I am trying to make a script work from both DrRacket and the repl, having this as my starting point: Racket calculator
Here is my current code:
#lang racket
(provide (all-defined-out))
(require parser-tools/lex
(prefix-in re: parser-tools/lex-sre)
parser-tools/yacc)
(define-tokens value-tokens (INT ANY))
(define-empty-tokens empty-tokens
(PLUS MINUS MULTIPLY DIVIDE NEWLINE EOF))
(define basic-lexer
(lexer
((re:+ numeric) (token-INT lexeme))
(#\+ (token-PLUS))
(#\- (token-MINUS))
(#\* (token-MULTIPLY))
(#\/ (token-DIVIDE))
((re:or #\tab #\space) (basic-lexer input-port))
(#\newline (token-NEWLINE))
((eof) (token-EOF))
(any-char (token-ANY lexeme))))
(define (display-plus expr)
(display "Result: ")
(let ((left (string->number (first expr)))
(right (string->number (last expr))))
(display (+ left right)))
(newline))
(define (display-minus expr)
(display "Result: ")
(let ((left (string->number (first expr)))
(right (string->number (last expr))))
(display (- left right)))
(newline))
(define (display-multiply expr)
(display "Result: ")
(let ((left (string->number (first expr)))
(right (string->number (last expr))))
(display (* left right)))
(newline))
(define (display-divide expr)
(display "Result: ")
(let ((left (string->number (first expr)))
(right (string->number (last expr))))
(display (/ left right)))
(newline))
(define basic-parser
(parser
(start start)
(end NEWLINE EOF)
(tokens value-tokens empty-tokens)
(error (lambda (ok? name value)
(printf "Couldn't parse: ~a\n" name)))
(grammar
(start ((expr) $1)
((expr start) $2))
(expr ((INT PLUS INT) (display-plus (list $1 $3)))
((INT MINUS INT) (display-minus (list $1 $3)))
((INT MULTIPLY INT) (display-multiply (list $1 $3)))
((INT DIVIDE INT) (display-divide (list $1 $3)))
((ANY) (displayln $1))))))
(define input1 (open-input-string "123 + 456"))
(define input2 (open-input-string "123 *456"))
(basic-parser (lambda() (basic-lexer input1)))
(basic-parser (lambda() (basic-lexer input2)))
;(define (my-repl)
; (display ">>> ")
; (let* ((input (read-line))
; (input-port (open-input-string
; (list->string
; (drop-right
; (string->list input) 1)))))
; (cond
; ((not (equal? "\r" input)
; (print (basic-parser
; (lambda () (basic-lexer input-port))))))))
; (my-repl))
(define (calc str)
(let* ([port (open-input-string str)]
[result (basic-parser (lambda() (basic-lexer port)))])
(displayln result)))
(define (repl)
(display ">>> ")
(let ((input (read-line)))
(print input)
(cond
((eof-object? input) (displayln "eof"))
((eq? input #\newline) (displayln "new line"))
(else (calc (read-line))))
(newline))
(repl))
A test from DrRacket is shown here:
Welcome to DrRacket, version 7.1 [3m].
Language: racket, with debugging; memory limit: 512 MB.
Result: 579
Result: 56088
> (repl)
>>> 1+1
"1+1"2+2
Result: 4
#<void>
>>> 3+3
"3+3"4+4
Result: 8
#<void>
And from the repl:
Welcome to Racket v7.1.
> (require "untitled7.rkt")
Result: 579
Result: 56088
> (repl)
>>> "\r"
#<void>
>>> 1+1
"1+1\r"2+2
Result: 4
#<void>
>>> 3+3
"3+3\r"4+4
Result: 8
#<void>
>>> #<eof>eof
>>> ; user break [,bt for context]
It only displays every second calculation. It appears that read-line returns a new line before waiting for user input, which I tried to check with (eof-object? input) and (eq? input #\newline) but now I get only every second result.
There are two problems:
First, you're reading a line, (let ((input (read-line))), but you're not sending that input to the calculator, you'r sending another one – (calc (read-line)).
You should pass input to calc for evaluation instead.
Second, you have a lot of #<void>s in your output.
This is because calc assumes that your parser produces a value that it can print:
(displayln result)
but the parser does not produce any value, it only prints one.
Either remove the output of result, or rewrite the parser to return the value to its caller.
