mutual-recursion | 易学教程

how to make HTML from a list in scheme, racket

阅读更多关于 how to make HTML from a list in scheme, racket

问题 This is a very long question ... I am new and joined, so please don't attack me. Apologies for my bad communications in English. I have some defintions: An HTML(H) is one of Str Tag A Tag is (cons Sym (listof H)) I want to use mutual recursion,make the HTML into real HTML code. For example, (list 'html (list 'head (list 'title "Hi")) (list 'body (list 'h1 "Welcome") "Text")) Turns into: "<html><head><title>Hi</title></head><body><h1>Welcome</h1>Text</body></html>" This should work for any

Mutually recursive evaluator in Haskell

阅读更多关于 Mutually recursive evaluator in Haskell

问题 Update: I've added an answer that describes my final solution (hint: the single Expr data type wasn't sufficient). I'm writing an evaluator for a little expression language, but I'm stuck on the LetRec construct. This is the language: type Var = String type Binds = [(Var, Expr)] data Expr = Var Var | Lam Var Expr | App Expr Expr | Con Int | Sub Expr Expr | If Expr Expr Expr | Let Var Expr Expr | LetRec Binds Expr deriving (Show, Eq) And this this the evaluator so far: data Value = ValInt Int

How to call two functions and use their results as arguments for each other?

阅读更多关于 How to call two functions and use their results as arguments for each other?

问题 I have code: let join a ~with':b ~by:key = let rec a' = link a ~to':b' ~by:key and b' = link b ~to':a' ~by:(Key.opposite key) in a' and compilation result for it is: Error: This kind of expression is not allowed as right-hand side of `let rec' build complete I can rewrite it to: let join a ~with':b ~by:key = let rec a'() = link a ~to':(b'()) ~by:key and b'() = link b ~to':(a'()) ~by:(Key.opposite key) in a'() It is compilable variant, but implemented function is infinitely recursive and it is

C++ Mutually Recursive Variant Type

阅读更多关于 C++ Mutually Recursive Variant Type

问题 I am trying to represent a PDF object type in c++ using variants. A PDF object is one of the following: Boolean Integer Real String Name Stream Array<Object> Map<Object, Object> As you can see, the Object type is mutually recursive because the Array type would require a declaration of the Map type which would require a declaration of the Array type. How could I go abouts representing this type in c++? If a variant isn't the best way, what is? Here is what I have tried so far but it doesn't

Mutual Recursion Question

阅读更多关于 Mutual Recursion Question

问题 How do I change two functions that are Mutual Recursive to each other to make them into a linear recursion? Do I have to have both the methods in a single method? 回答1: You should be able to simply "inline" the implementation of the second method, into the first method. That is, public static void methA() { // snippet 1 methB(); // snippet 2 } public static void methB() { // snippet 3 methA(); // snippet 4 } becomes public static void methAB() { // snippet 1 // snippet 3 methAB(); // snippet 2

Mutual Recursion Question

阅读更多关于 Mutual Recursion Question

Is it possible to define types that depend on each other and are defined in separated files?

阅读更多关于 Is it possible to define types that depend on each other and are defined in separated files?

问题 I am trying to implement a library with extended parsing capabilities. I decided that I will use fsyacc because I knew it from the university. Unfortunately I encountered following problem. I defined a class for the head of my grammar ( Head ) and place its implementation in a single file. Then I defined parser as: ... %start head %type <Head> head ... Fsyacc generates seeparated module ( Parser ). In order to succeed it has to be compiled in following order: Head.fs Parser.fs In order to

What is the standard way to optimise mutual recursion in F#/Scala?

阅读更多关于 What is the standard way to optimise mutual recursion in F#/Scala?

问题 These languages do not support mutually recursive functions optimization 'natively', so I guess it must be trampoline or.. heh.. rewriting as a loop) Do I miss something? UPDATE: It seems that I did lie about FSharp, but I just didn't see an example of mutual tail-calls while googling 回答1: First of all, F# supports mutually recursive functions natively, because it can benefit from the tailcall instruction that's available in the .NET IL (MSDN). However, this is a bit tricky and may not work

How does python implement mutual recursion?

阅读更多关于 How does python implement mutual recursion?

问题 Moving to python with C/Java background, I recently had to implement a mutual recursion, but something in python is bothering me: since a python program is interpreted line by line, if I have two functions one after another in the same python file: def A(n): B(n-1) # if I add A(1) here, it gives me an error def B(n): if n <= 0: return else: A(n-1) When the interpreter is reading A , B is not yet defined, however this code does not give me an error TL;DR My understanding is that, when def is

Unable to understand a mutual recursion

阅读更多关于 Unable to understand a mutual recursion

问题 I am reading Programming In Haskell, in the 8th chapter, the author gives an example of writing parsers. The full source is here: http://www.cs.nott.ac.uk/~gmh/Parsing.lhs I can't understand the following part: many permits zero or more applications of p , whereas many1 requires at least one successful application: many :: Parser a → Parser [a ] many p = many1 p +++ return [ ] many1 :: Parser a → Parser [a ] many1 p = do v ← p vs ← many p return (v : vs) How the recursive call happens at vs <