I am using Racket contract system, and I want to export a function of no arguments, which returns a lambda expression with no arguments, e. g.:
#lang racket
(define (foo)
(do-somthing)
(lambda ()
(do-other things)))
Does anyone know how to write contract for this kind of function?
I suspect it would look something along the lines of:
#lang racket/load
(module m racket
(provide/contract [foo (-> (-> any/c))])
(define (foo)
(+ 10 3) ; do something
(lambda ()
(+ 40 2) ; do other things
)))
(module n racket
(require 'm)
((foo)))
(require 'n)
(-> (-> any/c)) is a contract that matches functions that returns another function, which, when evaluated, returns a single integer value.
But if you'd like to relax return values of foo, you'd use just any instead of any/c, which allows any number of return values, not just a single value. Consider:
(module m racket
(provide/contract [foo (-> (-> any))])
(define (foo)
(+ 10 3) ; do something
(lambda ()
(values (+ 40 2) 666); do other things
)))
See Contracts on Higher-order Functions in Racket documentation.
Related
I'm having trouble figuring out how to go about creating a function that can take a series of the same function as arguments with the last argument as an operand. For example:
(func sqrt sqrt sqrt 390625)
The call above should return 5 as (sqrt 390625) > (sqrt 625) > (sqrt 25) > 5
I'm having trouble figuring out the exact way I should write this as any way I have tried has given me errors or achieved an infinite loop.
This the code is have so far:
(define func
(lambda L
(cond ( (equal? (length L) 2) ((car L) (cadr L)) ) ;; If the list consists of only 2 elements, carry out the function (element 1) onto the operand (element 2)
( #t (apply (car L) (func (cdr L))) ) ;; otherwise, apply the function (1st element) onto the rest of the list
)
)
)
The first condition works, for example returning 5 if i call (func sqrt 25), however the recursive call is throwing errors.
I would appreciate any help with this.
The OP doesn't provide a definition for chain, so that part is unclear, but I think that a fundamental problem here is that there is no recursive call to func; further, apply isn't used in the right position.
Instead of using (equal (length L) 2) as a base case, it might be nicer to make recursive calls as long as the first element in the input is a procedure, or otherwise just return the element:
#lang racket
(define multi-call
(lambda args
(let ((arg (car args)))
(if (procedure? arg)
(arg (apply multi-call (cdr args)))
arg))))
Here, when arg is a procedure, then it is applied to the result of calling multi-call recursively on the remaining arguments. Note that multi-call takes an arbitrary number of arguments, wrapping them in the list args. The reduction step provides (cdr args), which is a list of the remaining arguments. This means that apply should be used to call multi-call on those remaining arguments because multi-call expects an arbitrary number of arguments, not a list of arguments.
multi-call.rkt> (multi-call sqrt sqrt sqrt 390625)
5
I am trying to learn Common Lisp with the book Common Lisp: A gentle introduction to Symbolic Computation. In addition, I am using SBCL, Emacs and Slime.
In chapter 7, the author suggests there are three styles of programming the book will cover: recursion, iteration and applicative programming.
I am interested on the last one. This style is famous for the applicative operator funcall which is the primitive responsible for other applicative operators such as mapcar.
Thus, with an educational purpose, I decided to implement my own version of mapcar using funcall:
(defun my-mapcar (fn xs)
(if (null xs)
nil
(cons (funcall fn (car xs))
(my-mapcar fn (cdr xs)))))
As you might see, I used recursion as a programming style to build an iconic applicative programming function.
It seems to work:
CL-USER> (my-mapcar (lambda (n) (+ n 1)) (list 1 2 3 4))
(2 3 4 5)
CL-USER> (my-mapcar (lambda (n) (+ n 1)) (list ))
NIL
;; comparing the results with the official one
CL-USER> (mapcar (lambda (n) (+ n 1)) (list ))
NIL
CL-USER> (mapcar (lambda (n) (+ n 1)) (list 1 2 3 4))
(2 3 4 5)
Is there a way to implement mapcar without using recursion or iteration? Using only applicative programming as a style?
Thanks.
