parametric-polymorphism

Why does GHC infer a monomorphic type here, even with MonomorphismRestriction disabled?

本小妞迷上赌 提交于 2019-12-04 00:33:58
This was prompted by Resolving the type of `f = f (<*>) pure` , which discusses a more complicated example, but this one works too. The following definition compiles without problem: w :: Integral a => a w = fromInteger w ...Of course it doesn't work runtime-wise, but that's beside the question. The point is that the definition of w itself uses a specialised version of w :: Integer . Clearly that is a suitable instantiation, and therefore typechecks. However, if we remove the signature, then GHC infers not the above type, but only the concrete one: w' = fromInteger w' GHCi> :t w w :: Integral

difference between polymorphism and overloading

六眼飞鱼酱① 提交于 2019-12-03 21:35:54
I found there has many definition about polymorphism and overloading. Some people said that overloading is one type of polymorphism. While some people said they are not the same. Because only one function will be allocate in overloading. While the polymorphism need allocate the memory for each redefined member function. I really feel confusion about this, any one could explain this for me? Further, whether overloading happens at compile time while the polymorphism happens at running time? this was told by Bjarne Stroustrup in his book,polymorphism is one which a single thing that act in many

In Haskell, are “higher-kinded types” *really* types? Or do they merely denote collections of *concrete* types and nothing more?

筅森魡賤 提交于 2019-12-03 16:58:28
问题 Paramametrically polymorphic functions Consider the following function: f :: a -> Int f x = (1 :: Int) We might say that the type of f is a -> Int , and that f therefore is of a "polymorphic" type. Which of the following is the most accurate way to think about f ? There is in fact a single f of type a -> Int . It can be used, however, as an f :: Int -> Int , as an f :: Double -> Int , and so forth. Literally speaking, the type of f is NOT a -> Int . Indeed, that is just a shorthand way of

Scala: understanding parametric polymorphism

牧云@^-^@ 提交于 2019-12-03 16:53:27
What is the difference between def drop1[A](l: List[A]) = l.tail and def drop1(l: List[Int]) = l.tail provided the usage looks something like drop1(List(1,2,3)) ? When should one or the other be used and why? Whereas I can understand the second example, I don't really understand the purpose of the first one. It's very simple really. Your first example refers to the concept of generics . Generics have a simple goal, to make certain methods generic , e.g non-type dependant. Lets look at this trivial example. Say I want to write a drop1 method for List . I can either write one for every single

Polymorphism in OCaml - ad hoc,parametric, inclusion/subtyping

瘦欲@ 提交于 2019-12-03 16:21:53
I am having a problem understanding the different types of polymorphism, specifically in regards to OCaml. I understand that polymorphism allows for multiple types in OCaml denoted as 'a, but I do not understand what the different types of polymorphism are. If someone could give me an explanation with relatively low-level language that would be awesome! ad hoc, parametric, inclusion/subtyping Here's an approximation. Ad-hoc polymorphism usually refers to being able to declare the same name (usually a function) with different types, e.g. + : int -> int -> int and + : float -> float -> float in

Is there an automatic way to memoise global polymorphic values in Haskell?

早过忘川 提交于 2019-12-03 12:02:37
Polymorphic "constants", like 5 :: Num a => a , aren't really constants but functions of a dictionary argument. Hence, if you define primes :: Num n => [n] primes = ... Bad example of course, there's no good reason here to have it polymorphic... what I'm really interested is if you try to globally memoise a nontrivial polymorphic function, with e.g. memo-trie s. then this sequence won't be shared between calls from different sites, which isn't nice in terms of performance. (Isn't this the main reason the Haskell standard blessed us with the Dreaded Monomorphism Restriction?) The only way I can

Why is C++ said not to support parametric polymorphism?

北慕城南 提交于 2019-12-03 11:29:19
According to the wikipedia page for Parametric Polymorphism : Some implementations of type polymorphism are superficially similar to parametric polymorphism while also introducing ad hoc aspects. One example is C++ template specialization. Question: Why is C++ said to only implement something superficially similar to paramaterized polymorphism? In particular, aren't templates an example of full on parametric polymorphism? Why is C++ said to only implement something superficially similar to parameterized polymorphism? In particular, aren't templates an example of full on parametric polymorphism

What are type quantifiers?

久未见 提交于 2019-12-03 08:33:51
问题 Many statically typed languages have parametric polymorphism. For example in C# one can define: T Foo<T>(T x){ return x; } In a call site you can do: int y = Foo<int>(3); These types are also sometimes written like this: Foo :: forall T. T -> T I have heard people say "forall is like lambda-abstraction at the type level". So Foo is a function that takes a type (for example int), and produces a value (for example a function of type int -> int). Many languages infer the type parameter, so that

Scala converting recursively bounded type parameter (F-bounded) to type member

风格不统一 提交于 2019-12-03 08:24:16
问题 How would I convert: trait Foo[A <: Foo[A]] to a type member? I.e., I want something along the lines of the following: trait Foo { type A <: Foo {type A = ???} } but I am having difficulty because the name A is already taken within the type refinement. This question is similar (and spawned from): F-bounded quantification through type member instead of type parameter? 回答1: Use a self-type: scala> trait Foo { self => type A <: Foo {type A = self.A}} defined trait Foo scala> class Bar extends

In Haskell, are “higher-kinded types” *really* types? Or do they merely denote collections of *concrete* types and nothing more?

不想你离开。 提交于 2019-12-03 06:55:33
Paramametrically polymorphic functions Consider the following function: f :: a -> Int f x = (1 :: Int) We might say that the type of f is a -> Int , and that f therefore is of a "polymorphic" type. Which of the following is the most accurate way to think about f ? There is in fact a single f of type a -> Int . It can be used, however, as an f :: Int -> Int , as an f :: Double -> Int , and so forth. Literally speaking, the type of f is NOT a -> Int . Indeed, that is just a shorthand way of saying that there is a family of functions f whose type is concrete (i.e., there is an f :: Int -> Int ,