“模式匹配"指的是“模式匹配".Haskell中的类型 [英] "Pattern-matching" types in Haskell

问题描述

我想要一个函数 f ，该函数在 Int 上充当增量，并在所有其他类型上充当身份.我试图启用 TypeApplications 扩展名，并以最简单的方式做到这一点:

I want a function f that acts as an increment on Ints and as an identity on all the other types. I tried to enable the TypeApplications extension and do this the most straight-forward way possible:

f :: forall a. a -> a
f @Int x = x + 1
f @_   x = x

但是该扩展程序不启用此类模式:

But the extension does not enable such patterns:

<interactive>:6:1: error: Parse error in pattern: f @Int

Haskell中是否存在类型(或类似机制)的模式匹配?

Is there pattern-matching for types (or a similar mechanism) in Haskell?

推荐答案

如上所述，这在Haskell中是不可能实现的，因为它会违反称为"parametricity"的参数多态性的基本属性之一，从而确保了任何多态函数满足称为自由定理"的特定属性.

As stated, this is impossible to achieve in Haskell since it would violate one of the fundamental properties of parametric polymorphism known as "parametricity", which ensures that any polymorphic function satisfies a specific property knows as the "free theorem".

特别是类型为 forall a的终止函数.a->必须是身份.没有其他可能的实现方式.

In particular, a terminating function of type forall a. a -> a must be the identity. There are no other possible implementations.

话虽这么说，如果我们允许对 a 类型的约束，则这成为可能.通常，运行时类型测试是通过 Typeable 类型类在Haskell中完成的:

That being said, if we allow a constraint on type a, this becomes possible. Typically, run-time type-testing is done in Haskell through the Typeable type class:

f :: forall a. Typeable a => a -> a
f x = case eqT @a @Int of    -- Is a ~ Int ?
   Just Refl -> x + 1        -- If it is, we can add one.
   Nothing   -> x            -- Otherwise, return x as-is.

这需要大量GHC扩展，并导入 Data.Typeable .

This requires a bunch of GHC extensions, and to import Data.Typeable.

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