functor

Why are Promises Monads?

倾然丶 夕夏残阳落幕 提交于 2019-11-27 04:05:29
I've been learning about functional programming and have come across Monads, Functors and Applicatives. From my understanding the following definitions apply: a) ( A=>B ) => C[A] => C[B] | Functor b) ( A=>C[B] ) => C[A] => C[B] | Monad c) ( C[A=>B] ) => C[A] => C[B] | Applicative (reference: https://thedet.wordpress.com/2012/04/28/functors-monads-applicatives-can-be-so-simple/ ) Furthermore, I understand a Monad is a special case of a Functor. As in, it applies a function that returns a wrapped value to a wrapped value and returns a wrapped value. When we use Promise.then(func) , we are

In Functional Programming, what is a functor?

僤鯓⒐⒋嵵緔 提交于 2019-11-27 02:28:58
I've come across the term 'Functor' a few times while reading various articles on functional programming, but the authors typically assume the reader already understands the term. Looking around on the web has provided either excessively technical descriptions (see the Wikipedia article ) or incredibly vague descriptions (see the section on Functors at this ocaml-tutorial website ). Can someone kindly define the term, explain its use, and perhaps provide an example of how Functors are created and used? Edit : While I am interested in the theory behind the term, I am less interested in the

Generic functor for functions with any argument list

◇◆丶佛笑我妖孽 提交于 2019-11-27 02:01:57
问题 I need to implement a functor that takes any (!) function pointer when instantiated, analyses the argument types, stores the pointer and when operator() is called, does something with the pointer. The easiest case being, calling the function with its arguments. I tried converting the function pointer to something like std::function and I get the error: error: invalid use of incomplete type ‘struct std::function<void (*)(...)>’ /usr/include/c++/4.6/functional:1572:11: error: declaration of

why a js map on an array modify the original array?

送分小仙女□ 提交于 2019-11-27 00:34:13
问题 I'm quite confuse by the behaviour of map(). I have an array of objects like this : const products = [{ ..., 'productType' = 'premium', ... }, ...] and i'm passing this array to a function that should return the same array but with all product made free : [{ ..., 'productType' = 'free', ... }, ...] The function is : const freeProduct = function(products){ return products.map(x => x.productType = "free") } Which returns the following array : ["free", "free", ...] So i rewrote my function to be

Are Functor instances unique?

老子叫甜甜 提交于 2019-11-26 23:03:25
问题 I was wondering to what extent Functor instances in Haskell are determined (uniquely) by the functor laws. Since ghc can derive Functor instances for at least "run-of-the-mill" data types, it seems that they must be unique at least in a wide variety of cases. For convenience, the Functor definition and functor laws are: class Functor f where fmap :: (a -> b) -> f a -> f b fmap id = id fmap (g . h) = (fmap g) . (fmap h) Questions: Can one derive the definition of map starting from the

How do I get the argument types of a function pointer in a variadic template class?

放肆的年华 提交于 2019-11-26 22:05:57
This is a follow up of this problem: Generic functor for functions with any argument list I have this functor class (full code see link above): template<typename... ARGS> class Foo { std::function<void(ARGS...)> m_f; public: Foo( std::function<void(ARGS...)> f ) : m_f(f) {} void operator()(ARGS... args) const { m_f(args...); } }; In operator() I can access the args... easily with a recursive "peeling" function as described here http://www2.research.att.com/~bs/C++0xFAQ.html#variadic-templates My problem is: I want to access the types of the arguments of f, i.e. ARGS..., in the constructor.

Stateful functors & STL : Undefined behaviour

久未见 提交于 2019-11-26 21:20:23
问题 I am following this Function objects tutorial Copy-pasta below: I am unable to understand the following: Predicates should always be implemented as stateless function objects to avoid unexpected results. There is no guarantee how often an algorithm might internally copy the predicate. Thus, having predicates that are implemented as stateful function objects might have unexecpted results. The example is as follows: #include <iostream> #include <vector> #include <algorithm> #include <iterator>

How do functors work in haskell?

对着背影说爱祢 提交于 2019-11-26 19:26:01
I'm trying to learn Haskell and I'm through all the basics. But now I'm stuck, trying to get my head around functors. I've read that "A functor transforms one category into another category". What does this mean? I know it's a lot to ask, but could anyone give me a plain english explanation of functors or maybe a simple use case ? A fuzzy explanation would be that a Functor is some sort of container and an associated function fmap that allows you to alter whatever is contained, given a function that transforms the contained. For instance, lists are this kind of container, such that fmap (+1)

Monads as adjunctions

南笙酒味 提交于 2019-11-26 18:49:25
问题 I've been reading about monads in category theory. One definition of monads uses a pair of adjoint functors. A monad is defined by a round-trip using those functors. Apparently adjunctions are very important in category theory, but I haven't seen any explanation of Haskell monads in terms of adjoint functors. Has anyone given it a thought? 回答1: Edit : Just for fun, I'm going to do this right. Original answer preserved below The current adjunction code for category-extras now is in the

Why use functors over functions?

僤鯓⒐⒋嵵緔 提交于 2019-11-26 18:26:00
Compare double average = CalculateAverage(values.begin(), values.end()); with double average = std::for_each(values.begin(), values.end(), CalculateAverage()); What are the benefits of using a functor over a function? Isn't the first a lot easier to read (even before the implementation is added)? Assume the functor is defined like this: class CalculateAverage { private: std::size_t num; double sum; public: CalculateAverage() : num (0) , sum (0) { } void operator () (double elem) { num++; sum += elem; } operator double() const { return sum / num; } }; Oliver Charlesworth At least four good