Obs.: I tried to see how it was implemented. But it was not possible
CL-USER> (function-lambda-expression #'mapcar)
NIL
T
MAPCAR
I also used Emacs M-. to look for the documentation. However, the points below did not help me. I used this to find the files below:
/usr/share/sbcl-source/src/code/list.lisp
(DEFUN MAPCAR)
/usr/share/sbcl-source/src/compiler/seqtran.lisp
(:DEFINE-SOURCE-TRANSFORM MAPCAR)
/usr/share/sbcl-source/src/compiler/fndb.lisp
(DECLAIM MAPCAR SB-C:DEFKNOWN)
mapcar is by itself a primitive applicative operator (pag. 220 of Common Lisp: A gentle introduction to Symbolic Computation). So, if you want to rewrite it in an applicative way, you should use some other primitive applicative operator, for instance map or map-into. For instance, with map-into:
CL-USER> (defun my-mapcar (fn list &rest lists)
(apply #'map-into (make-list (length list)) fn list lists))
MY-MAPCAR
CL-USER> (my-mapcar #'1+ '(1 2 3))
(2 3 4)
CL-USER> (my-mapcar #'+ '(1 2 3) '(10 20 30) '(100 200 300))
(111 222 333)
Technically, recursion can be implemented as follows:
(defun fix (f)
(funcall (lambda (x) (funcall x x))
(lambda (x) (funcall f (lambda (&rest y) (apply (funcall x x) y))))))
Notice that fix does not use recursion in any way. In fact, we could have only used lambda in the definition of f as follows:
(defconstant fix-combinator
(lambda (g) (funcall
(lambda (x) (funcall x x))
(lambda (x) (funcall
g
(lambda (&rest y) (apply (funcall x x)
y)))))))
(defun fix-2 (f)
(funcall fix-combinator f))
The fix-combinator constant is more commonly known as the y combinator.
It turns out that fix has the following property:
Evaluating (apply (fix f) list) is equivalent to evaluating (apply (funcall f (fix f)) list). Informally, we have (fix f) = (funcall f (fix f)).
Thus, we can define map-car (I'm using a different name to avoid package lock) by
(defun map-car (func lst)
(funcall (fix (lambda (map-func) (lambda (lst) ; We want mapfunc to be (lambda (lst) (mapcar func lst))
(if (endp lst)
nil
(cons (funcall func (car lst))
(funcall map-func (cdr lst)))))))
lst))
Note the lack of recursion or iteration.
That being said, generally mapcar is just taken as a primitive notion when using the "applicative" style of programming.
Another way you can implement mapcar is by using the more general reduce function (a.k.a. fold). Let's name the user-provided function f and define my-mapcar.
The reduce function carries an accumulator value that builds up the resulting list, here it is going take a value v, a sublist rest, and call cons with (funcall f v) and rest, so as to build a list.
More precisely, here reduce is going to implement a right-fold, since cons is right-associative (e.g. the recursive list is the "right" hand side, ie. the second argument of cons, e.g. (cons a (cons b (cons nil)))).
In order to define a right-fold with reduce, you pass :from-end t, which indicates that it builds-up a value from the last element and the initial accumulator to obtain a new accumulator value, then the second to last element with that new accumulator to build a new accumulator, etc. This is how you ensure that the resulting elements are in the same order as the input list.
In that case, the reducing function takes its the current element as its first argument, and the accumulator as a second argument.
Since the type of the elements and the type of the accumulator are different, you need to pass an :initial-value for the accumulator (the default behavior where the initial-value is taken from the list is for functions like + or *, where the accumulator is in the same domain as the list elements).
With that in mind, you can write it as follows:
(defun my-map (f list)
(reduce (lambda (v rest) (cons (funcall f v) rest))
list
:from-end t
:initial-value nil))
For example:
(my-map #'prin1-to-string '(0 1 2 3))
; => ("0" "1" "2" "3")
We find this function builder to realize composition in P.Graham's "ANSI Common Lisp" (page 110).
The arguments are n>0 quoted function names. I don't understand it completely, so I'll quote the code here and specify my questions underneath it:
(defun compose (&rest fns)
(destructuring-bind (fn1 . rest) (reverse fns)
#'(lambda (&rest args)
(reduce #'(lambda (v f) (funcall f v))
rest
:initial-value (apply fn1 args)))))
The argument list to compose is reversed and unpacked, its (now first) element bound to 'fn1' and the rest to 'rest'.
The body of the outermost lambda is a reduce: (funcall fi (funcall fi-1 ... ) ), with operands in inverted order to restore the initial one.
1) What is the role of the outermost lambda expression? Namely, where does it get its 'args' from? Is it the data structure specified as the first argument of destructuring-bind?
2) Where does the innermost lambda take its two arguments from?
I mean I can appreciate what the code does but still the lexical scope is a bit of a mystery to me.
Looking forward to any and all comments!
