I'm attempting to write a scheme function that prints each entry of a list on a new line. This function and sample input works as expected, then gives an "illegal function" error and exits. I'm using tinyscheme from the debian repos.
(define print-list
(lambda (l)
(if (null? l)
(display "done\n")
(
(display (car l))
(newline)
(print-list (cdr l))
)
)
)
)
(print-list '(1 2 3 4 5) )
A parenthesis followed by an expression means function application.
(expr1 expr2 ...)
Means evaluate expr1, expr2, .... Then apply the result of expr1 to the result of expr2 ... . If the result of expr1 is not a function, then you'll see the "illegal function" error.
The fix is as leppie states to add a begin:
(define print-list
(lambda (l)
(if (null? l)
(display "done\n")
(begin
(display (car l))
(newline)
(print-list (cdr l))))))
(print-list '(1 2 3 4 5) )
Here
(begin expr1 expr2 ....)
means evalute expr1, expr2, ... in order. Finally return the value of the last expression.
Block Structure
In scheme, the branches of the if special form do not have an implicit block structure. As Soegaard suggests, begin may be used. Or you can use cond to create a block—in which case, expanding to more than two branches comes along for free.
(define (print-list a-list)
(cond ((null? a-list)(display "done\n"))
(else
(display (car a-list))
(newline)
(print-list (cdr a-list)))))
Problem with Return Type
Note that print-list is problematic because it returns...well something undefined. For example in MIT Scheme:
1 ]=> (print-list '(a b c))
a
b
c
done
;Unspecified return value
Or in Gauche Scheme:
gosh> (print-list '(a b c))
a
b
c
done
#<undef>
While in Racket's implementation of R5RS the return value is not printed.
> (print-list '(a b c))
a
b
c
done
Returning a Value
One common approach to signalling the end of a series of side effects is to return a value. null is one possibility. A symbol is another.
(define (print-list a-list)
(cond ((null? a-list) 'done) ; changed this line.
(else
(display (car a-list))
(newline)
(print-list (cdr a-list)))))
Calling it now will in a Scheme REPL removes the problematic return value:
gosh> (print-list '(a b c))
a
b
c
done
Related
Here is my code about postfix in scheme:
(define (stackupdate e s)
(if (number? e)
(cons e s)
(cons (eval '(e (car s) (cadr s))) (cddr s))))
(define (postfixhelper lst s)
(if (null? lst)
(car s)
(postfixhelper (cdr lst) (stackupdate (car lst) s))))
(define (postfix list)
(postfixhelper list '()))
(postfix '(1 2 +))
But when I tried to run it, the compiler said it takes wrong. I tried to check it, but still can't find why it is wrong. Does anyone can help me? Thanks so much!
And this is what the compiler said:
e: unbound identifier;
also, no #%app syntax transformer is bound in: e
eval never has any information about variables that some how are defined in the same scope as it is used. Thus e and s does not exist. Usually eval is the wrong solution, but if you are to use eval try doing it as as little as you can:
;; Use eval to get the global procedure
;; from the host scheme
(define (symbol->proc sym)
(eval sym))
Now instead of (eval '(e (car s) (cadr s))) you do ((symbol->proc e) (car s) (cadr s)). Now you should try (postfix '(1 2 pair?))
I've made many interpreters and none of them used eval. Here is what I would have done most of the time:
;; Usually you know what operators are supported
;; so you can map their symbol with a procedure
(define (symbol->proc sym)
(case sym
[(+) +]
[(hyp) (lambda (k1 k2) (sqrt (+ (* k1 k1) (* k2 k2))))]
[else (error "No such operation" sym)]))
This fixes the (postfix '(1 2 pair?)) problem. A thing that I see in your code is that you always assume two arguments. But how would you do a double? eg something that just doubles the one argument. In this case symbol->proc could return more information:
(define (symbol->op sym)
(case sym
[(+) (cons + 2)]
[(double) (cons (lambda (v) (* v v)) 1)]
[else (error "No such operation" sym)]))
(define op-proc car)
(define op-arity cdr)
And in your code you could do this if it's not a number:
(let* ([op (symbol->op e)]
[proc (op-proc op)]
[arity (op-arity op)])
(cons (apply proc (take s arity)
(drop s arity)))
take and drop are not R5RS, but they are simple to create.
