner (lst n &optional (i 0))
(unless (= i len)
(cons (if (= i n) (funcall fctn (car lst)) (car lst))
(inner (cdr lst) n (1+ i)))))
(outer (&optional (i 0))
(unless (= i len)
(cons (inner lst i) (outer (1+ i))))))
(outer))))
Part of the implementation strategy that you choose here will depend on whether you want to support structure sharing or not. Some of the answers have provided solutions where you get completely new lists, which may be what you want. If you want to actually share some of the common structure, you can do that too, with a solution like this. (Note: I'm using first/rest/list* in preference to car/car/cons, since we're working with lists, not arbitrary trees.)
(defun foo (operation list)
(labels ((foo% (left right result)
(if (endp right)
(nreverse result)
(let* ((x (first right))
(ox (funcall operation x)))
(foo% (list* x left)
(rest right)
(list* (revappend left
(list* ox (rest right)))
result))))))
(foo% '() list '())))
The idea is to walk down list once, keeping track of the left side (in reverse) and the right side as we've gone through them, so we get as left and right:
() (1 2 3 4)
(1) (2 3 4)
(2 1) (3 4)
(3 2 1) (4)
(4 3 2 1) ()
At each step but the last, we take the the first element from the right side, apply the operation, and create a new list use revappend with the left, the result of the operation, and the rest of right. The results from all those operations are accumulated in result (in reverse order). At the end, we simply return result, reversed. We can check that this has the right result, along with observing the structure sharing:
CL-USER> (foo '- '(1 2 3 4 5))
((-1 2 3 4 5) (1 -2 3 4 5) (1 2 -3 4 5) (1 2 3 -4 5) (1 2 3 4 -5))
By setting *print-circle* to true, we can see the structure sharing:
CL-USER> (setf *print-circle* t)
T
CL-USER> (let ((l '(1 2 3 4 5)))
(list l (foo '- l)))
((1 . #1=(2 . #2=(3 . #3=(4 . #4=(5))))) ; input L
((-1 . #1#)
(1 -2 . #2#)
(1 2 -3 . #3#)
(1 2 3 -4 . #4#)
(1 2 3 4 -5)))
Each list in the output shares as much structure with the original input list as possible.
I find it easier, conceptually, to write some of these kind of functions recursively, using labels, but Common Lisp doesn't guarantee tail call optimization, so it's worth writing this iteratively, too. Here's one way that could be done:
(defun phoo (operation list)
(do ((left '())
(right list)
(result '()))
((endp right)
(nreverse result))
(let* ((x (pop right))
(ox (funcall operation x)))
(push (revappend left (list* ox right)) result)
(push x left))))
The base case of a recursion can be determined by asking yourself "When do I want to stop?".
As an example, when I want to compute the sum of an integer and all positive integers below it, I can do this recusively with a base case determined by answering "When do I want to stop?" with "When the value I might add in is zero.":
(defun sumdown (val)
(if (zerop val)
0
(+ (sumdown (1- val)) val)))
With regard to 'preserve the list in each call', rather than trying to preserve anything I would just build up a result as you go along. Using the 'sumdown' example, this can be done in various ways that are all fundamentally the same approach.
The approach is to have an auxiliary function with a result argument that lets you build up a result as you recurse, and a function that is intended for the user to call, which calls the auxiliary function:
(defun sumdown1-aux (val result)
(if (zerop val)
result
(sumdown1-aux (1- val) (+ val result))))
(defun sumdown1 (val)
(sumdown1-aux val 0))
You can combine the auxiliary function and the function intended to be called by the user by using optional arguments:
(defun sumdown2 (val &optional (result 0))
(if (zerop val)
result
(sumdown2 (1- val) (+ val result))))
You can hide the fact that an auxiliary function is being used by locally binding it within the function the user would call:
(defun sumdown3 (val)
(labels ((sumdown3-aux (val result)
(if (zerop val)
result
(sumdown3-aux (1- val) (+ val result)))))
(sumdown3-aux val 0)))
A recursive solution to your problem can be implemented by answering the question "When do I want to stop when I want to operate on every element of a list?" to determine the base case, and building up a result list-of-lists (instead of adding as in the example) as you recurse. Breaking the problem into smaller pieces will help - "Make a copy of the original list with the nth element replaced by the result of calling the function on that element" can be considered a subproblem, so you might want to write a function that does that first, then use that function to write a function that solves the whole problem. It will be easier if you are allowed to use functions like mapcar and substitute or substitute-if, but if you are not, then you can write equivalents yourself out of what you are allowed to use.
