combinators

问题 I'm trying to figure out how to terminate a repetition of words using a keyword. An example: class CAQueryLanguage extends JavaTokenParsers { def expression = ("START" ~ words ~ "END") ^^ { x => println("expression: " + x); x } def words = rep(word) ^^ { x => println("words: " + x) x } def word = """\w+""".r } When I execute val caql = new CAQueryLanguage caql.parseAll(caql.expression, "START one two END") It prints words: List(one, two, END) , indicating the words parser has consumed the END

问题 Is there some substitute of map which evaluates the list in parallel? I don't need it to be lazy. Something like: pmap :: (a -> b) -> [a] -> [b] letting me pmap expensive_function big_list and have all my cores at 100%. 回答1: Yes, see the parallel package: ls `using` parList rdeepseq will evaluate each element of the list in parallel via the rdeepseq strategy. Note the use of parListChunk with a good chunk value might give better performance if your elements are too cheap to get a benefit

问题 Suppose I have two functions f :: [a] -> b and g :: [a] -> c . I have the following two questions: If I perform (f &&& g) xs where xs :: [a] , and if both f and g involve loops, is it possible for the compiler to optimize these two loops into one? (Please note that I am not asking whether some specific Haskell compiler implements this. I want to know whether such a thing is possible .) Can the traverse function from Traverse type class help me have such an optimization with something along

问题 I've been using this json combinator for several basic / standard cases without really understanding how it works. All was fine. Now I want to get myself prepared for whatever advanced cases might come; I need to understand the code. Ref.: https://www.playframework.com/documentation/2.3.x/ScalaJsonCombinators I think I can understand the Reads: implicit val locationReads: Reads[Location] = ( (JsPath \ "lat").read[Double] and (JsPath \ "long").read[Double] )(Location.apply _) It creates a

问题 I want to make a CoffeeScript function that even if it is invoked multiple times, has its effects only run once. Is one of these, or another way a good way to make a once-invokable function ? Is the extra do an issue or actually better ? once_maker_a = (f)-> done=false -> f.call() unless done done=true once_maker_b = (f)-> do(done=false)-> -> f.call() unless done done=true oa = once_maker_a(-> console.log 'yay A') ob = once_maker_b(-> console.log 'yay B') oa() yay A #runs the function passed

问题 Hello i'm so new in the programming and want to learn some from you :) I'm doing a program in .c and i am stucked in a part. I want to get 3 or more inputs max size of 5 characters. (For example: HELLO, HI, GOOD, BYE) And i want to stack them in a new string which holds same letters only once from those 4 strings (Example: H,E,L,L,O,I,G,D,B,Y) #include <stdio.h> #include <string.h> int main(void) { char first[5], second[5], third[5], fourth[5]; printf("Enter 1st word: \n"); scanf(" %5s",

问题 Here's the code (also here): #lang racket (define poorY ((lambda length (lambda (ls) (cond [(null? ls) 0] [else (add1 ((length length) (cdr ls)))]))) (lambda length (lambda (ls) (cond [(null? ls) 0] [else (add1 ((length length) (cdr ls)))]))))) When I run it: > (poorY '(9 7 8)) . . application: not a procedure; expected a procedure that can be applied to arguments given: '(#<procedure>) arguments...: '(#<procedure>) The screenshot looks like this: I'm using DrRacket as the repl. What's wrong

问题 I am having some troubles with function composition and types. I would like to compose filter (which returns a list) with len , which takes a list as an argument (technically a Foldable but I am simplifying here). Looking at the types everything is as expected: > :t length length :: Foldable t => t a -> Int > :t filter filter :: (a -> Bool) -> [a] -> [a] So now I would expect the type of (len . filter) to be (length . filter) :: (a -> Bool) -> [a] -> Int while in reality is > :t (length .

问题 I'd like to know what this code means in Scheme: (define ((K x) y) x) (define (((S x) y) z) ((x z) (y z))) The whole file is here. Is this legal Scheme? Is (K x) a parametrized function, something like generic functions in Java? I looked up the MIT Scheme reference, there seems to be nothing mentioned for definition of this kind. 回答1: Trying it in MIT Scheme works (define ((K x) y) x) ;Value: k ((k 3) 4) ;Value: 3 Apparently, these are the definitions for K and S combinators from a

问题 Suppose I have an ambiguous language expressed in combinator parser. Is there a way to make certain expressions locally greedy? Here's an example of what I mean. import scala.util.parsing.combinator._ object Example extends JavaTokenParsers { def obj: Parser[Any] = (shortchain | longchain) ~ anyrep def longchain: Parser[Any] = zero~zero~one~one def shortchain: Parser[Any] = zero~zero def anyrep: Parser[Any] = rep(any) def any: Parser[Any] = zero | one def zero: Parser[Any] = "0" def one: