I have an expression: (map some-proc some-list)
which evaluates to, say, '(#f #f #f).
I want to check whether all booleans in this list are true. However,
(and '(#f #f #f)) returns '(#f #f #f), while I heed #f.
Now, if I (apply and '(#f #f #f)), I get the error:
and: bad syntax in: and in the DrRacket environment. That is confusing because the Racket Reference gives the example of (apply + '(1 2 3)) which seems to be identical to my problem.
What am I doing wrong and how I get my #f out of '(#f #f #f)?
you can use andmap for this:
> (andmap (lambda (x) x) '(#f #f #f))
#f
The problem stems from the fact that and is not a procedure, but a macro, in order to avoid the evaluation of all of its arguments.
It would work in Lazy Racket, though, where and is a procedure.
Just another way of doing it:
(foldr (lambda(x y) (and x y)) #t '(#f #f #f))
or I will rewrite uselpa's solution (andmap):
(andmap identity '(#f #f #f))
Related
I have code like this:
(define-syntax macron
(syntax-rules ()
((_ name)
(lambda (x)
(eval (cons 'name x) (interaction-environment))))))
(define x (map (macron lambda)
'(((x) (display x)) ((a b) (+ a b)))))
(let ((square (car x))
(sum (cadr x)))
(display (square 10))
(newline)
(display (sum 1 2 3))
(newline))
the code is working it use macro as value by wrapping it with lambda. My question is how can I put inside syntax-rule macro literal symbol 'name instead of (cons 'lambda ...) so the output code is:
(lambda (x)
(eval (cons 'name x) (interaction-environment)))
so it work with code like this:
(define (name x)
(display x)
(newline))
(for-each (macron lambda) ;; lambda can be anything
'((1) (2) (3)))
and it print all the numbers.
I know that I can change the name in pattern into something else, but I want to know more about syntax-rules and it's edge cases. So is it possible to have name if I use it as input pattern?
I'm looking for answers with R7RS, that have more of this type of edge cases covered.
All macros happens in compile time so runtime stuff might not exist. That means that you should think of it as syntax sugar and use it as susch. eg.
(for-each (macron something) '((1) (2) (3)))
Should then have an expansion based on that. Your current expansion is that it turns into this:
(for-each (lambda (x)
(eval (cons 'someting x) (interaction-environment))
'((1) (2) (3)))
For something being a macro this will apply the macro in runtime. It is bad. It also removes the need for the macro in the first place. You could do this instead:
(define (macron-proc name)
(lambda (x)
(eval (cons name x) (interaction-environment))))
(for-each (macron-proc 'something) '((1) (2) (3)))
I made a programming language that had passable macros:
(define xor (flambda (a b) `(if ,a (not ,b) ,b)))
(define (fold comb init lst)
(if (null? lst)
init
(fold comb (comb (car lst) init) (cdr lst))))
(fold xor #f '(#t #t)) ; ==> #f
It's not a very good approach if you are targeting an efficient compiled end product. The first macros were indeed like this and they removed it in LISP 1.5 before Common Lisp. Scheme avoided macros for many years and opted for syntax-rules in R4RS as an optional feature. R6RS is the only version that has full power macros.
With a procedure instead of macros this is actually the same as the following code with the bad eval removed:
(for-each (lambda (x)
(apply something x))
'((1) (2) (3)))
Which means you can implement macron much easier:
(define-syntax macron
(syntax-rules ()
((_ name)
(lambda (x)
(apply name x)))))
But from looking at this now you don't need a macro at all. This is partial application.
(define (partial proc arg)
(lambda (lst)
(apply proc arh lst)))
(map (partial + 3) '((1 2) (3 4) (4 5)))
; ==> (6 10 12)
There is actually a SRFI-26 called cut/cute which allows us to do something similar where it wraps it in a lambda:
(map (cut apply + 3 <>) '((1 2) (3 4) (4 5)))
The syntax-rules are the macros with the least power. You cannot do anything unhygienic and you cannot make new identifiers based on other ones. Eg. it' impossible to implement a racket style struct where you can do (struct complex [real imag]) and have the macro create complex?, complex-real, and complex-imag as procedures. You need to do as SRFI-57 does and require th euser to specify all the names such that you don't need to concatenate to new identifiers.
Right now R7RS-small only has syntax-rules. I think it was a mistake not to have a more powerful macro as an alternative since now the R7RS-large cannot be implemented with R7RS-small.
I'm trying to implement the Towers of Hanoi.I'm not printing out anything between my recursive calls yet, but I keep getting an error saying
'('(LIST) 'NIL 'NIL) should be a lambda expression
I've read that the reason this happens is because of a problem with the parenthesis, however I cannot seem to find what my problem is. I think it's happening in the pass-list function when I am trying to call the hanoi function. My code:
(defun pass-list(list)
(hanoi('('(list)'()'())))
)
(defun hanoi ('('(1) '(2) '(3)))
(hanoi '('(cdr 1) '(cons(car 1) 2) '(3)))
(hanoi '('(cons(car 3)1) '(2)'(cdr 3)))
)
This code has many syntax problems; there are erroneous quotes all over the place, and it looks like you're trying to use numbers as variables, which will not work. The source of the particular error message that you mentioned comes from
(hanoi('('(list)'()'())))
First, understand that the quotes in 'x and '(a b c) are shorthand for the forms (quote x) and (quote (a b c)), and that (quote anything) is the syntax for getting anything, without anything being evaluated. So '(1 2 3) gives you the list (1 2 3), and '1 gives you 1. quote is just a symbol though, and can be present in other lists, so '('(list)'()'()) is the same as (quote ((quote (list)) (quote ()) (quote ()))) which evaluates to the list ((quote (list)) (quote ()) (quote ())). Since () can also be written nil (or NIL), this last is the same as ('(list) 'NIL 'NIL). In Common Lisp, function calls look like
(function arg1 arg2 ...)
where each argi is a form, and function is either a symbol (e.g., list, hanoi, car) or a list, in which case it must be a lambda expression, e.g., (lambda (x) (+ x x)). So, in your line
(hanoi('('(list)'()'())))
we have a function call. function is hanoi, and arg1 is ('('(list)'()'())). But how will this arg1 be evaluated? Well, it's a list, which means it's a function application. What's the function part? It's
'('(list)'()'())
which is the same as
'('(list 'NIL 'NIL))
But as I just said, the only kind of list that can be function is a lambda expression. This clearly isn't a lambda expression, so you get the error that you're seeing.
I can't be sure, but it looks like you were aiming for something like the following. The line marked with ** is sort of problematic, because you're calling hanoi with some arguments, and when it returns (if it ever returns; it seems to me like you'd recurse forever in this case), you don't do anything with the result. It's ignored, and then you go onto the third line.
(defun pass-list(list)
(hanoi (list list) '() '()))
(defun hanoi (a b c)
(hanoi (rest a) (cons (first a) b) c) ; **
(hanoi (cons (first c) a) b (rest c)))
If hanoi is supposed to take a single list as an argument, and that list is supposed to contain three lists (I'm not sure why you'd do it that way instead of having hanoi take just three arguments, but that's a different question, I suppose), it's easy enough to modify; just take an argument abc and extract the first, second, and third lists from it, and pass a single list to hanoi on the recursive call:
(defun hanoi (abc)
(let ((a (first abc))
(b (second abc))
(c (third abc)))
(hanoi (list (rest a) (cons (first a) b) c))
(hanoi (list (cons (first c) a) b (rest c)))))
I'd actually probably use destructuring-bind here to simplify getting a, b, and c out of abc:
(defun hanoi (abc)
(destructuring-bind (a b c) abc
(hanoi (list (rest a) (cons (first a) b) c))
(hanoi (list (cons (first c) a) b (rest c)))))
I'm currently working on a LISP exercise for a small project and need severe help. This may be more or less of a beginner's question but I'm absolutely lost on writing a certain function that takes in two unevaluated functions and spits out the result dependent on if the variables were given an assignment or not.
An example would be
(setq p1 '(+ x (* x (- y (/ z 2)))))
Where
(evalexp p1 '( (x 2) (z 8) ))
returns (+ 2 (* 2 (- y 4)))
My goal is to write the evalexp function but I can't even think of where to start.
So far I have
(defun evalexp (e b) )
.. not very much. If anyone could please help or lead me in a good direction I'd be more than appreciative.
Here's a full solution. It's pretty straightforward, so I'll leave out a full explanation. Ask me in the comments if there's anything you can't figure out yourself.
(Using eval to do the actual evaluation might not be what you want in your exercise/project. Look up "meta-circular interpreter" for another way.)
(defun apply-env (exp env)
(reduce (lambda (exp bdg) (subst (cadr bdg) (car bdg) exp))
env :initial-value exp))
(defun try-eval (exp)
(if (atom exp)
exp
(let ((exp (mapcar #'try-eval exp)))
(if (every #'numberp (cdr exp))
(eval exp)
exp))))
(defun evalexp (exp env)
(try-eval (apply-env exp env)))
Here's a hint, this is how you might do it (in pseudocode):
function replace(vars, list):
for each element of list:
if it's an atom:
if there's an association in vars:
replace atom with value in vars
else:
leave atom alone
else:
recursively apply replace to the sublist
There will certainly be some details to work out as you convert this to Lisp code.
What is the exclusive or functions in scheme? I've tried xor and ^, but both give me an unbound local variable error.
Googling found nothing.
I suggest you use (not (equal? foo bar)) if not equals works. Please note that there may be faster comparators for your situiation such as eq?
As far as I can tell from the R6RS (the latest definition of scheme), there is no pre-defined exclusive-or operation. However, xor is equivalent to not equals for boolean values so it's really quite easy to define on your own if there isn't a builtin function for it.
Assuming the arguments are restricted to the scheme booleans values #f and #t,
(define (xor a b)
(not (boolean=? a b)))
will do the job.
If you mean bitwise xor of two integers, then each Scheme has it's own name (if any) since it's not in any standard. For example, PLT has these bitwise functions, including bitwise-xor.
(Uh, if you talk about booleans, then yes, not & or are it...)
Kind of a different style of answer:
(define xor
(lambda (a b)
(cond
(a (not b))
(else b))))
Reading SRFI-1 shed a new light upon my answer. Forget efficiency and simplicity concerns or even testing! This beauty does it all:
(define (xor . args)
(odd? (count (lambda (x) (eqv? x #t)) args)))
Or if you prefer:
(define (true? x) (eqv? x #t))
(define (xor . args) (odd? (count true? args)))
(define (xor a b)
(and
(not (and a b))
(or a b)))
Since xor could be used with any number of arguments, the only requirement is that the number of true occurences be odd. It could be defined roughly this way:
(define (true? x) (eqv? x #t))
(define (xor . args)
(odd? (length (filter true? args))))
No argument checking needs to be done since any number of arguments (including none) will return the right answer.
However, this simple implementation has efficiency problems: both length and filter traverse the list twice; so I thought I could remove both and also the other useless predicate procedure "true?".
The value odd? receives is the value of the accumulator (aka acc) when args has no remaining true-evaluating members. If true-evaluating members exist, repeat with acc+1 and the rest of the args starting at the next true value or evaluate to false, which will cause acc to be returned with the last count.
(define (xor . args)
(odd? (let count ([args (memv #t args)]
[acc 0])
(if args
(count (memv #t (cdr args))
(+ acc 1))
acc))))
> (define (xor a b)(not (equal? (and a #t)(and b #t))))
> (xor 'hello 'world)
$9 = #f
> (xor #f #f)
$10 = #f
> (xor (> 1 100)(< 1 100))
$11 = #t
I revised my code recently because I needed 'xor in scheme and found out it wasn't good enough...
First, my earlier definition of 'true? made the assumption that arguments had been tested under a boolean operation. So I change:
(define (true? x) (eqv? #t))
... for:
(define (true? x) (not (eqv? x #f)))
... which is more like the "true" definition of 'true? However, since 'xor returns #t if its arguments have an 'odd? number of "true" arguments, testing for an even number of false cases is equivalent. So here's my revised 'xor:
(define (xor . args)
(even? (count (lambda (x) (eqv? x #f)) args)))
I have a list of elements '(a b c) and I want to find if (true or false) x is in it, where x can be 'a or 'd, for instance. Is there a built in function for this?
If you need to compare using one of the build in equivalence operators, you can use memq, memv, or member, depending on whether you want to look for equality using eq?, eqv?, or equal?, respectively.
> (memq 'a '(a b c))
'(a b c)
> (memq 'b '(a b c))
'(b c)
> (memq 'x '(a b c))
#f
As you can see, these functions return the sublist starting at the first matching element if they find an element. This is because if you are searching a list that may contain booleans, you need to be able to distinguish the case of finding a #f from the case of not finding the element you are looking for. A list is a true value (the only false value in Scheme is #f) so you can use the result of memq, memv, or member in any context expecting a boolean, such as an if, cond, and, or or expression.
> (if (memq 'a '(a b c))
"It's there! :)"
"It's not... :(")
"It's there! :)"
What is the difference between the three different functions? It's based on which equivalence function they use for comparison. eq? (and thus memq) tests if two objects are the same underlying object; it is basically equivalent to a pointer comparison (or direct value comparison in the case of integers). Thus, two strings or lists that look the same may not be eq?, because they are stored in different locations in memory. equal? (and thus member?) performs a deep comparison on lists and strings, and so basically any two items that print the same will be equal?. eqv? is like eq? for almost anything but numbers; for numbers, two numbers that are numerically equivalent will always be eqv?, but they may not be eq? (this is because of bignums and rational numbers, which may be stored in ways such that they won't be eq?)
> (eq? 'a 'a)
#t
> (eq? 'a 'b)
#f
> (eq? (list 'a 'b 'c) (list 'a 'b 'c))
#f
> (equal? (list 'a 'b 'c) (list 'a 'b 'c))
#t
> (eqv? (+ 1/2 1/3) (+ 1/2 1/3))
#t
(Note that some behavior of the functions is undefined by the specification, and thus may differ from implementation to implementation; I have included examples that should work in any R5RS compatible Scheme that implements exact rational numbers)
If you need to search for an item in a list using an equivalence predicate different than one of the built in ones, then you may want find or find-tail from SRFI-1:
> (find-tail? (lambda (x) (> x 3)) '(1 2 3 4 5 6))
'(4 5 6)
Here's one way:
> (cond ((member 'a '(a b c)) '#t) (else '#f))
#t
> (cond ((member 'd '(a b c)) '#t) (else '#f))
#f
member returns everything starting from where the element is, or #f. A cond is used to convert this to true or false.
You are looking for "find"
Basics - The simplest case is just (find Entry List), usually used as a predicate: "is Entry in List?". If it succeeds in finding the element in question, it returns the first matching element instead of just "t". (Taken from second link.)
http://www.cs.cmu.edu/Groups/AI/html/cltl/clm/node145.html
-or-
http://www.apl.jhu.edu/~hall/Lisp-Notes/Higher-Order.html
I don't know if there is a built in function, but you can create one:
(define (occurrence x lst)
(if (null? lst) 0
(if (equal? x (car lst)) (+ 1 (occurrence x (cdr lst)))
(occurrence x (cdr lst))
)
)
)
Ỳou will get in return the number of occurrences of x in the list. you can extend it with true or false too.
(define (member? x list)
(cond ((null? list) #f)
((equal? x (car list)) #t)
(else (member? x (cdr list)))))
The procedure return #t (true) or #f (false)
(member? 10 '(4 2 3))
output is #f