Replace (calc (read-line)) with (calc input).
Related
Exercise 3.52,
(define sum 0)
(define (accum x)
(set! sum (+ x sum))
sum)
;1: (define seq (stream-map accum (stream-enumerate-interval 1 20)))
;2: (define y (stream-filter even? seq))
;3: (define z (stream-filter (lambda (x) (= (remainder x 5) 0))
; seq))
;4: (stream-ref y 7)
;5: (display-stream z)
Step 1:
;1: ==> (cons-stream 1 (stream-map proc (stream-cdr s)) (Assume stream-cdr is evaluated only when we force the cdr of this stream)
sum is now 1
Step 2:
1 is not even, hence (also memoized so not added again), it calls (stream-filter pred (stream-cdr stream)).
This leads to
evaluation of cdr hence materializing 2 which is even, hence it should call: (cons-stream 2 (stream-cdr stream)).
According to this answer should be 1+2 = 3 , but it is 6
Can someone help with why the cdr's car is materialized before the current cdr is called?
Using Daniel P. Friedman's memoizing tail
#lang r5rs
(define-syntax cons-stream
(syntax-rules ()
((_ h t) (cons h (lambda () t)))))
(define (stream-cdr s)
(if (and (not (pair? (cdr s)))
(not (null? (cdr s))))
(set-cdr! s ((cdr s))))
(cdr s))
we observe:
> sum
0
> (define seq (stream-map accum (stream-enumerate-interval 1 20)))
> sum
1
> seq
(mcons 1 #<procedure:friedmans-tail.rkt:21:26>)
> (define y (stream-filter even? seq))
> sum
6
> seq
(mcons
1
(mcons
3
(mcons 6 #<procedure:friedmans-tail.rkt:21:26>)))
> y
(mcons 6 #<procedure:friedmans-tail.rkt:21:26>)
>
stream-filter? needs to get to the first element of the stream it is constructing in order to construct it. A stream has its head element already forced, calculated, so it must be already present.
In the list of accumulated sums of the enumerated interval from 1 to 20, the first even number is 6:
1 = 1
1+2 = 3
1+2+3 = 6
...
I'm currently implementing a small parser in Clojure that takes an input string like:
aaa (bbb(ccc)ddd(eee)) fff (ggg) hhh
and returns the string without characters that are not in brackets, i.e.
(bbb(ccc)ddd(eee))(ggg)
I've written the following function:
(defn- parse-str [input]
(let [bracket (atom 0)
output (atom [])]
(doseq [ch (seq input)]
(case ch
\( (swap! bracket inc)
\) (swap! bracket dec)
nil)
(if (or (> #bracket 0) (= ch \)))
(swap! output conj ch)))
(apply str #output)))
which works for me:
(parse-str "aaa (bbb(ccc)ddd(eee)) fff (ggg) hhh")
"(bbb(ccc)ddd(eee))(ggg)"
I am however concerned that my approach is a too object oriented since it uses atoms as some kind of local variables to keep the current state of the parser.
Is it possible to write the same function from a more functional programming perspective? (avoiding the atoms?)
Any comments to improve my code are appreciated as well.
Two ways: You can use explicit recursion or reduce.
(defn parse-str [input]
(letfn [(parse [input bracket result]
(if (seq input)
(let [[ch & rest] input]
(case ch
\( (recur rest (inc bracket) (conj result ch))
\) (recur rest (dec bracket) (conj result ch))
(recur rest bracket (if (> bracket 0)
(conj result ch)
result))))
result))]
(clojure.string/join (parse input 0 []))))
(defn parse-str [input]
(clojure.string/join
(second (reduce (fn [acc ch]
(let [[bracket result] acc]
(case ch
\( [(inc bracket) (conj result ch)]
\) [(dec bracket) (conj result ch)]
[bracket (if (> bracket 0)
(conj result ch)
result)])))
[0 []]
input))))
In a lot of cases where you would use local variables, you just put any variable that changes as a parameter to loop, thereby using recursion instead of mutation.
