为访问者记录语法默认值

问题描述

限时删除！！

当我刚刚时，我遇到了一个有趣的问题：

  data Gender = Male |女性
派生（方程式，显示）
 
数据年龄=婴儿|孩子| PreTeen |成人
派生（Eq，Show，Ord）
 
 data衣物=裤子性别年龄
 |衬衫性别年龄
 |裙子年龄 - 假定为女性
派生（显示，公式）
  

假设我希望使用记录语法编写最终数据类型：

  data衣服=裤子{gender :: Gender，age :: Age} 
 |衬衫{性别::性别，年龄::年龄} 
 |裙子{age :: Age} 
导出（Show，Eq）
  

问题是，我希望 gender $ skirt foo 始终评估为女（不管 foo ，这是一个年龄）。我可以想到几种方法来实现这一点，但他们要求我使用智能构造函数，理论上允许





1. >裙子男foo 但不暴露构造函数


2. 定义我自己的性别函数

使用＃1，通过不暴露模块中的构造函数，我可以有效地防止模块的用户利用记录语法。对于＃2，我必须完全放弃记录语法，或者定义一个额外的函数 gender'，这又会破坏记录语法。

有没有办法既利用记录语法，又为我的构造函数提供了一个默认，不可更改的值？我也接受非记录语法解决方案（镜头，也许是？），只要它们一样优雅（或更多）。

解决方案

如果没有令人信服的反例，答案似乎是否。

As I was writing up an answer just now, I ran across an interesting problem:

data Gender = Male | Female
            deriving (Eq, Show)

data Age = Baby | Child | PreTeen | Adult
         deriving (Eq, Show, Ord)

data Clothing = Pants Gender Age
              | Shirt Gender Age
              | Skirt Age         -- assumed to be Female
              deriving (Show, Eq)

Suppose I wish to write the final data type with record syntax:

data Clothing = Pants {gender :: Gender, age :: Age}
              | Shirt {gender :: Gender, age :: Age}
              | Skirt {age :: Age}
              deriving (Show, Eq)

The problem is, I want gender $ Skirt foo to always evaluate to Female (regardless of foo, which is an Age). I can think of a few ways to accomplish this, but they require that I either

1. use smart constructors, theoretically allowing Skirt Male foo but not exposing Constructors
2. define my own gender function

With #1, by not exposing the constructor in the module, I effectively prevent users of the module from taking advantage of record syntax. With #2, I have to forego record syntax entirely, or define an additional function gender', which again defeats record syntax.

Is there a way to both take advantage of record syntax, and also provide a "default", unchangeable value for one of my constructors? I am open to non-record-syntax solutions as well (lenses, perhaps?) as long as they are just as elegant (or moreso).

解决方案

In the absence of a convincing counterexample, the answer seems to be "no".

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