I am trying to write by myself the cons function in scheme. I have written this code:
(define (car. z)
(z (lambda (p q) p)))
and I am trying to run :
(car. '(1 2 3))
I expect to get the number 1, but it does not work properly.
When you implement language data structures you need to supply constructors and accessors that conform to the contract:
(car (cons 1 2)) ; ==> 1
(cdr (cons 1 2)) ; ==> 2
(pair? (cons 1 2)) ; ==> 2
Here is an example:
(define (cons a d)
(vector a d))
(define (car p)
(vector-ref p 0))
(define (cdr p)
(vector-ref p 1))
Now if you make an implementation you would implement read to be conformant to this way of doing pairs so that '(1 2 3) would create the correct data structure the simple rules above is still the same.
From looking at car I imagine cons looks like this:
(define (cons a d)
(lambda (p) (p a d)))
It works with closures. Now A stack machine implementation of Scheme would analyze the code for free variables living passed their scope and thus create them as boxes. Closures containing a, and d aren't much different than vectors.
I urge you to implement a minimalistic Scheme interpreter. First in Scheme since you can use the host language, then a different than a lisp language. You can even do it in an esoteric language, but it is very time consuming.
Sylwester's answer is great. Here's another possible implementation of null, null?, cons, car, cdr -
(define null 'null)
(define (null? xs)
(eq? null xs))
(define (cons a b)
(define (dispatch message)
(match message
('car a)
('cdr b)
(_ (error 'cons "unsupported message" message))
dispatch)
(define (car xs)
(if (null? xs)
(error 'car "cannot call car on an empty pair")
(xs 'car)))
(define (cdr xs)
(if (null? xs)
(error 'cdr "cannot call cdr on an empty pair")
(xs 'cdr)))
It works like this -
(define xs (cons 'a (cons 'b (cons 'c null))))
(printf "~a -> ~a -> ~a\n"
(car xs)
(car (cdr xs))
(car (cdr (cdr xs))))
;; a -> b -> c
It raises errors in these scenarios -
(cdr null)
; car: cannot call car on an empty pair
(cdr null)
; cdr: cannot call cdr on an empty pair
((cons 'a 'b) 'foo)
;; cons: unsupported dispatch: foo
define/match adds a little sugar, if you like sweet things -
(define (cons a b)
(define/match (dispatch msg)
(('car) a)
(('cdr) b)
(('pair?) #t)
((_) (error 'cons "unsupported dispatch: ~a" msg)))
dispatch)
((cons 1 2) 'car) ;; 1
((cons 1 2) 'cdr) ;; 2
((cons 1 2) 'pair?) ;; #t
((cons 1 2) 'foo) ;; cons: unsupported dispatch: foo
I have found a recursive problem in one page that says the following:
If a person enter a string with two consecutive letters that are the same, it should put a 5 between them. For example if I enter "hello"
it should print "hel5lo"
I have done the following program in Scheme:
(define (function listT)
(if (empty? listT)
'()
(begin
(if (eq? (car listT) (car (cdr listT)))
(display 5)
(display (car listT))
)))
(function (cdr listT)))
and tested with:
(function'( 'h 'e 'l 'l 'o))
and the problem I got is
car: contract violation
expected: pair?
given: ()
I suppose that is because at one moment (car (cdr listT)) will face an empty list, have tried with a conditional before, but still with some issues.
Is it possible to do it only using recursion over the list of characters with cdr and car? I mean not with new variables, strings, using reverse or loops?
Any help?
Thanks
This happens when there is only one character left in the list; (cdr listT) will be the empty list '() and the car of the empty list is undefined.
