since yesterday I've been trying to program a special case statement for scheme that would do the following:
(define (sort x)
(cond ((and (list? x) x) => (lambda (l)
(sort-list l)))
((and (pair? x) x) => (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else "here")))
instead of using all the and's and cond's statement, I would have:
(define (sort x)
(scase ((list? x) => (lambda (l)
(sort-list l)))
((pair? x) => (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else "here")))
What I could do so far, was this:
(define (sort x)
(scase (list? x) (lambda (l)
(sort-list l)))
(scase (pair? x) (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p))))
with this code:
(define-syntax scase
(syntax-rules ()
((if condition body ...)
(if condition
(begin
body ...)))))
What I wanted to do now, is just allow the scase statement to have multiple arguments like this:
(scase ((list? (cons 2 1)) 'here)
((list? '(2 1)) 'working))
but I can't seem to figure out how I can do that. Maybe you guys could give me a little help?
Thanks in advance ;)
If this is an exercise in learning how to use syntax-rules, then disregard this answer.
I see a way to simplify your code that you are starting with.
(define (sort x)
(cond ((list? x)
(sort-list x))
((pair? x)
(if (> (car x) (cdr x))
(cons (cdr x) (car x))
x)))
(else "here")))
Since all the (and (list? x) x) => (lambda l ... does is see if x is a list, and then bind l to x, (since #f is not a list, and '() is not false, at least in Racket), you can just skip all that and just use x. You do not need to use => in case, and in this case it doesn't help. => is useful if you want to do an test that returns something useful if successful, or #f otherwise.
Now, if you want to use a macro, then you're going to need to clarify what you want it to do a bit better. I think that case already does what you want. Your existing macro is just if, so I'm not sure how to extend it.
I found the solution for my question, here it goes:
(define-syntax cases
(syntax-rules ()
((_ (e0 e1 e2 ...)) (if e0 (begin e1 e2 ...)))
((_ (e0 e1 e2 ...) c1 c2 ...)
(if e0 (begin e1 e2 ...) (cases c1 c2 ...)))))
Thank you all anyway :)
Here's a solution :
#lang racket
(require mzlib/defmacro)
(define-syntax scase
(syntax-rules (else)
((_ (else body1)) body1)
((_ (condition1 body1) (condition2 body2) ...)
(if condition1
body1
(scase (condition2 body2) ...)))))
(define (sort1 x)
((scase ((list? x) (lambda (l)
(sort l <)))
((pair? x) (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else (lambda (e) "here")))
x))
It works in DrRacket. I made three changes to your solution. First, i renamed your sort procedure to sort1 since sort is inbuilt in scheme ( I have used it inside sort1). Second, I have changed the sort1 itself so that the input given will be passed to the procedure returned by scase and you will directly get the sorted result. Third, I have modified the scase syntax extension, so that it will accept the else condition.
>(sort1 (list 3 1 2))
'(1 2 3)
> (sort1 (cons 2 1))
'(1 . 2)
> (sort1 'here)
"here"
I suggest you read "The Scheme Programming Language" by Kent Dybvig. There is an entire chapter on syntactic extensions.
Related
So for a college assignment we've been asked to work with macros and I'm finding it hard to understand how to implement code in scheme (we went from reversing a string to building an interpreter in one lecture).
(define macro-alist
`((and ,(λ (e)
(let ((forms (cdr e)))
(cond ((null? forms) '#t)
((null? (cdr forms)) (car forms))
(else `(if ,(car forms) (and ,#(cdr forms)) #f))))))
;(or ,error)
;(let ,error)
;(cond ,error)
(if ,(λ (e) (let ((guard (cadr e))
(then-part (caddr e))
(else-part (cadddr e)))
`((%if ,guard (λ () ,then-part) (λ () ,else-part))))))
))
We were asked to 'fill in the error holds in macro-alist' for the weekend and I'm finding it difficult.
