多态函数中的GADT

Question

我有一个函数有问题，我想使其具有多态性。我想以分析或数字方式整合功能。在分析整合时，我提供结果。在数字情况下，我想使用各种方法，目前，tanhsinh 中的 tanhsinh 例程。我也想了解更多关于gadts的知识，所以我试图找到一个使用它们的解决方案。

到目前为止我有以下内容：

import qualified Data.VectorSpace as DV
import Numeric.Integration.TanhSinh

data IntegrationType a b  where
      MkAnalytic :: (DV.AdditiveGroup b) =>  (c -> b) -> c -> c -> IntegrationType Analytic b
      MkNumeric ::  NumericType  -> IntegrationType  Numeric [Result]

data Analytic = Analytic
data Numeric = Numeric
data Method = Trapez | Simpson
data IntBounds = Closed | NegInfPosInf | ZeroInf

data NumericType = MkSingleCoreTanhSinh  IntBounds Method  (Double -> Double)  Double Double
              | MkParallelTanhSinhExplicit IntBounds (Strategy [Double])  Method  (Double -> Double)  Double  Double
              | MkParallelTanhSinh IntBounds  Method  (Double -> Double)  Double  Double 

integrate :: IntegrationType a b -> b
integrate (MkAnalytic f l h) = f h DV.^-^ f l
integrate (MkNumeric (MkSingleCoreTanhSinh Closed Trapez f l h )) =  trap f l h
integrate (MkNumeric (MkSingleCoreTanhSinh Closed Simpson f l h )) =  simpson f l h

此代码可以编译，因为我在构造函数 MkNumeric 中明确声明类型变量 b 是

[Result]

为什么我必须这样做？我可以不保留类型变量 b 如

data IntegrationType a b  where
     MkNumeric ::  NumericType  -> IntegrationType  Numeric b

当我执行此操作时出现错误：

Could not deduce (b ~ [Result])
from the context (a ~ Numeric)
  bound by a pattern with constructor
             MkNumeric :: forall b. NumericType -> IntegrationType Numeric b,
           in an equation for `integrate'
  at test-classes-new-programm.hs:139:12-64
  `b' is a rigid type variable bound by
      the type signature for integrate :: IntegrationType a b -> b
      at test-classes-new-programm.hs:137:14
Relevant bindings include
  integrate :: IntegrationType a b -> b
    (bound at test-classes-new-programm.hs:138:1)
In the expression: trap f l h
In an equation for `integrate':
    integrate (MkNumeric (MkSingleCoreTanhSinh Closed Trapez f l h))
      = trap f l h

Answer 1

类型

integrate :: IntegrationType a b -> b

表示，对于我选择的 any a 和 b，如果我调用 integrate 类型为 IntegrationType a b，我会得到一个 b 类型的值。当你定义

MkNumeric ::  NumericType -> IntegrationType Numeric b

你让应用构造函数的人决定 b 是什么。所以我可以使用 MkNumeric 来制造一个值，比如说，类型 IntegrationType Numeric Int。但是你的 integrate 知道如何生成 [Result]，而不是 Int。

在working代码中，只要integrate"opens up the evidence box"通过匹配MkNumeric，它就知道b ~ [Result]和因此它可以 return 那种类型的东西。

多态函数中的GADT

GADT in polymorphic function

haskell

gadt