Hello I am following the quick introduction to racket. In chapter 5 they propose that this code (which worked fine):
(define (four p)
(define two-p (hc-append p p))
(vc-append two-p two-p))
would tipically be written by using let, so I wrote my code:
#lang slideshow
(define (square n)
(filled-rectangle n n))
(define (four p)
(let ([two-p (hc-apppend p p)])
(vc-append two-p two-p)))
but it doesn't work.
The error message is:
expand: unbound identifier in module in: hc-apppend
Just checking: Did you remember
#lang slideshow
as the first line?
Also, count the number of p characters! It is hc-append. The error mentions hc-apppend: it should be hc-append.
Related
I'm trying to experiment with what I can do in Racket, and I want to suffix numbers with letters.
For this example, I'd simply like to represent 10000 as 10K, and 1000000 as 1M.
Is there way (with macros or otherwise) that I can expand 1M to:
(* 1 1000000)
Or something to that effect?
In Racket, things like 10K are identifiers, which normally would refer to variables. There are two ways to make them into numbers:
1: redefine what "undefined" identifiers mean
You can redefine what to do on an undefined identifier by defining a #%top macro.
#lang racket
(require syntax/parse/define
(only-in racket [#%top old-#%top]))
(define-syntax-parser #%top
[(_ . x:id)
#:when (id-has-a-k-at-the-end? #'x)
(transform-id-into-number #'x)]
[(_ . x)
#'(old-#%top . x)])
However, this has a subtle problem. If there are any identifiers or variables in your program with K's on the end, they could override any numbers that were written that way. You would need to be careful not to accidentally override something that was intended to be a number.
2: make a reader extension that turns them into numbers instead of identifiers
This will take more time, but it's closer to the "right way" to do this, since it avoids conflicts when variables happen to have K's on the end.
One of the easier ways to extend the reader is with a readtable. You can make a function that extends a readtable like this:
;; Readtable -> Readtable
(define (extend-readtable orig-rt)
;; Char InputPort Any Nat Nat Nat -> Any
(define (rt-proc char in src ln col pos)
....)
...
(make-readtable orig-rt
#f 'non-terminating-macro rt-proc
...))
To use this to define a #lang language, you need to put the reader implementation in your-language/lang/reader.rkt. Here that's number-with-k/lang/reader.rkt, where the number-with-k directory is installed as a single-collection package (raco pkg install path/to/number-with-k).
number-with-k/lang/reader.rkt
#lang racket
(provide (rename-out [-read read]
[-read-syntax read-syntax]
[-get-info get-info]))
(require syntax/readerr
syntax/module-reader)
;; Readtable -> Readtable
(define (extend-readtable orig-rt)
;; Char InputPort Any Nat Nat Nat -> Any
(define (rt-proc char in src ln col pos)
....)
...
(make-readtable orig-rt
#f 'non-terminating-macro rt-proc))
;; [X ... -> Y] -> [X ... -> Y]
(define ((wrap-reader rd) . args)
(parameterize ([current-readtable (extend-readtable (current-readtable))])
(apply rd args)))
(define-values [-read -read-syntax -get-info]
(make-meta-reader 'number-with-k
"language path"
lang-reader-module-paths
wrap-reader
wrap-reader
identity))
The main work goes into filling in the .... holes in the extend-readtable function. For example, you can make it recognize identifiers that end with K like this:
;; Readtable -> Readtable
(define (extend-readtable orig-rt)
;; Char InputPort Any Nat Nat Nat -> Any
(define (rt-proc char in src ln col pos)
(define v (read-syntax/recursive src in char orig-rt #f))
(cond
[(and (identifier? v) (id-has-a-k-at-the-end? v))
(transform-id-into-number v)]
[else
v]))
(make-readtable orig-rt
#f 'non-terminating-macro rt-proc))
Once this is done, and you have the number-with-k directory installed as a package, you should be able to use #lang number-with-k like this:
#lang number-with-k racket
(+ 12K 15)
; => 12015
The simplest to is to define the suffixes you need.
(define K 1000)
(define M 1000000)
Then write (* 3.14 M) for 3.14 millions.
As others mention, Racket supports scientific notation 3.14E6 is also 3.14 million.
Yet another alternative is to define functions K, M etc like:
(define (M x) (* x 1000000))
Then you can write
(M 3.14)
to mean 3.14 million.
Racket does already have built in support for this, kind of, via scientific notation:
1e6 ; 1000000.0 ("1M")
2e7 ; 20000000.0
For example,
(require racket/generator)
(define f add1)
(define init 0)
(in-producer (generator () (let loop ([x init]) (yield x) (loop (f x)))))
Is there any better way to do this? I don't quite like generators since they have hidden states.