(defn- parse-str [input]
;; Instead of using atoms to hold the state, use parameters in loop
(loop [output []
bracket 0
;; The [ch & tail] syntax is called destructuring,
;; it means let ch be the first element of (seq input),
;; and tail the rest of the elements
[ch & tail] (seq input)]
;; If there's no elements left, ch will be nil, which is logical false
(if ch
(let [bracket* (case ch
\( (inc bracket)
\) (dec bracket)
bracket)
output* (if (or (> bracket* 0) (= ch \)))
(conj output ch)
output)]
;; Recurse with the updated values
(recur output* bracket* tail))
;; If there's no characters left, apply str to the output
(apply str output))))
This is an iterative version of your function; but it's still functionally pure. I find having the code laid out like this makes it easy to read. Remember, when using recursion, always check your termination condition first.
(defn parse-str [s]
(loop [[x & xs] (seq s), acc [], depth 0]
(cond
(not x) (clojure.string/join acc)
(= x \() (recur xs (conj acc x) (inc depth))
(= x \)) (recur xs (conj acc x) (dec depth))
(<= depth 0) (recur xs acc depth)
:else (recur xs (conj acc x) depth))))
I can print n-numbers as list with this code below:
(define (print-first-n stream1 n)
(cond((= n 0) '())
(else(cons(stream-car stream1) (print-first-n (stream-cdr stream1) (- n 1))))))
But I have no idea about how to add commas.
You can't print a comma in a normal list, but we can build a string with the contents of the stream, separated by commas. This will work, assuming that the string contains numbers:
(define (print-first-n stream1 n)
(cond ((= n 1)
(number->string (stream-car stream1)))
(else
(string-append
(number->string (stream-car stream1)) ", "
(print-first-n (stream-cdr stream1) (- n 1))))))
The above solution is fine for a small value of n, but terribly inefficient for large values (lots of temporary strings will be created, with O(n^2) complexity for the append operation). For a more efficient implementation, consider using SRFI-13's concatenation procedures, like this:
(require srfi/13)
(define (print-first-n stream1 n)
(let loop ((strm stream1) (n n) (acc '()))
(if (= n 1)
(string-concatenate-reverse
(cons (number->string (stream-car strm)) acc))
(loop (stream-cdr strm)
(sub1 n)
(list* ", " (number->string (stream-car strm)) acc)))))
Either way: let's say that integers is an infinite stream of integers starting at 1, this is how it would look:
(print-first-n integers 5)
=> "1, 2, 3, 4, 5"
If the stream contains some other data type, use the appropriate procedure to convert each element to a string.
If your function just prints the stream contents, and doesn't need to build a string (like Óscar's answer), here's my take on it (uses SRFI 41 streams):
(define (print-first-n stream n)
(stream-for-each (lambda (delim item)
(display delim)
(display item))
(stream-cons "" (stream-constant ", "))
(stream-take n stream)))
Example:
> (define natural (stream-cons 1 (stream-map (lambda (x) (+ x 1)) natural)))
> (print-first-n natural 10)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
To output to a string (like Óscar's answer), just wrap the whole thing in a string port:
(define (print-first-n stream n)
(call-with-output-string
(lambda (out)
(stream-for-each (lambda (delim item)
(display delim out)
(display item out))
(stream-cons "" (stream-constant ", "))
(stream-take n stream)))))
I have the code
(define alternate
(letrec ([f (lambda (x) (cons x (lambda () (f (+ x 1)))))])
(lambda () (f 1))))
The result is 1,2,3..
How i could change it to take 1,2,1,2,1,2..
I tried cons inside the f but didn't work.
Any ideas?
You might also find generators useful: docs
Welcome to DrRacket, version 5.3.3.5 [3m].
Language: racket [custom].
> (require racket/generator)
> (define g (generator () (let LOOP () (yield 1) (yield 2) (LOOP))))
> (g)
1
> (g)
2
> (g)
1
> (g)
2
UPDATE:
Even better, use an infinite-generator:
Welcome to DrRacket, version 5.3.3.5 [3m].
Language: racket [custom].
> (require racket/generator)
> (define g (infinite-generator (yield 1) (yield 2)))
> (g)
1
> (g)
2
> (g)
1
> (g)
2
This is straightforward to implement using streams:
(define (alternate)
(stream-map (lambda (x)
(if (even? x) 1 2))
(in-naturals)))
The trick here is that a stream is built using stream-cons, which basically does what you're implementing by hand: it creates a list where its elements are "promises" that get evaluated only when needed.
stream-cons produces a lazy stream for which stream-first forces the evaluation of first-expr to produce the first element of the stream, and stream-rest forces the evaluation of rest-expr to produce a stream for the rest of the returned stream.
