functor

This question already has an answer here: Why Functor class has no return function? 4 answers Reading this Wikibook about Haskell and Category Theory basics , I learn about Functors: A functor is essentially a transformation between categories, so given categories C and D, a functor F : C -> D maps any object A in C to F(A), in D. maps morphisms f : A -> B in C to F(f) : F(A) -> F(B) in D. ... which sounds all nice. Later an example is provided: Let's have a sample instance, too: instance Functor Maybe where fmap f (Just x) = Just (f x) fmap _ Nothing = Nothing Here's the key part: the type

I found that binary_function is removed from C++11. I am wondering why. C++98: template <class T> struct less : binary_function <T,T,bool> { bool operator() (const T& x, const T& y) const {return x<y;} }; C++11: template <class T> struct less { bool operator() (const T& x, const T& y) const {return x<y;} typedef T first_argument_type; typedef T second_argument_type; typedef bool result_type; }; MODIFIED ---------------------------------------------------------------------------- template<class arg,class result> struct unary_function { typedef arg argument_type; typedef result result_type; };

Apparently, every Arrow is a Strong profunctor. Indeed ^>> and >>^ correspond to lmap and rmap . And first' and second' are just the same as first and second . Similarly every ArrowChoice is also Choice . What profunctors lack compared to arrows is the ability to compose them. If we add composition, will we get an arrow? In other words, if a (strong) profunctor is also a category , is it already an arrow? If not, what's missing? What profunctors lack compared to arrows is the ability to compose them. If we add composition, will we get an arrow? MONOIDS This is exactly the question tackled in

How can I deduce statically if an argument is a C++ function object (functor)? template <typename F> void test(F f) {} I tried is_function<F>::value , but this doesn't work. It also seems there is no is_functor trait, so perhaps it's not possible. I appear to be only looking for a specific member function, in this case the function call operator: F::operator() . It is possible to create such a trait, with two restrictions: For the compiler, a free function is something fundamentally different from a class functor that overloads operator() . Thus we have to treat both cases seperately when

I Just can't seem to wrap my head around them. As I understand it's dynamically adding logic to a class. Are classes within the framework prepared for this? Why should I just extend the class and add the functionality to it in the extension. I would be globally accessible and afaik much easier to maintain. I've Read there are 4 functor types: Comparer Closure Predicate Transformer We should probably Handle each one of them. p.s. is there something like it in vb? So I can state I think that lambda expressions are functors. This clears up things for me a bit :) (hehe) Lambda expressions are

The type blows my mind: class Contravariant (f :: * -> *) where contramap :: (a -> b) -> f b -> f a Then I read this , but contrary to the title, I wasn't any more enlightened. Can someone please give an explanation of what a contravariant functor is and some examples? From a programmer's point of view the essence of functor-ness is being able to easily adapt things. What I mean by "adapt" here is that if I have an f a and I need an f b , I'd like an adaptor that will fit my f a in my f b -shaped hole. It seems intuitive that if I can turn an a into a b , that I might be able to turn a f a