I found some resources and combining them with my own brief knowledge I have :
`((or ,(lambda (e)
(and (list-strictly-longer-than? e 0)
(equal? (list-ref e 0) 'or)
(letrec ([visit (lambda (i)
(if(null? i)
#t
(and (is-exression? (car i))
(visit (cdr i)))))])
(visit (cdr e)))))))
`((let ,(lambda (e)
(and (proper-list-of-given-length? e 3)
(equal? (car e) 'let)
(list? (cadr e))
(is-expression? (list-ref e 2))
(lectrec ([visit (trace-lambda visit (i a)
(if(null? i)
#t
(and (proper-list-of-given-length? (car i) 2)
(is-identifier? (caar i))
(is-expression? (cadar i))
(not (member (caar i) a))
(visit (cdr i) (cons (caar i) a)))))])
(visit (cadr e) '()))))))
`((cond ,(lambda (e)
(and (list-strictly-longer-than? e 1)
(equal? (car v) 'cond)
(lectrec ([visit (lambda (i)
(if (null? (cdr i))
(is-else-clause? (car i))
(if (pair? (cdr i))
(and (cond? (car i))
(visit (cdr i))))))])
(visit (cdr e)))))))
For or, let and cond. I'm wondering if these are correct or if I'm close. I don't understand much about macros or scheme in general so some information/help on what to do would be appreciated.
If you look at the implementation of and:
(define expand-and
(λ (e)
(let ((forms (cdr e)))
(cond ((null? forms) '#t)
((null? (cdr forms)) (car forms))
(else `(if ,(car forms) (and ,#(cdr forms)) #f))))))
(expand-and '(and)) ; ==> #t
(expand-and '(and a)) ; ==> a
(expand-and '(and a b)) ; ==> (if a (and b) #f)
I notice two things. It doesn't really double check that the first element is and or if it's a list. Perhaps the interpreter doesn't use this unless it has checked this already?
Secondly it doesn't seem like you need to expand everything. As you see you might end up with some code + and with fewer arguments. No need for recursion since the evaluator will do that for you.
I think you are overthinking it. For or it should be very similar:
(expand-or '(or)) ; ==> #f
(expand-and '(or a b c)) ; ==> (let ((unique-var a)) (if unique-var unique-var (or b c)))
The let binding prevents double evaluation of a but if you have no side effects you might just rewrite it to (if a a (or b)). As with and or might expand to use or with fewer arguments than the original. This trick you can do with cond as well:
(cond (a b c)
...) ; ==>
(if a
(begin b c)
(cond ...))
let does not need this since it's perhaps the simplest one if you grasp map:
(let ((a x) (c y))
body ...) ; ==>
((lambda (a c) body ...) x y)
The report has examples of how the macros for these are made, but they might not be the simplest to rewrite to functions that takes code as structure like your interpeter. However using the report to understand the forms would perhaps worked just as well as posting a question here on SO.
I'm learning Common Lisp and I've been trying to write a function to reverse a list. This is the function but every time I try to run it I get "Segmentation Fault (Core Dumped)"
(defun reverseList(x)
(if (cdr x)
(cons (reverseList x) (car x))
(car x)))
(reverseList '(1 2 3))
I think this will work:
(define reverseList(lambda(x)(
if (null? (cdr x))
(cons (car x) '())
(append (reverseList (cdr x)) (cons (car x) '()))
)
)
)
I am learning Lisp and I had to write a function whose return value was a list containing the odd integers (if any) from the given input. In code I have this:
(defun f3 (a)
(cond
((null a) nil )
((and (numberp (car a)) (oddp (car a))) (cons (car a) (f3 (cdr a))))
(T (f3 (cdr a)))
) ; end cond
)
I originally wanted to use the append function, but I kept getting errors.
