I am writing a program which needs communication between processes.
my code:
#lang racket
(define-values (sp o i e) (subprocess #f #f #f "c://player1.exe" ))
(define count 10)
(for ([c (in-naturals)])
(cond
[(equal? count 0) (error "Province is empty!") ]
[else
(write "server" i)
(set! count (sub1 count))
(flush-output i)
(display (read o))]))
and the player1.exe code:
#lang racket
(define (interact notification)
(cond
[(eq? notification "server") (write "true" (current-output-port))]
[else (write "false" (current-output-port))]))
(for ([c (in-naturals)])
(interact (read (current-input-port)))
(write "player" (current-output-port))
(sleep 0.1)
flush-output (current-output-port))
I am getting output if I run without loops. I am also getting output when only player is sending messages. But with both server and player sending messages the program gets hanged.
What do you think the problem is?
The last line in your player1.exe file looks suspicious. flush-output is not actually being applied as a function. Rather than
flush-output (current-output-port)
you probably mean:
(flush-output (current-output-port))
From a style point of view: the functions read, write, and flush-output all work on the current input and output ports by default, so you don't need to provide them. Take a look at the documentation for those functions, such as flush-output, and you'll see that it mentions that the current-output-port is its default.
So the line that we just looked at can be written as:
(flush-output)
More issues: don't use eq? to compare strings. Use string=? instead. The reason is that there can be two strings that have the same textual content, but for which eq? will still be able to distinguish the two. e.g.:
kui $ racket
Welcome to Racket v5.2.1.
> (eq? "a" (string-copy "a"))
#f
> (string=? "a" (string-copy "a"))
#t
Related
I'm attempting to program a simple function that adds integers to a list descending from a range of "high" and "low", incremented by "step"
For example,
if the input is (3 12 3), the expected output is '(12 9 6 3)
Below is the following code:
(define (downSeries step high low [(define ret '())])
(if (< high low)
ret
(cons ret (- high step))
(downSeries (step (- high step) low))))
I'm pretty new to racket, but I'm really not sure why this isn't compiling. Any tips? Thank you.
Since only racket is tagged and no special languages are describes it is expeted the first line in the definition window is #lang racket. Answer will be different for student languages.
1 The last argument is nested in two parentheses and is illegal syntax. Default arguments only have one set:
(define (test mandatory (optional '()))
(list mandatory optional))
(test 1) ; ==> (1 ())
(test 1 2) ; ==> (1 2)
2 You have 4 operands in your if form. It takes maximum 3!
(if prediate-expression
then-expression
else-expression)
Looking at the code you should have the cons expression in the position of ret argument. Having it before the recursion makes it dead code. ret will always be (). Eg this loks similar to a typical fold implementation:
(define (fold-1 combine init lst)
(if (null? lst)
init ; fully grown init returned
(fold-1 combine
(combine (car lst) init) ; init grows
(cdr lst))))
I'm reading a file char by char and constructing a list which is consist of list of letters of words. I did that but when it comes to testing it prints out NIL. Also outside of test function when i print out list, it prints nicely. What is the problem here? Is there any other meaning of LET keyword?
This is my read fucntion:
(defun read-and-parse (filename)
(with-open-file (s filename)
(let (words)
(let (letter)
(loop for c = (read-char s nil)
while c
do(when (char/= c #\Space)
(if (char/= c #\Newline) (push c letter)))
do(when (or (char= c #\Space) (char= c #\Newline) )
(push (reverse letter) words)
(setf letter '())))
(reverse words)
))))
This is test function:
(defun test_on_test_data ()
(let (doc (read-and-parse "document2.txt"))
(print doc)
))
This is input text:
hello
this is a test
You're not using let properly. The syntax is:
(let ((var1 val1)
(var2 val2)
...)
body)
If the initial value of the variable is NIL, you can abbreviate (varN nil) as just varN.
You wrote:
(let (doc
(read-and-parse "document2.txt"))
(print doc))
Based on the above, this is using the abbreviation, and it's equivalent to:
(let ((doc nil)
(read-and-parse "document2.txt"))
(print doc))
Now you can see that this binds doc to NIL, and binds the variable read-and-parse to "document2.txt". It never calls the function. The correct syntax is:
(let ((doc (read-and-parse "document2.txt")))
(print doc))
Barmar's answer is the right one. For interest, here is a version of read-and-parse which makes possibly-more-idiomatic use of loop, and also abstracts out the 'is the character white' decision since this is something which is really not usefully possible in portable CL as the standard character repertoire is absurdly poor (there's no tab for instance!). I'm sure there is some library available via Quicklisp which deals with this better than the below.
I think this is fairly readable: there's an outer loop which collects words, and an inner loop which collects characters into a word, skipping over whitespace until it finds the next word. Both use loop's collect feature to collect lists forwards. On the other hand, I feel kind of bad every time I use loop (I know there are alternatives).
By default this collects the words as lists of characters: if you tell it to it will collect them as strings.