This shows how alternate returns an infinite stream of elements of the form 1 2 1 2 1 2 ...
(define alt (alternate))
(stream-ref alt 0)
=> 1
(stream-ref alt 1)
=> 2
(stream-ref alt 2)
=> 1
(stream-ref alt 3)
=> 2
Alternatively, if you need a list of n elements of the sequence use this procedure, which by the way should be part of Racket in the first place:
(define (stream-take s n)
(if (zero? n)
'()
(cons (stream-first s)
(stream-take (stream-rest s) (sub1 n)))))
Now it works as expected:
(define alt (alternate))
(stream-take alt 0)
=> '()
(stream-take alt 1)
=> '(1)
(stream-take alt 2)
=> '(1 2)
(stream-take alt 3)
=> '(1 2 1)
Here's a way to do it as a small modification of your existing code:
(define alternate
(letrec ([f (lambda (x) (cons x (lambda () (f (if (= x 1) 2 1)))))])
(lambda () (f 1))))
I am trying to write a simple parser which creates a sxml-expression from a string, e. g.
"This is a [Test]" ===> (item "This is a" (subitem "Test"))
Anybody who is wondering about the square brackets within the given example may have a look at the so called Leiden conventions.
This is the code I have written so far:
(define my-sequence '("this" "[" "is" "a" "]" "test"))
(define (left-square-bracket? item)
(or (equal? item "[")
(eq? item #\x005b)))
(define (right-square-bracket? item)
(or (equal? item "]")
(eq? item #\x005d)))
(define (parse-sequence sequence)
(cond ((null? sequence) '())
((left-square-bracket? (car sequence))
(let ((subsequence (get-subsequence (cdr sequence))))
(list subsequence)))
(else
(cons (car sequence)
(parse-sequence (cdr sequence))))))
(define (get-subsequence sequence)
(if (right-square-bracket? (car sequence))
'()
(cons (car sequence)
(get-subsequence (cdr sequence)))))
Evaluating (parse-sequence my-sequence) yields ("this" ("is" "a")). A nested expression has been created, but the program finished without having evaluated the last item "test". The question is, how do I return from get-subsequence to parse-sequence?
Any help is appreciated, many thanks in advance! :)
To address your initial questions, how to return multiple values: use the "values" form. Here is an example implementation where the inner procedure returns both the remaining list to be processed and the result so far. It recurses on opening brackets.
(define (parse-sequence lst)
(define (parse-seq lst)
(let loop ((lst lst) (res null))
(cond
((null? lst) (values null res))
((string=? (car lst) "[")
(let-values ([(lst2 res2) (parse-seq (cdr lst))])
(loop lst2 (append res (list res2)))))
((string=? (car lst) "]")
(values (cdr lst) res))
(else
(loop (cdr lst) (append res (list (car lst))))))))
(let-values ([(lst res) (parse-seq lst)])
res))
then
(parse-sequence '("this" "is" "a" "test"))
(parse-sequence '("this" "[" "is" "a" "]" "test"))
(parse-sequence '("this" "[" "is" "[" "a" "]" "]" "test"))
will yield
'("this" "is" "a" "test")
'("this" ("is" "a") "test")
'("this" ("is" ("a")) "test")
I made some progress by using open-input-string in combination with read-char:
(define my-sequence (open-input-string "this [is a] test"))
(define (parse-sequence sequence)
`(item
,#(let loop ((next-char (read-char sequence)))
(cond ((eof-object? next-char) '())
((left-square-bracket? next-char)
(let ((subsequence (get-subsequence sequence)))
(cons subsequence
(loop (read-char sequence)))))
(else
(cons next-char
(loop (read-char sequence))))))))
(define (get-subsequence sequence)
`(subitem
,#(let loop ((next-char (read-char sequence)))
(if (right-square-bracket? next-char)
'()
(cons next-char
(loop (read-char sequence)))))))
(parse-sequence my-sequence)
===> (item #\t #\h #\i #\s #\space (subitem #\i #\s #\space #\a) #\space #\t #\e #\s #\t)
Now work goes on, step by step. :)
Any comments and suggestions are still appreciated. :)