It was recommended to me to use cons function. When I did this my function started working (code is above). I originally had this:
(defun f3 (a)
(cond
((null a) ())
((and (numberp (car a)) (oddp (car a))) (append (f3 (cdr a)) (car a))))
(T (append () (f3 (cdr a))))
)
)
but kept getting errors. For example, if I called (f3 '(1 2 3)) it would say "error 3 is not type LIST". So, my questions are why does cons work here and why did append not work? How does cons work? Thanks in advance.
append wants list arguments, and (car a) is not a list. Instead of (car a) you'd need (list (car a)). In other words, (append (f3 (cdr a)) (list (car a))).
That will basically work, but you'll get the result in reverse order. So that should be (append (list (car a)) (f3 (cdr a))).
Also note that your (append () (f3 (cdr a))) is equivalent to just (f3 (cdr a)).
The resulting changes in your original would be:
(defun f3 (a)
(cond
((null a) ())
((and (numberp (car a)) (oddp (car a)))
(append (list (car a)) (f3 (cdr a)))))
(T (f3 (cdr a)))))
But, you wouldn't normally use append to prepend a single element to a list. It would more naturally be done using cons. So
(append (list (car a)) (f3 (cdr a)))
Is more appropriately done by:
(cons (car a) (f3 (cdr a)))
Which finally takes you right to the working version you showed.
While something like mbratch's answer will help you in learning about list manipulation (and so is probably a more useful answer for you at this point in your study), it's also important to learn about the standard library of the language that you're using. In this case, you're trying to filter out everything except odd numbers. Using remove-if-not, that's just:
(defun keep-odd-numbers (list)
(remove-if-not (lambda (x)
(and (numberp x) (oddp x)))
list))
CL-USER> (keep-odd-numbers '(1 a 2 b 3 c 4 d 5 e))
;=> (1 3 5)
While this isn't a fix to your actual problem, which #mbratch provided, here's the way I would implement something like this using the LOOP macro (another part of the standard library):
(defun keep-odd-numbers (list)
(loop for x in list collecting x when (and (numberp x) (oddp x))))
i worte this function to remove numbers from a list x
(defun rm-nums (x)
(cond
((null x) nil)
(t (mapcar 'numberp x))))
however when i enter (rm-nums '(32 A T 4 3 E))
returns (T NIL NIL T T NIL)
i want it instead of returning T or Nil, i want it to return the values that caused NIL only [which are not numbers]
so this example should return (A T E)
i am supposed to use mapcar WITHOUT recursion or iteration or the bultin function "remove-if"
i think it is related to something called apply-append but i know nothing about it. any help?
I think your course had this in mind:
(defun my-remove-if (pred lst)
(apply #'append (mapcar (lambda (x)
(and (not (funcall pred x))
(list x)))
lst)))
It does use apply and append and mapcar, like you said. Example usage:
(my-remove-if #'numberp '(32 a t 4 3 e))
=> (a t e)
More idiomatic solution suggested by Rörd:
(defun my-remove-if (pred lst)
(mapcan (lambda (x)
(and (not (funcall pred x))
(list x)))
lst))
Need to write a union function in lisp that takes two lists as arguments and returns a list that is the union of the two with no repeats. Order should be consistent with those of the input lists
For example: if inputs are '(a b c) and '(e c d) the result should be '(a b c e d)
Here is what I have so far
(defun stable-union (x y)
(cond
((null x) y)
((null y) x))
(do ((i y (cdr i))
(lst3 x (append lst3
(cond
((listp i)
((null (member (car i) lst3)) (cons (car i) nil) nil))
(t (null (member i lst3)) (cons i nil) nil)))))
((null (cdr i)) lst3)))
My error is that there is an "illegal function object" with the segment (null (member (car i) lst3))
Advice?