Thanks in advance,
//Marco
It's probably easier if you consider first a couple of practical examples:
(defun compose1 (a)
(lambda (&rest args)
(apply a args)))
(defun compose2 (a b)
(lambda (&rest args)
(funcall a (apply b args))))
(defun compose3 (a b c)
(lambda (&rest args)
(funcall a (funcall b (apply c args)))))
So the outermost lambda is the return value: a function that takes any arguments, what it does with it is applying the last function and chaining all the others in reverse order on the result got from last function.
Note: compose1 could be defined more simply as (defun compose1 (a) a).
A somewhat equivalent but less efficient version could be
(defun compose (&rest functions)
(if (= (length functions) 1)
(car functions)
(lambda (&rest args)
(funcall (first functions)
(apply (apply #'compose (rest functions))
args)))))
1) The outermost lambda creates a closure for you, because the result of (combine ...) is a function that calulates the composition of other functions.
2) The innermost lambda gets ists argument from the function reduce. Reduce takes a function (the innermost lambda) of two arguments and applies it stepwise to a list, e.g.
(reduce #'- '(1 2 3 4)) is (- (- (- 1 2) 3) 4)
How do I do recursion in an anonymous function, without using tail recursion?
For example (from Vanderhart 2010, p 38):
(defn power
[number exponent]
(if (zero? exponent)
1
(* number (power number (- exponent 1)))))
Let's say I wanted to do this as an anonymous function. And for some reason I didn't want to use tail recursion. How would I do it? For example:
( (fn [number exponent] ......))))) 5 3)
125
Can I use loop for this, or can loop only be used with recur?
The fn special form gives you the option to provide a name that can be used internally for recursion.
(doc fn)
;=> (fn name? [params*] exprs*)
So, add "power" as the name to complete your example.
(fn power [n e]
(if (zero? e)
1
(* n (power n (dec e)))))
Even if the recursion happened in the tail position, it will not be optimized to replace the current stack frame. Clojure enforces you to be explicit about it with loop/recur and trampoline.
I know that in Clojure there's syntactic support for "naming" an anonymous function, as other answers have pointed out. However, I want to show a first-principles approach to solve the question, one that does not depend on the existence of special syntax on the programming language and that would work on any language with first-order procedures (lambdas).
In principle, if you want to do a recursive function call, you need to refer to the name of the function so "anonymous" (i.e. nameless functions) can not be used for performing a recursion ... unless you use the Y-Combinator. Here's an explanation of how it works in Clojure.
Let me show you how it's used with an example. First, a Y-Combinator that works for functions with a variable number of arguments:
(defn Y [f]
((fn [x] (x x))
(fn [x]
(f (fn [& args]
(apply (x x) args))))))
Now, the anonymous function that implements the power procedure as defined in the question. Clearly, it doesn't have a name, power is only a parameter to the outermost function:
(fn [power]
(fn [number exponent]
(if (zero? exponent)
1
(* number (power number (- exponent 1))))))
Finally, here's how to apply the Y-Combinator to the anonymous power procedure, passing as parameters number=5 and exponent=3 (it's not tail-recursive BTW):
((Y
(fn [power]
(fn [number exponent]
(if (zero? exponent)
1
(* number (power number (- exponent 1)))))))
5 3)
> 125
fn takes an optional name argument that can be used to call the function recursively.
E.g.:
user> ((fn fact [x]
(if (= x 0)
1
(* x (fact (dec x)))))
5)
;; ==> 120
Yes you can use loop for this. recur works in both loops and fns
user> (loop [result 5 x 1] (if (= x 3) result (recur (* result 5) (inc x))))
125
an idomatic clojure solution looks like this:
user> (reduce * (take 3 (repeat 5)))
125
or uses Math.pow() ;-)
user> (java.lang.Math/pow 5 3)
125.0
loop can be a recur target, so you could do it with that too.
I'm trying to find the "best" implementation of a multi-argument "compose" in Scheme (I know it's a builtin in some implementations, but assume for the moment I am using one that doesn't have this).
For a 2-argument compose function I have this:
(define compose
(lambda (f g)
(lambda x
(f (apply g x)))))
This has the advantage that if the right-most function needs additional arguments, these can still be passed through the combined function. This has the pleasing property that composing the identity function on top of something does not change the function.
For example:
(define identity
(lambda (x) x))
(define list1
(compose identity list))
(define list2
(compose identity list1))
(list2 1 2 3)
> (1 2 3)
Now to do an "n-argument" compose I could do this:
(define compose-n
(lambda args
(foldr compose identity args)))
((compose-n car cdr cdr) '(1 2 3))
> 3
But this no longer preserves that nice "identity" property:
((compose-n identity list) 1 2 3)
> procedure identity: expects 1 argument, given 3: 1 2 3
The problem is that "initial" function used for the foldr command. It has built:
(compose identity (compose list identity))
So... I'm not sure the best way around this. "foldl" would seem to be the natural better alternative, because I want to it start with "identity" on the left not the right...
