How to express inheritance in Coq？

Content Index :

How to express inheritance in Coq？
Different notations to express inheritance
In Scala is there a nicer a way to express the following inheritance?
Multiple inheritance in C++: What is the good way to express the diagram in C++ with multiple inheritance?
Jade template inheritance without Express
OOP Inheritance in Express router

How to express inheritance in Coq？

Tag : development , By : Dasharath Yadav

Date : November 24 2020, 03:01 PM

I think the issue was by ths following , You seem to be asking about how to compute the closure of a function under repeated function application. The key to the problem is to find a way to ensure termination, i.e., a way to determine the maximum number of times the function might be called. In this case, an easy upper bound is List.length l; an element cannot have more transitive-parents than there are generations. Using this insight, we can define a function that takes a list of numbers, and outputs a list of those numbers together with all of their parents, and then we apply this function List.length l times to itself, starting with parents of c:

Require Import Coq.Lists.List. Import ListNotations.
Require Import Coq.Sorting.Mergesort. Import NatSort.
Scheme Equality for nat.
Inductive Gen : Set :=
| BGen : nat -> nat -> Gen.

Definition g1 := BGen 1 2.
Definition g2 := BGen 2 3.


Fixpoint parents (l : list Gen) (c : nat) :=
  match l with
  | [] => []
  | (BGen p c') :: l' => if nat_beq c c'
                         then [p]
                         else parents l' c
  end.

Fixpoint deduplicate' (ls : list nat) :=
  match ls with
  | [] => []
  | x :: [] => [x]
  | x :: ((y :: ys) as xs)
    => if nat_beq x y
       then deduplicate' xs
       else x :: deduplicate' xs
  end.
Definition deduplicate (ls : list nat) := deduplicate' (sort ls).

Definition parents_step (l : list Gen) (cs : list nat) :=
  deduplicate (cs ++ List.flat_map (parents l) cs).

Fixpoint all_parents' (l : list Gen) (cs : list nat) (fuel : nat) :=
  match fuel with
  | 0 => cs
  | S fuel'
    => all_parents' l (parents_step l cs) fuel'
  end.
Definition all_parents (l : list Gen) (c : nat) :=
  deduplicate (all_parents' l (parents l c) (List.length l)).

Definition gs := (g1::g2::nil).

Compute all_parents gs 3. (* [1; 2] *)

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How to express inheritance in Coq？ , Multiple inheritance in C++: What is the good way to express the diagram in C++ with multiple inheritance , Different notations to express inheritance , OOP Inheritance in Express router , c# inheritance ,

Different notations to express inheritance

Tag : generics , By : noboruwatanabe

Date : March 29 2020, 07:55 AM

I wish did fix the issue. Well, there is nothing stopping Scala from doing so, but, as a matter of fact, they do not express the same thing at all. And, in fact, you can see that in Java, where you can write X super Y, but you can't say class X super Y.
The keyword extends express a relationship between classes, one of inheritance. On the other hand, <: and >: express a relationship between types, one of boundaries. When I say X <: Y, then it is valid for both X and Y to be String, for example, while String extends String would be meaningless. It is also the case that List[String] <: List[AnyRef], though, again, List[String] extends List[AnyRef] is meaningless. And, just to make the point, it is not true that Set[String] <: Set[AnyRef]. In all these examples I just gave we are talking about the same class, but not, necessarily, about the same type.

In Scala is there a nicer a way to express the following inheritance?

Tag : scala , By : tanminivan

Date : March 29 2020, 07:55 AM

To fix this issue In my base class I have , For example,

def singleword(fOnStart: Start => Unit) =
  new EventReceiver {
    override def onStart(start: Start) { fOnStart(start) }
  }

singleword { start =>
  println("benchmark started")
}

Multiple inheritance in C++: What is the good way to express the diagram in C++ with multiple inheritance?

Tag : cpp , By : user98986

Date : March 29 2020, 07:55 AM

it helps some times This is pseudo code but should illustrate the hierarchy:
Solution 1:

class IBaseInterface {}

class Base : IBaseInterface {}

class Child1 : Base {}

class Child2 : Base {}

class IChildInterface {}

class Base {}

class Child1 : Base, IChildInterface  {}

class Child2 : Base, IChildInterface  {}

Jade template inheritance without Express

Tag : node.js , By : CodeOfficer

Date : March 29 2020, 07:55 AM

I wish this help you You don't need Express at all to use Jade's Template inheritance; you only need Jade:

// app.js
var jade = require('jade');

var options = { pretty: true, locals: {} };

jade.renderFile(__dirname + '/home.jade', options, function (err, html) {
    console.log(html);
});

// home.jade
extends core

block body
  h1 Home

// core.jade
doctype html
html
  head
    meta(charset='utf-8')
    title Foo
  body
    block body

OOP Inheritance in Express router

Tag : javascript , By : Mike

Date : March 29 2020, 07:55 AM

wish helps you The idiomatic way to do this with Express is to use middleware for the common bits and reuse that middleware in your routes. For example, this middleware can take care of parsing the project data:

function parseProject(req, res, next) {
  res.locals.project = JSON.parse(req.body.data);
  req.files.forEach((item) => {
    res.locals.project[item.fieldname] = '/' + item.path;
  })
  next();
}

const projectRouter = express.Router();
projectRouter
  .use(upload.any())
  .use(parseProject)
  .put('/:id', putProject)
  .post('/:id', postProject);

app.use('/projects', projectRouter);