So you either need to check that the cdr isn't empty, for example:
(define (f str)
(let loop ((lst (string->list str)) (res '()))
(if (null? lst)
(list->string (reverse res))
(let ((c (car lst)))
(loop (cdr lst)
(cons c
(if (and (not (null? res)) (char=? c (car res)))
(cons #\5 res)
res)))))))
or, instead of looking one character ahead, turn around your logic and keep track of the last character, which is initialised to some value that will be different in every case (not as elegant as the first solution though IMO):
(define (f str)
(list->string
(let loop ((prev #f) (lst (string->list str)))
(if (null? lst)
'()
(let ((c (car lst)))
(if (equal? c prev)
(cons #\5 (cons c (loop c (cdr lst))))
(cons c (loop c (cdr lst)))))))))
[EDIT alternatively, with an explicit inner procedure:
(define (f str)
(define (inner prev lst)
(if (null? lst)
'()
(let ((c (car lst)))
(if (equal? c prev)
(cons #\5 (cons c (inner c (cdr lst))))
(cons c (inner c (cdr lst)))))))
(list->string (inner #f (string->list str))))
]
Testing:
> (f "hello")
"hel5lo"
> (f "helo")
"helo"
> (f "heloo")
"helo5o"
Side note: don't double quote:
> '('h 'e 'l 'l 'o)
'('h 'e 'l 'l 'o)
> (car '('h 'e 'l 'l 'o))
''h
This is probably not what you expected. Instead:
> '(h e l l o)
'(h e l l o)
> (car '(h e l l o))
'h
or
> (list 'h 'e 'l 'l 'o)
'(h e l l o)
> (car (list 'h 'e 'l 'l 'o))
'h
Also note that these are symbols, whereas, since you start from a string, you want characters:
> (string->list "hello")
'(#\h #\e #\l #\l #\o)
EDIT 2
I see you are still struggling with my answer. Here's a solution that should be as minimal as you requested, I hope this is it:
(define (f lst (prev #f))
(unless (null? lst)
(when (equal? (car lst) prev) (display "5"))
(display (car lst))
(f (cdr lst) (car lst))))
or even
(define (f lst)
(unless (null? lst)
(display (car lst))
(when (and (not (null? (cdr lst))) (equal? (car lst) (cadr lst)))
(display "5"))
(f (cdr lst))))
Testing:
> (f '(h e l l o))
hel5lo
> (f '(h e l o))
helo
> (f '(h e l o o))
helo5o
I have found a solution:
(define (func lisT)
(if (empty? (cdr lisT))
(display (car lisT))
(begin
(if (eq? (car lisT) (car (cdr lisT)))
(begin
(display (car lisT))
(display 5)
)
(display (car lisT))
)
(func (cdr lisT))
)
))
Here's a solution including just one, top-level recursive function:
(define (insert list item)
(if (< (length list) 2) ;; not enough elements to compare?
list ;; then just return the input
(let ((first (car list)) ;; capture the first element,
(second (cadr list)) ;; the second element,
(rest (insert (cdr list) item))) ;; and the recursively processed tail
(cons first ;; construct a list with the first element
(if (eq? first second) ;; compare the first two and return either
(cons item rest) ;; the item before the rest
rest))))) ;; or just the rest
It takes as input a list and an item to be inserted between each two consecutive identical elements. It does not display anything, but rather returns another list with the result of the insertion. For example,
(insert '(1 2 2 3 3 3 2 2 1) 0)
results in
(1 2 0 2 3 0 3 0 3 2 0 2 1)
This hopefully solves your problem and seeds further experimentation.
Here is a straightforward function from a list to a list:
(define (add5s s)
(cond ((null? s) s)
((null? (cdr s)) s)
((equal? (car s) (cadr s)) (cons (car s) (cons 5 (add5s (cdr s)))))
(else (cons (car s) (add5s (cdr s))))
)
)
A list either:
is null
has one element
begins with two equal elements
begins with two unequal elements
A list with a 5 put between consecutive equal elements is respectively:
the list
the list
the first element followed by a 5 followed by the rest of it with a 5 put between consecutive equal elements
the first element followed by the rest of it with a 5 put between consecutive equal elements
A Scheme string is not a list of characters or a list of symbols. If you want to input and output strings then you should use the corresponding string operators. Or write a function that defines this one, calls it with string->list of an input string and outputs list->string of this one's result list. Or a function like this one but that branches on string->list of its input string and outputs list->string of what this one returns.