But a naive implementation:
(define compose-n
(lambda args
(foldl compose identity args)))
which works (have to reverse the order of function applications):
((compose-n cdr cdr car) '(1 2 3))
> 3
doesn't solve the problem because now I end up having to put the identity function on the left!
((compose-n cdr cdr car) '(1 2 3))
> procedure identity: expects 1 argument, given 3: 1 2 3
It's like, I need to use "foldr" but need some different "initial" value than the identity function... or a better identity function? Obviously I'm confused here!
I'd like to implement it without having to write an explicit tail-recursive "loop"... it seems there should be an elegant way to do this, I'm just stuck.
You might want to try this version (uses reduce from SRFI 1):
(define (compose . fns)
(define (make-chain fn chain)
(lambda args
(call-with-values (lambda () (apply fn args)) chain)))
(reduce make-chain values fns))
It's not rocket science: when I posted this on the #scheme IRC channel, Eli noted that this is the standard implementation of compose. :-) (As a bonus, it also worked well with your examples.)
The OP mentioned (in a comment to my answer) that his implementation of Scheme does not have call-with-values. Here's a way to fake it (if you can ensure that the <values> symbol is never otherwise used in your program: you can replace it with (void), (if #f #f), or whatever you like that's not used, and that's supported by your implementation):
(define (values . items)
(cons '<values> items))
(define (call-with-values source sink)
(let ((val (source)))
(if (and (pair? val) (eq? (car val) '<values>))
(apply sink (cdr val))
(sink val))))
What this does is that it fakes a multi-value object with a list that's headed by the <values> symbol. At the call-with-values site, it checks to see if this symbol is there, and if not, it treats it as a single value.
If the leftmost function in your chain can possibly return a multi-value, your calling code has to be prepared to unpack the <values>-headed list. (Of course, if your implementation doesn't have multiple values, this probably won't be of much concern to you.)
The issue here is that you're trying to mix procedures of different arity. You probably want to curry list and then do this:
(((compose-n (curry list) identity) 1) 2 3)
But that's not really very satisfying.
You might consider an n-ary identity function:
(define id-n
(lambda xs xs))
Then you can create a compose procedure specifically for composing n-ary functions:
(define compose-nary
(lambda (f g)
(lambda x
(flatten (f (g x))))))
Composing an arbitrary number of n-ary functions with:
(define compose-n-nary
(lambda args
(foldr compose-nary id-n args)))
Which works:
> ((compose-n-nary id-n list) 1 2 3)
(1 2 3)
EDIT: It helps to think in terms of types. Let's invent a type notation for our purposes. We'll denote the type of pairs as (A . B), and the type of lists as [*], with the convention that [*] is equivalent to (A . [*]) where A is the type of the car of the list (i.e. a list is a pair of an atom and a list). Let's further denote functions as (A => B) meaning "takes an A and returns a B". The => and . both associate to the right, so (A . B . C) equals (A . (B . C)).
Now then... given that, here's the type of list (read :: as "has type"):
list :: (A . B) => (A . B)
And here's identity:
identity :: A => A
There's a difference in kind. list's type is constructed from two elements (i.e. list's type has kind * => * => *) while identity's type is constructed from one type (identity's type has kind * => *).
Composition has this type:
compose :: ((A => B).(C => A)) => C => B
See what happens when you apply compose to list and identity. A unifies with the domain of the list function, so it must be a pair (or the empty list, but we'll gloss over that). C unifies with the domain of the identity function, so it must be an atom. The composition of the two then, must be a function that takes an atom C and yields a list B. This isn't a problem if we only give this function atoms, but if we give it lists, it will choke because it only expects one argument.
Here's how curry helps:
curry :: ((A . B) => C) => A => B => C
Apply curry to list and you can see what happens. The input to list unifies with (A . B). The resulting function takes an atom (the car) and returns a function. That function in turn takes the remainder of the list (the cdr of type B), and finally yields the list.
Importantly, the curried list function is of the same kind as identity, so they can be composed without issue. This works the other way as well. If you create an identity function that takes pairs, it can be composed with the regular list function.
While it would have been nice for the "empty" list to devolve to the identity function, surrendering this appears to result in the following, which isn't too bad:
(define compose-n
(lambda (first . rest)
(foldl compose first rest)))
((compose-n cdr cdr car) '(1 2 3))
((compose-n list identity identity) 1 2 3)