Streams
Using streams is probably the easiest:
(require racket/stream)
;; X [X -> X] -> [Streamof X]
(define (repeated-fn-stream init f)
(stream-cons init (repeated-fn-stream (f init) f)))
(repeated-fn-stream 0 add1)
Sequences
Alternatively, using sequences and make-do-sequence:
(require racket/sequence)
;; X [X -> X] -> [Sequenceof X]
(define (repeated-fn-sequence init f)
;; A "Pos" is an X that's the result of applying f repeatedly to init
(define (pos->element pos) pos)
(define (next-pos pos) (f pos))
(define init-pos init)
(make-do-sequence
(λ ()
(values pos->element
next-pos
init-pos
#false
#false
#false))))
(repeated-fn-sequence 0 add1)
If you wanted to use sequences, and you wanted to use define-sequence-syntax to make for loops specialize it:
(this is completely unnecessary for "pure" functionality, but it may have different performance characteristics)
(require (for-syntax syntax/parse))
(define-sequence-syntax in-repeated-fn-sequence
(λ () #'repeated-fn-sequence) ; when used as a normal expression
(syntax-parser ; when used *directly* as a for-loop clause
[[(x) (_ init-expr f-expr)]
#'[(x) (:do-in
([(init) init-expr] [(f) f-expr])
#true
([x init])
#true
()
#true
#true
[(f x)])]]))
(for/list ([x (in-repeated-fn-sequence 0 add1)]
[i (in-range 10)])
x)
When using define-sequence-syntax, you should make sure that for everything there is a "single point of truth". Because of that you often see this pattern:
(define-sequence-syntax in-___
(λ () #'in-___/proc) ; when used as a normal expression
(syntax-parser
....everything that defines the actual functionality....))
;; This is completely determined by the sequence-syntax above,
;; that way there is NO duplicated functionality and NO chance for
;; it to get "out of sync".
(define (in-___/proc parameter ...)
(for/stream ([elem (in-___ parameter ...)])
elem))
What that means for this is that once you decide you want to use define-sequence-syntax, you should define the repeated-fn-sequence function in terms of it:
(define (repeated-fn-sequence init f)
(for/stream ([elem (in-repeated-fn-sequence init f)])
elem))
That way if the in-repeated-fn-sequence needs to be changed to fix a bug or switch representations, the function version changes with it automatically.
The best function for this job is an unfold… but unfortunately, Racket does not provide a built-in sequence-unfold or stream-unfold operation. However, there is a stream-unfold operation in the srfi/41 library, which will meet your needs. You can see this in action with the following program:
#lang racket
(require (only-in srfi/41 stream-unfold))
(define nats (stream-unfold identity (const #t) add1 0))
(for/list ([i (in-range 20)] [n nats]) n)
This produces the following output:
'(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19)
If you don’t want to use srfi/41, you can write stream-unfold yourself in terms of the racket/stream API without too much difficulty, and without any statefulness:
(define (stream-unfold mapper pred? gen base)
(let loop ([base base])
(if (pred? base)
(stream-cons (mapper base) (loop (gen base)))
empty-stream)))
I am new to Lisp-scheme and fairly new to the functional paradigm as a whole, and am currently doing an assignment which requires me to overload a function with the same name, but different sets of parameters in racket. Below is an example of what I'm trying to achieve:
#lang racket
(define (put-ball-in-box two-by-fours nails ball)
... )
(define (put-ball-in-box box ball)
... )
These are not the actual functions, but close enough. As implied, both functions would put a ball in a box, but one would assemble the box from its components first, then call the other. Obviously, when I try the above in DrRacket or using the command line, I get a module: duplicate definition for identifier ... error.
Is there a way to achieve this in racket?
Maybe the answer is right in front of me, but I have spent the last two hours searching for this and couldn't find anything, so would appreciate any pointers.
Thank you.
It doesn't in the usual sense of "writing another definition somewhere else."
It allows shadowing, which is defining a procedure with the same name as an imported procedure. Thus you can (define + ...) and your definition of + will hide the + from racket/base. If you want the original procedure, then you can do something like the following, where I define + to be either addition or string-appending.
#lang racket/base
(require (rename-in racket/base (+ base:+)))
(define (+ . args)
(if (andmap string? args)
(apply string-append args)
(apply base:+ args)))
Another thing you can do is use racket/match to have different behavior based on the shape of the argument.
#lang racket/base
(require racket/match)
(define (fib . arg)
(match arg
[(list n) (fib n 1 0)]
[(list 1 a b) a]
[(list 0 a b) b]
[(list n a b) (fib (sub1 n) (+ a b) a)]))
This second example still doesn't quite do what you want since you have to go to the original definition point and modify the match clauses. But it might be sufficient for your purposes.
A more complicated example would be to use custom syntax to create a define/overload form. But I think you'll find the racket/match solution to be best.
You have the concept of default values as in JS and PHP:
(define (fib n (a 0) (b 1))
(if (zero? n)
a
(fib (sub1 n) b (+ a b))))
(fib 10) ; ==> 55
Now if you had 5 optional parameters you need to order them and even pass some values just to be able to add a later one. To avoid that you can use keywords:
(define (test name #:nick [nick name] #:job [job "vacant"])
(list name nick job))
(test "sylwester" #:job "programmer")
; ==> ("sylwester" "sylwester" "programmer")
Now Racket has classes. You can call a method like (send object method args ...).