You've got your parens all jumbled-up:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x) ) END OF COND form - has no effect
(do ((i y (cdr i))
^^
(lst3 x (append lst3
(cond
((listp i)
( (null (member (car i) lst3))
^^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ called as a function
(cons (car i) nil) with two arguments
nil ) )
^^
(t NEXT 3 forms have no effect
(null (member i lst3))
(cons i nil)
nil )))) )
^^
((null (cdr i)) lst3)))
Here's your code as you probably intended it to be, with corrected parenthesization and some ifs added where needed:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x)
(t
(do ((i y (cdr i))
(lst3 x (append lst3
(cond
((listp i)
(if (null (member (car i) lst3))
(cons (car i) nil)
nil))
(t
(if (null (member i lst3))
(cons i nil)
nil))))))
((null (cdr i)) lst3)))))
There are still problems with this code. Your do logic is wrong, it skips the first element in y if it contains just one element. And you call append all the time whether it is needed or not. Note that calling (append lst3 nil) makes a copy of top-level cons cells in lst3, entirely superfluously.
Such long statements as you have there are usually placed in do body, not inside the update form for do's local variable.
But you can use more specialized forms of do, where appropriate. Here it is natural to use dolist. Following "wvxvw"'s lead on using hash-tables for membership testing, we write:
(defun stable-union (a b &aux (z (list nil)))
(let ((h (make-hash-table))
(p z))
(dolist (i a)
(unless (gethash i h)
(setf (cdr p) (list i) p (cdr p))
(setf (gethash i h) t)))
(dolist (i b (cdr z))
(unless (gethash i h)
(setf (cdr p) (list i) p (cdr p))
(setf (gethash i h) t)))))
using a technique which I call "head-sentinel" (z variable pre-initialized to a singleton list) allows for a great simplification of the code for the top-down list building at a cost of allocating one extra cons cell.
The error is because you're trying to execute the result of evaluating (null (member (car i) lst3)). In your cond expression, if i is a list, then it attempts to evaluate the expression
((null (member (car i) lst3)) (cons (car i) nil) nil))
And return the result. The first element in an expression should be a function, but
(null (member (car i) lst3))
Is going to return a boolean value. Hence the failure. The structure of your code needs some attention. What you've missed is that you need an inner cond, there.
Incidentally, this would be a much cleaner function if you did it recursively.
I'm a Schemer rather than a Lisper, but I had a little think about it. Here's the skeleton of a recursive implementation:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x)
((listp y)
(cond
((member (car y) x) (stable-union ??? (???)))
(t (stable-union (append x (??? (???))) (cdr y)))))
((not (member y x)) (append x (list y)))
(t x)))
(Edited to correct simple tyop in second-last line, thanks to Will Ness for spotting it)
(remove-duplicates (append '(a b c) '(e c d)) :from-end t)
Because you started off with do, and because a recursive solution would be even worse, here's what you could've done:
(defun union-stable (list-a list-b)
(do ((i list-b (cdr i))
filtered back-ref)
((null i) (append list-a back-ref))
(unless (member (car i) list-a)
(if back-ref
(setf (cdr filtered) (list (car i))
filtered (cdr filtered))
(setf back-ref (list (car i))
filtered back-ref)))))
This is still quadratic time, and the behaviour is such that if the first list has duplicates, or the second list has duplicates, which are not in the first list - they will stay. I'm not sure how fair it is to call this function a "union", but you'd have to define what to do with the lists if they have duplicates before you try to unify them.
And this is what you might've done if you were interested in the result, rather than just exercising. Note that it will ensure that elements are unique, even if the elements repeat in the input lists.
(defun union-stable-hash (list-a list-b)
(loop for c = (car (if list-a list-a list-b))
with back-ref
with hash = (make-hash-table)
for key = (gethash c hash)
with result
do (unless key
(if back-ref
(setf (cdr result) (list c)
result (cdr result))
(when (or list-a list-b)
(setf back-ref (list c)
result back-ref)))
(setf (gethash c hash) t))
do (if list-a (setf list-a (cdr list-a))
(setf list-b (cdr list-b)))
do (unless (or list-a list-b)
(return back-ref))))