(It is really not clear what code is to be written. You say "enters a string", but your "tested" code is a function that takes a list as argument, rather than reading from a port. And you say "put a 5" but you print argument list elements or a 5 via display to a port, rather than returning a value of the type of the argument. And you give an example passing an argument that is a list of quoted symbols rather than just symbols let alone characters. (If you want to pass a list of symbols then use '(h e l l o) or (list 'h 'e 'l 'l 'o).) Say exactly what is to be produced, eg, a function with what arguments, return value and effect on ports.)
This question already has answers here:
"application: not a procedure" in binary arithmetic procedures
(1 answer)
Getting every nth atom using scheme does not pick up the last atom
(3 answers)
Closed 8 years ago.
Hi I'm writing something that grabs every third element after the first element. However I can't test the logic due to ";The object (a b c d e f g) is not applicable." The code is below, atom? checks if it's a list, listful is defined as an empty list.
(DEFINE (threes var)
(if(atom? var)
((newline) (DISPLAY "Bad input") ))
(APPEND(listful (CAR var)))
(if(> 3 (length var))
(threes (cdddr(listful)))
(listful))
)
Can anyone give me some tips? Here is how I'm calling the method in the Scheme environment.
>(threes (list1))
>(threes '(A B C D E F))
An if can only have one expression as the consequent, and one as the alternative. If you need more than one you have to use a begin to enclose the sequence of expressions - surrounding the expressions with () won't work, and that's causing the error reported, because Scheme interprets () as function application. This would be the correct syntax:
(define (threes var)
(if (atom? var)
(begin
(newline)
(display "Bad input")))
(append (listful (car var)))
(if (> 3 (length var))
(begin
(threes (cdddr (listful)))
(listful))))
… However, that's unlikely to work. What you want to do has been asked a couple of times in the last days, in particular here is my own previous answer:
(define (threes lst)
(cond ((or (null? lst) (null? (cdr lst))) lst)
((null? (cdr (cdr lst))) (list (car lst)))
(else (cons (car lst)
(threes (cdr (cdr (cdr lst))))))))
For example:
(threes '(a b c d e f g h i j k l m n o p))
=> '(a d g j m p)
See the original question for other ways to solve the problem, with detailed explanations.
I wrote this program:
(define find-combination
{lambda (a b)
(if (eq? ((quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b)))))
(display "1*v1" + ((quotient (car a) (car b))*"v2"))
(display "0*v1" + "0*v2"))})
(find-combination (list 2 2) (list 2 1))
a and b are two lists. Its give me the next problem: procedure application: expected procedure, given: 1; arguments were: 2.
I didn't get what is the problem. Someone can help me? Thank u.
First of all you have too much brackets after eq? - what you wrote means evaluating (quotient (car a) (car b)) and treating it as a function with argument (quotient (car (cdr a)) (car (cdr b))). The error means that first thing was evaluated to 1 and your interpreter expected it to be a procedure, not an integer. This line should be:
(if (eq? (quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b))))
or even:
(if (eq? (quotient (car a) (car b)) (quotient (cadr a) (cadr b)))
Apart from that, lines with display calls are wrong - Scheme doesn't have an infix notation, so + and * are out of place.
First of all you have a set of curly braces in your code(the one before lambda)
Also you have another set of paranthesis around the parameters you passed to eq?
It should be something like this:
(eq? (quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b))))
In Scheme and Racket, parentheses change the meaning of things.
1
is a number, but
(1)
is a call to 1 as a function... but 1 is a number, not a function, so this will cause the error you describe.
Your use of curly braces is also a little unsettling to me.