(defun char-white-p (c)
;; Is a character white? The fallback for this is horrid, since
;; tab &c are not a standard characters. There must be a portability
;; library with a function which does this.
#+LispWorks (lw:whitespace-char-p c)
#+CCL (ccl:whitespacep c) ;?
#-(or LispWorks CCL)
(member char (load-time-value
(mapcan (lambda (n)
(let ((c (name-char n)))
(and c (list c))))
'("Space" "Newline" "Page" "Tab" "Return" "Linefeed"
;; and I am not sure about the following, but, well
"Backspace" "Rubout")))))
(defun read-and-parse (filename &key (as-strings nil))
"Parse a file into a list of words, splitting on whitespace.
By default the words are returned as lists of characters. If
AS-STRINGS is T then they are coerced to strings"
(with-open-file (s filename)
(loop for maybe-word = (loop with collecting = nil
for c = (read-char s nil)
;; carry on until we hit EOF, or we
;; hit whitespace while collecting a
;; word
until (or (not c) ;EOF
(and collecting (char-white-p c)))
;; if we're not collecting and we see
;; a non-white character, then we're
;; now collecting
when (and (not collecting) (not (char-white-p c)))
do (setf collecting t)
when collecting
collect c)
while (not (null maybe-word))
collect (if as-strings
(coerce maybe-word 'string)
maybe-word))))
I'm trying to print a list of pairs of values (representing key/value pairs) in Racket.
Here's the code I have right now:
#lang racket
(define (write-json keyvalues)
(displayln "{")
(for-each
(lambda (kv) (
(displayln (format "~a: ~a," (car kv) (cdr kv)))))
keyvalues)
(displayln "}"))
(write-json (list (cons "a" 1) (cons "b" 2)))
When I run the example, it prints:
{
a: 1,
Then, it crashes with this error message:
application: not a procedure;
expected a procedure that can be applied to arguments
given: #<void>
arguments...: [none]
context...:
/.../racket/collects/racket/private/map.rkt:58:19: loop
"test.rkt": [running body]
for-loop
run-module-instance!125
perform-require!78
Any idea what's going on?
Thanks!
This is a paren issue. You have an extra set of parentheses around your lambda's body, ie:
( (displayln (format "~a: ~a;" (car kv) (cdr kv))) )
Since displayln is used for side effect, its output is void, hence why your error message states that you're trying to run (#<void>).
In general, whenever you get an error stating "expected a procedure that can be applied to arguments", see if you have parentheses issues in your code block. Editors like Dr. Racket would highlight that region for you.
I am working on problem to get the occurence of Prime in a list in lisp.
Input:
Write a function (OccurencesOfPrimes < list >) which counts the number of primes in a (possibly nested) list.
Output: Example: (OccurencesOfPrimes (((1)(2))(5)(3)((8)3)) returns 4.
I am using the below code but getting the error like:
(
defun OccurencesOfPrimes (list)
(loop for i from 2 to 100
do ( setq isPrime t)
(loop for j from 2 to i
never (zerop (mod i j))
(setq isPrime f)
(break)
)
)
(if (setq isPrime t)
(append list i)
)
)
)
LOOP: illegal syntax near (SETQ ISPRIME F) in
(LOOP FOR J FROM 2 TO I NEVER (ZEROP (MOD I J)) (SETQ ISPRIME F) (BREAK)
)
Any help.
It is important to keep the format consistent with the expected conventions of the language. It helps when reading the code (in particular with other programmers), and can help you see errors.
Also, you should use an editor which, at the minimum, keep tracks of parentheses. In Emacs, when you put the cursor in the first opening parenthesis, the matching parenthesis is highlighted. You can spot that you have one additional parenthesis that serves no purpose.
(
defun OccurencesOfPrimes (list)
(loop for i from 2 to 100
do ( setq isPrime t)
(loop for j from 2 to i
never (zerop (mod i j))
(setq isPrime f)
(break)
)
)
(if (setq isPrime t)
(append list i)
)
) ;; <- end of defun
) ;; <- closes nothing
In Lisp, parentheses are for the computer, whereas indentation is for humans. Tools can automatically indent the code according to the structure (the parenthesis), and any discrepancy between what indentation you expect and the one being computed is a hint that your code is badly formed. If you look at the indentation of your expressions, you can see how deep you are in the form, and that alone helps you understand the code.
Symbol names are dash-separated, not camlCased.
Your code, with remarks:
(defun occurences-of-primes (list)
;; You argument is likely to be a LIST, given its name and the way
;; you call APPEND below. But you never iterate over the list. This
;; is suspicious.
(loop
for i from 2 to 100
do
(setq is-prime t) ;; setting an undeclared variable
(loop
for j from 2 to i
never (zerop (mod i j))
;; the following two forms are not expected here according
;; to LOOP's grammar; setting IS-PRIME to F, but F is not
;; an existing variable. If you want to set to false, use
;; NIL instead.
