存在量词默默地破坏了模板Haskell（makeLenses）。为什么？ [英] Existential quantifier silently disrupts Template Haskell (makeLenses). Why?

问题描述

  { - ＃LANGUAGE TemplateHaskell＃ - } 
 { - ＃LANGUAGE ExistentialQuantification＃ - } 
 
模块玩具其中
 
导入Control.Lens 
 
数据Bar = Bar {_barish :: String} 
 data Foo =全部显示a => Foo {_fooish :: a} 
 
 $（makeLenses''Bar）
 $（makeLenses''Foo）
 
x = barish 
y = fooish 
  

我收到以下错误消息：

  Toy.hs：15：5：
不在范围内：`fooish'
也许你的意思是'_fooish'（第9行）
  

这是我第一次尝试使用存在量词;我不知道为什么这些功能的组合会打破。更令人担忧的是，为什么我不会收到有关makeLenses失败的错误消息？我跑了 runhaskell Toy.hs

解决方案

你的功能_fooish。如果你试图这样做，你会得到这样的错误：

$ $ $ pre $ $ code>由于转义类型，不能使用记录选择器_fooish作为函数变量
可能的修正：使用模式匹配语法来代替
在表达式中：_fooish

<所以镜头不能为你生成镜头。为什么它没有提供错误？那么，有时候你有更多的领域可以生成镜头。在这里看起来情况并非如此，但我认为通常makeLine只是跳过一些不可能做的事情，并尝试生成其余的东西。

I have this file:

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE ExistentialQuantification #-}

module Toy where

import Control.Lens

data Bar = Bar { _barish :: String }
data Foo = forall a. Show a => Foo { _fooish :: a }

$(makeLenses ''Bar)
$(makeLenses ''Foo)

x = barish
y = fooish

and I get the following error message:

Toy.hs:15:5:
    Not in scope: `fooish'
    Perhaps you meant `_fooish' (line 9)

This is my first time attempting to use existential quantifiers; I have no idea why this combination of features breaks. Even more worryingly, why do I get no error message about makeLenses failing? I ran runhaskell Toy.hs

解决方案

You can't actually use your function _fooish. If you try to do that, you get the error:

Cannot use record selector `_fooish' as a function due to escaped type variables
Probable fix: use pattern-matching syntax instead
In the expression: _fooish

So lens can't generate a lens for you. Why doesn't it give an error? Well, sometimes you have additional fields for which it's possible to generate lenses. It seems this not the case here, but I think in general makeLenses just skips everything that is impossible to do and tries to generate the rest.

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