Does someone know why the following produces the expected result - (2 4 6)
(defmacro mult2 (lst)
(define (itter x)
(list '* 2 x))
`(list ,#(map itter lst)))
(mult2 (1 2 3))
while I expected that this one would (with the list identifier)
(defmacro mult2 (lst)
(define (itter x)
(list '* 2 x))
`(list ,#(map itter lst)))
(mult2 '(1 2 3))
Macro "arguments" are not evaluated. So, when you pass in '(1 2 3), i.e., (quote (1 2 3)), that is exactly what the macro sees.
P.S. You are much better off using hygienic macros in Scheme. Here's an example using syntax-case:
(define-syntax mult2
(lambda (stx)
(define (double x)
#`(* 2 #,x))
(syntax-case stx ()
((_ lst)
#`(list #,#(map double (syntax-e #'lst)))))))
(That's still not how such a macro is idiomatically written, but I tried to mirror your version as closely as possible.)
That's because the '(1 2 3) is expanded by the reader into (quote (1 2 3)). Since you only destructure one list in your macro, it won't work as expected.
Some general advice: if you're working in Racket you probably want to avoid using defmacro. That is definitely not the idiomatic way to write macros. Take a look at syntax-rules and, if you want to define more complicated macros, syntax-parse. Eli also wrote an article explaining syntax-case for people used to defmacro.
I have a Scheme macro and a long list, and I'd like to map the macro across the list, just as if it were a function. How can I do that using R5RS?
The macro accepts several arguments:
(mac a b c d)
The list has
(define my-list ((a1 b1 c1 d1)
(a2 b2 c2 d2)
...
(an bn cn dn)))
And I'd like to have this:
(begin
(mac a1 b1 c1 d2)
(mac a2 b2 c2 d2)
...
(mac an bn cn dn))
(By the way, as you can see I'd like to splice the list of arguments too)
Expanding on z5h's answer of using eval, the methods below show how a map-macro macro can be written if interaction-environment in implemented in the version of R5RS in use:
(define test-list '((1 2 3 4)
(5 6 7 8)))
;Or if your version of scheme implments interaction-environment then:
(define-syntax trade
(syntax-rules ()
((_ a b c d) (display (list b a d c)))))
;!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
;Careful this is not really mapping. More like combined map and apply.
;!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(define-syntax map-macro
(syntax-rules ()
((_ mac ls) (let ((mac-list (map (lambda (lst) (cons 'trade lst)) ls)))
(eval
`(begin
,#mac-list)
(interaction-environment))))
))
(map-macro trade test-list)
;outputs: (2 1 4 3)(6 5 8 7)
So that last map-macro call evaluates the following:
What ends up getting evaluated from (map-macro trade test-list) is:
(begin
(trade 1 2 3 4)
(trade 5 6 7 8))
Which is not quite a map, but I believe it does answers your question.
Syntactic extensions are expanded into
core forms at the start of evaluation
(before compilation or interpretation)
by a syntax expander. -Dybvig, "The
Scheme Programming Language:
A macro operates on syntax. This happens before compilation or execution. It gives you another way of writing the same code.
A function operates on variables and values (which might be lists, atoms, numbers, etc) who's value is known when the function is invoked.
So mapping a macro doesn't make sense. You're asking to invoke something (macro expansion) that already happened long ago.
If you need something to write code for you and evaluate it at runtime, then might be one of those cases where you need eval.
Would something like
(map (lambda (l) (mac (car l) (caar l) (caaar l) (caaaar l))) mylist)
work?