Haskell中的指针相等性？ [英] Pointer equality in Haskell?

问题描述

Haskell中有指针质量的概念吗？ ==需要东西导出Eq，并且我有一些包含（Value - > IO Value）的东西，并且既没有 - >也没有IO派生Eq。

Is there any notion of pointer quality in Haskell? == requires things to be deriving Eq, and I have something which contains a (Value -> IO Value), and neither -> nor IO derive Eq.

我正在创建另一种语言的解释器， 有指针相等，所以我试图模仿这种行为，同时仍然能够使用Haskell函数来建模闭包。

I'm creating an interpreter for another language which does have pointer equality, so I'm trying to model this behavior while still being able to use Haskell functions to model closures.

编辑：示例：我想要一个函数 special 可以这样做：

Example: I want a function special that would do this:

> let x a = a * 2
> let y = x
> special x y
True
> let z a = a * 2
> special x z
False

推荐答案

：根据您的示例，您可以使用IO monad对其进行建模。只需将你的函数分配给IORef并比较它们。

EDIT: Given your example, you could model this with the IO monad. Just assign your functions to IORefs and compare them.

Prelude Data.IORef> z <- newIORef (\x -> x)
Prelude Data.IORef> y <- newIORef (\x -> x)
Prelude Data.IORef> z == z
True
Prelude Data.IORef> z == y
False

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