(define circle%
(class object%
(super-new)
(init-field radius)
(define/public (area)
(* radius radius 3.1415))))
(define cube%
(class object%
(super-new)
(init-field side)
(define/public (area)
(* side side))))
(define circle (new circle% [radius 7]))
(define cube (new cube% [side 7]))
(map
(lambda (o) (send o area))
(list circle cube))
; ==> (153.9335 49)
Notice that the two classes hasn't really commited to a joint interface with area so this is pure duck typing. Thus you can make a function that expects a class that implements a message and it doesn't need to worry about other aspects of the class at all.
(define generalized-triangular
(lambda (input n)
(if (= n 1)
1
(+ (input n) (generalized-triangular (- n 1))))))
This program is designed to take a number and a function as inputs and do the following..
f(1) + f(2) + f(3)+ … + f(N).
An example input would be:
(generalized-triangular square 3)
The Error message:
if: bad syntax;
has 4 parts after keyword in: (if (= n 1) 1 (+ (input n) (generalized-triangular (- n 1))) input)
The error is quite explicit - an if form can only have two parts after the condition - the consequent (if the condition is true) and the alternative (if the condition is false). Perhaps you meant this?
(if (= n 1)
1
(+ (input n) (generalized-triangular input (- n 1))))
I moved the input from the original code, it was in the wrong place, as the call to generalized-triangular expects two arguments, in the right order.
For the record: if you need to execute more than one expression in either the consequent or the alternative (which is not the case for your question, but it's useful to know about it), then you must pack them in a begin, for example:
(if <condition> ; condition
(begin ; consequent
<expression1>
<expression2>)
(begin ; alternative
<expression3>
<expression4>))
Alternatively, you could use a cond, which has an implicit begin:
(cond (<condition> ; condition
<expression1> ; consequent
<expression2>)
(else ; alternative
<expression3>
<expression4>))
Literal answer
The code you posted in your question is fine:
(define generalized-triangular
(lambda (input n)
(if (= n 1)
1
(+ (input n) (generalized-triangular (- n 1))))))
The error message in your question would be for something like this code:
(define generalized-triangular
(lambda (input n)
(if (= n 1)
1
(+ (input n) (generalized-triangular (- n 1)))
input)))
The problem is input. if is of the form (if <cond> <then> <else>). Not counting if itself, it has 3 parts. The code above supplies 4.
Real answer
Two tips:
Use DrRacket to write your code, and let it help you with the indenting. I couldn't make any sense of your original code. (Even after someone edited it for you, the indentation was a bit wonky making it still difficult to parse mentally.)
I don't know about your class, but for "real" Racket code I'd recommend using cond instead of if. Racket has an informal style guide that recommends this, too.
here's the tail-recursive
(define (generalized-triangular f n-max)
(let loop ((n 1) (sum 0))
(if (> n n-max)
0
(loop (+ n 1) (+ sum (f n))))))
Since you're using the racket tag, I assume the implementation of generalized-triangular is not required to use only standard Scheme. In that case, a very concise and efficient version (that doesn't use if at all) can be written with the racket language:
(define (generalized-triangular f n)
(for/sum ([i n]) (f (+ i 1))))
There are two things necessary to understand beyond standard Scheme to understand this definition that you can easily look up in the Racket Reference: how for/sum works and how a non-negative integer behaves when used as a sequence.
I've been attempting to learn programming with the book "Structures and Interpretation of Computer Programs. To do the exercises I've been using DrRacket (I couldn't find a scheme interpreter for Windows 7, and DrRacket seems pretty good), and haven't had any problems so far. But while doing exercise 1.22 I've ran into an issue. I've wrote a procedure that gives a given number (n) of prime numbers larger than a:
(define (search-for-primes a n)
(define (sfp-iter a n counter)
(cond ((and (prime? a) (= counter n))
((newline) (display "end")))
((prime? a)
((newline)
(display a)
(sfp-iter (+ a 1) n (+ counter 1))))
(else (sfp-iter (+ a 1) n counter))))
(sfp-iter a n 0))
The procedure works as intended, displaying all that it should, but after displaying end it shows the following error message:
application: not a procedure;
expected a procedure that can be applied to arguments
given: #
arguments...:
#
And highlights the following line of code:
((newline) (display "end"))
What is the problem?
(I apologize for any mistakes in spelling and so, English isn't my native language, I also apologize for any error in formatting or tagging, I'm new here)
You have a couple of parenthesis problems, this fixes it:
(define (search-for-primes a n)
(define (sfp-iter a n counter)
(cond ((and (prime? a) (= counter n))
(newline) (display "end"))
((prime? a)
(newline)
(display a)
(sfp-iter (+ a 1) n (+ counter 1)))
(else (sfp-iter (+ a 1) n counter))))
(sfp-iter a n 0))
In the first and second conditions of the cond, you were incorrectly surrounding the code with (). That's unnecessary, in a cond clause all the expressions that come after the condition are implicitly surrounded by a (begin ...) form, so there's no need to group them together.