(setq is-prime f)
;; BREAK enters the debugger, maybe you wanted to use
;; LOOP-FINISH instead, but the NEVER clause above should
;; already be enough to exit the loop as soon as its
;; sub-expression evaluates to NIL.
(break)))
;; The return value of (SETQ X V) is V, so here your test would
;; always succeed.
(if (setq is-prime t)
;; Append RETURNS a new list, without modifying its
;; arguments. In particular, LIST is not modified. Note that "I"
;; is unknown at this point, because the bindings effective
;; inside the LOOP are not visible in this scope. Besides, "I"
;; is a number, not a list.
(append list i)))
Original question
Write one function which counts all the occurrences of a prime number in a (possibly nested) list.
Even though the homework questions says "write one function", it does not say that you should write one big function that compute everything at once. You could write one such big function, but if you split your problem into sub-problems, you will end with different auxiliary functions, which:
are simpler to understand (they do one thing)
can be reused to build other functions
The sub-problems are, for example: how to determine if a number is a prime? how to iterate over a tree (a.k.a. a possibly nested list)? how to count
the occurrences?
The basic idea is to write an "is-prime" function, iterate over the tree and call "is-prime" on each element; if the element is prime and was never seen before, add 1 to a counter, local to your function.
You can also flatten the input tree, to obtain a list, then sort the resulting
list; you iterate over the list while keeping track of the last
value seen: if the value is the same as the previous one, you
already know if the number is prime; if the previous number differs, then
you have to test if the number is prime first.
You could also abstract things a little more, and define a higher-order tree-walker function, which calls a function on each leaf of the tree. And write another higher-order function which "memoizes" calls: it wraps around a
function F so that if you call F with the same arguments as before,
it returns the result that was stored instead of recomputing it.
Example
I'll combine the above ideas because if you give that answer to a teacher you are likely to have to carefully explain what each part does (and if you can, great for you); this is not necessarily the "best" answer, but it covers a lot of things.
(defun tree-walk-leaves (tree function)
(typecase tree
(null nil)
(cons
(tree-walk-leaves (car tree) function)
(tree-walk-leaves (cdr tree) function))
(t (funcall function tree))))
(defun flatten (tree &optional keep-order-p)
(let ((flat nil))
(tree-walk-leaves tree (lambda (leaf) (push leaf flat)))
(if keep-order-p
(nreverse flat)
flat)))
(defun prime-p (n)
(or (= n 2)
(and (> n 2)
(oddp n)
(loop
for d from 3 upto (isqrt n) by 2
never (zerop (mod n d))))))
(defun count-occurences-of-prime (tree)
(count-if #'prime-p (remove-duplicates (flatten tree))))
(count-occurences-of-prime '(((1)(2))(5)(3)((8)3)))
=> 4
If, instead, you don't want to remove duplicates but count the multiple times a prime number occurs, you can do:
(count-if (memoize #'prime-p) (flatten tree))
... where memoize is:
(defun memoize (function &key (test #'equalp) (key #'identity))
(let ((hash (make-hash-table :test test)))
(lambda (&rest args)
(let ((args (funcall key args)))
(multiple-value-bind (result exists-p) (gethash args hash)
(values-list
(if exists-p
result
(setf (gethash args hash)
(multiple-value-list (apply function args))))))))))
(memoize is useless if there are no duplicates)
I have the following Racket code:
#lang racket
(define (for-each proc items)
(cond ((not (null? items))
(proc (car items))
(for-each proc (cdr items)))))
(for-each (lambda (x) (newline) (display x))
(list 57 321 88))
Because the condition isn't satisfied after 88 is printed, and the Racket's documentation says:
If none of the question-expressions evaluates to true, cond’s value is the answer-expression of the else clause. If there is no else, cond reports an error.
So the expected result is an error, however I get the correct output:
57
321
88
Can somebody tell me why? (ps: I'm using DrRacket 6.2)
Changing the indentation a little:
(define (for-each proc items)
(cond
[(not (null? items)) (proc (car items))
(for-each proc (cdr items))]))
A call (for-each f (list 1 2 3))
will call the function and bind proc to f and items to '(1 2 3).
The test (not (null? items)) is true, since '(1 2 3) is non-empty.
The right hand side is therefore evaluated. The right hand side does two things:
1.) (proc (car items)) calls f with the first element in the list (here 1) and then 2). calls (for-each proc (cdr items)). Here (cdr items) becomes '(2 3).
Eventually for-each will be called with (for-each f '()).
At that point (not (null? items)) will evaluate to #f and the cond will try the next clause. Since there are no more clauses cond will return void.
The documentation says:
If no cond-clauses are present, the result is #<void>.
Note: The cond in the teaching languages will return an error. The constructs in the teaching language generally reports errors, where there is a chance of shooting ones own foot.
I don't know where you got that information. To quote the spec:
The last test-expr in a cond can be replaced by else. In terms of evaluation, else serves as a synonym for #t, but it clarifies that the last clause is meant to catch all remaining cases. If else is not used, then it is possible that no test-exprs produce a true value; in that case, the result of the cond expression is #void.