I asked a similar question before, I just want to make sure I understand the idea, in the -lambda(x)- line 4, what is the x, where it is coming from?
(define (cached-assoc xs n)
(letrec ([memo (make-vector n #f)]
[acc 0]
[f (lambda(x)
(let ([ans (vector-assoc x memo)])
(if ans
(cdr ans)
(let ([new-ans (assoc x xs)])
(begin
(vector-set! memo acc (cons x new-ans))
(set! acc (if (= (+ acc 1)) 0 (+ acc 1)))
new-ans)))))])
f))
Your cached-assoc procedure returns f, a function which has x as an unbound parameter. It doesn’t have a value yet, the value will be whatever you pass to it when you finally invoke it:
((cached-assoc xs n) 10)
In the above example x will be bound to 10, and xs and n will be whatever you defined for them before calling cached-assoc. I believe the source of the confusion is the fact that cached-assoc is returning a lambda, once you understand this it'll be clear.
Related
You can notice the v in the lambda in the function body, where is the v coming from, what it is based on?
(define (cached-assoc xs n)
(letrec ([memo (make-vector n #f)]
[acc 0]
[f (lambda(x)
(let ([ans (vector-assoc x memo)])
(if ans
(cdr ans)
(let ([new-ans (assoc x xs)])
(begin
(vector-set! memo acc (cons x new-ans))
(set! acc (if (= (+ acc 1)) 0 (+ acc 1)))
new-ans)))))])
(lambda (v) (f v))))
The whole expression is returning a lambda as a result, and in that lambda there's a formal parameter named v. It doesn't have a value yet, you'll need to call the lambda to bind a value to v and produce a result (assuming the code is working):
((letrec ([memo (make-vector n #f)] ; notice the extra opening `(`, we want to call the returned lambda
[acc 0]
[f (lambda(x)
(let ([ans (vector-assoc x memo)])
(if ans
(cdr ans)
(let ([new-ans (assoc x xs)])
(begin
(vector-set! memo acc (cons x new-ans))
(set! acc (if (= (+ acc 1)) 0 (+ acc 1)))
new-ans)))))])
(lambda (v) (f v)))
10) ; <- the value 10 gets bound to `v`
However, your code isn't right. You are referring to variables named n and xs, but they are not defined anywhere and need a value of their own. The procedure vector-assoc doesn't exist. Also, the lambda at the end is redundant, you could simply return f, there's no need to wrap it in an additional lambda. Finally: you should define the whole expression with a name, it'll make it easier to call it.
I won't go into more details because first you need to fix the function and make it work, and is not clear at all what you want to do - but that should be a separate question.
I am going through a practice exam for my programming languages course. One of the problems states:
Define a function named function+ that “adds” two functions together and returns this composition. For example:
((function+ cube double) 3)
should evaluate to 216, assuming reasonable implementations of the functions cube and double.
I am not sure how to approach this problem. I believe you are supposed to use the functionality of lambdas, but I am not entirely sure.
If you need a procedure which allows you two compose to unary procedures (procedure with only 1 parameter), you'll smack yourself in the head after seeing how simple the implementation is
(define (function+ f g)
(λ (x) (f (g x))))
(define (cube x)
(* x x x))
(define (double x)
(+ x x))
((function+ cube double) 3)
;=> 216
Basically if you need to do that you just do (x (y args ...)) so if you need to have a procedure that takes two arguments proc1 and proc2 returns a lambda that takes any number of arguments. You just use apply to give proc1 arguments as a list and pass the result to proc2. It would look something like this:
(define (compose-two proc2 proc1)
(lambda args
...))
The general compose is slightly more complicated as it takes any number of arguments:
#!r6rs
(import (rnrs))
(define my-compose
(let* ((apply-1
(lambda (proc value)
(proc value)))
(gen
(lambda (procs)
(let ((initial (car procs))
(additional (cdr procs)))
(lambda args
(fold-left apply-1
(apply initial args)
additional))))))
(lambda procs
(cond ((null? procs) values)
((null? (cdr procs)) (car procs))
(else (gen (reverse procs)))))))
(define (add1 x) (+ x 1))
((my-compose) 1) ;==> 1
((my-compose +) 1 2 3) ; ==> 6
((my-compose sqrt add1 +) 9 15) ; ==> 5
How do I assign anonymous functions to local variables in either cl, emacs lisp or clojure?
I've tried the following with no success.
(let ((y (lambda (x) (* x x)) )) (y 2))
and
((lambda (x) 10) (lambda (y) (* y y)))
In CL, you could use flet or labels.
(defun do-stuff (n)
(flet ((double (x) (* 2 x)))
(double n)))
(do-stuff 123) ;; = 246
As Chris points out, since double is not recursive, we should use flet, as the difference between the two is that labels can handle recursive functions.
Check out docs for info on labels, or this question for the difference between labels and flet.
I'm trying to find the first missing key in a hash table, that should contain keys [1 ... N], using Scheme.
So far, I have the following code:
(define (find-missing n ht)
(define c 1)
(let forVertices ()
(cond (not (hash-has-key? ht c))
(c)
)
(set! c (+ c 1))
(when (>= n c) (forVertices))
)
)
When I execute the function to test it, nothing is returned. What am I doing wrong?
You have a couple of problems with the parentheses, and the else case is missing. This should fix the mistakes:
(define (find-missing n ht)
(define c 1)
(let forVertices ()
(cond ((not (hash-has-key? ht c))
c)
(else
(set! c (+ c 1))
(when (>= n c)
(forVertices))))))
… But you should be aware that the way you wrote the procedure is not idiomatic, at all. In general, in Scheme you should avoid mutating state (the set! operation), you can write an equivalent procedure by correctly setting up the recursion and passing the right parameters. Also, it's a good idea to return something whenever the recursion exits (for instance: #f), otherwise your procedure will return #<void> if there are no keys missing. Here, this is what I mean:
(define (find-missing n ht)
(let forVertices ((c 1))
(cond ((> c n) #f)
((not (hash-has-key? ht c)) c)
(else (forVertices (+ c 1))))))
Is it possible to call a Scheme function using only the function name that is available say as a string in a list?
Example
(define (somefunc x y)
(+ (* 2 (expt x 2)) (* 3 y) 1))
(define func-names (list "somefunc"))
And then call the somefunc with (car func-names).
In many Scheme implementations, you can use the eval function:
((eval (string->symbol (car func-names))) arg1 arg2 ...)
You generally don't really want to do that, however. If possible, put the functions themselves into the list and call them:
(define funcs (list somefunc ...))
;; Then:
((car funcs) arg1 arg2 ...)
Addendum
As the commenters have pointed out, if you actually want to map strings to functions, you need to do that manually. Since a function is an object like any other, you can simply build a dictionary for this purpose, such as an association list or a hash table. For example:
(define (f1 x y)
(+ (* 2 (expt x 2)) (* 3 y) 1))
(define (f2 x y)
(+ (* x y) 1))
(define named-functions
(list (cons "one" f1)
(cons "two" f2)
(cons "three" (lambda (x y) (/ (f1 x y) (f2 x y))))
(cons "plus" +)))
(define (name->function name)
(let ((p (assoc name named-functions)))
(if p
(cdr p)
(error "Function not found"))))
;; Use it like this:
((name->function "three") 4 5)