functor Questions
3
Solved
Control.Lens.Tutorial says:
type Traversal' a b = forall f . Applicative f => (b -> f b) -> (a -> f a)
type Lens' a b = forall f . Functor f => (b -> f b) -> (a -> f a)
...
Avulsion asked 21/2 at 14:5
3
Solved
During play around objective package, I noticed following type has interesting property.
> {-# LANGUAGE RankNTypes #-}
> data N f r = N { unN :: forall x. f x -> (x, r) }
It is a Functo...
Compress asked 29/4, 2016 at 3:23
6
I am getting into Haskell and found the book "learn you a Haskell" most helpful. I am up to the section on applicative functors.
I am puzzled by the following as it appears in the book:
(\x y z ...
Separator asked 27/1, 2014 at 8:31
1
Recall the Applicative class:
class Functor f => Applicative f where
pure :: a -> f a
liftA2 :: (a -> b -> c) -> f a -> f b -> f c
(<*>) :: f (a -> b) -> f a -&g...
Rybinsk asked 26/8, 2023 at 6:28
5
Solved
The free monoids are often being regarded as "list monoids". Yet, I am interested in other possible structures which might give us free monoids.
Firstly, let us go over the definition of ...
Sheya asked 16/9, 2020 at 20:57
19
Solved
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...
Deroo asked 8/1, 2010 at 21:26
3
Solved
After looking up the Control.Monad documentation, I'm confused about
this passage:
The above laws imply:
fmap f xs = xs >>= return . f
How do they imply that?
5
Solved
Suppose I have a method def doSomething: String which can raise a DoSomethingException if something goes wrong.
If I write Try(doSomething), is there a simple way to map the exception without reco...
2
std::transform from the <algorithm> header applies to ranges, and it is what "enables" us to use ranges as the functors they are (in the sense of category theory(¹)). std::transform...
Igloo asked 9/1, 2022 at 9:58
8
Solved
Why is operator () of stateless functor not allowed to be static? Stateless lambda objects are convertible to pointers to free functions having the same signature as their operator ().
Stephan T. ...
Armory asked 25/9, 2015 at 6:53
5
Solved
From categorical point of view, functor is pair of two maps (one between objects and another between arrows of categories), following some axioms.
I have assumed, what every Functor instance is si...
Stipulation asked 8/2, 2014 at 15:12
2
Solved
I was kinda surprised when I read the source code of instances of Applicative Complex and Monad Complex from GHC Data.Complex module:
-- | @since 4.9.0.0
instance Applicative Complex where
pure a...
Stclair asked 8/11, 2017 at 13:32
4
This has me stumped. How do you write a Functor instance for newtype Mu f = InF {outF :: f (Mu f)}
11
Solved
I'm trying to use a C library in a C++ app and have found my self in the following situation (I know my C, but I'm fairly new to C++). On the C side I have a collection of functions that takes a fu...
1
Solved
I started to learn Prolog and I just read that the atom at the beginning of an structure is usually called functor.
I'm also familiar with the term functor from Category Theory and Functional Progr...
Assertive asked 27/5, 2022 at 21:19
1
Let's say I have a type that looks like this:
data Term a = Terminal a
| Application (Term a) (Term a)
| Abstraction String (Term a)
Now, I want to map Term a to Term b. Ideally, I would be able t...
Kumasi asked 14/5, 2022 at 22:55
12
Solved
In the Boost Signals library, they are overloading the () operator.
Is this a convention in C++? For callbacks, etc.?
I have seen this in code of a co-worker (who happens to be a big Boost fan). ...
Nuggar asked 25/11, 2008 at 14:7
5
Solved
Why does the for_each call on functor doesn't update sum::total at the end?
struct sum
{
sum():total(0){};
int total;
void operator()(int element)
{
total+=element;
}
};
int main()
{
s...
5
Solved
I'm quite confused by the behavior of map().
I have an array of objects like this:
const products = [{
...,
'productType' = 'premium',
...
}, ...]
And I'm passing this array to a function that ...
Mihrab asked 10/3, 2016 at 16:46
1
Solved
I recently discovered how to simulate higher order types in Java in a somewhat roundabout way like so
interface H<F, T> { }
Here H encodes a higher order type that takes a type parameter F w...
Gerhan asked 12/1, 2022 at 12:19
2
Solved
What is the point of making Functor a super class of Applicative and Monad. Both Applicative and Monad immediately imply the only implementation of Functor that abides by the laws as far as I...
Hipster asked 2/1, 2022 at 7:7
5
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 s...
Kizzie asked 31/1, 2012 at 16:56
2
Bifunctors have a map function with this signature:
bimap :: (a -> b) -> (c -> d) -> p a c -> p b d
You could also have a map like this:
othermap :: ((a, c) -> (b, d)) -> p a...
Jerboa asked 28/12, 2021 at 0:6
1
Solved
I recently defined a type whose fields I might fail to compute:
data Foo = Foo {x, y :: Int, others :: NonEmpty Int}
data Input
computeX, computeY :: Input -> Maybe Int
computeOthers :: Input ...
Tabard asked 7/12, 2021 at 10:21
3
Solved
There are similar questions here but they are attached to a particular programming language and I am looking for an answer on the conceptual level.
As I understand, Functors are essentially immuta...
Cnidus asked 22/7, 2017 at 9:3
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