了解Haskell的布尔派生的奥德 [英] Understanding Haskell's Bool Deriving an Ord

问题描述

了解你哈斯克尔 presents的布尔类型：

Learn You a Haskell presents the Bool type:

数据布尔=假|真正的推导（ORD）

我不明白其中的道理比较布尔的

I don't understand the reason for comparing Bool's.

> False `compare` True
LT
> True `compare` False
GT

如果布尔并没有从奥德？

推荐答案

布尔形成的界格 *，其中假是底和真是最常见的。这界格的定义（总）订购，其中假真的是严格小于真少。 （他们也是这个格子的唯一元素。）

Bool forms a bounded lattice* where False is bottom and True is top. This bounded lattice defines a (total) ordering where False really is strictly less than True. (They are also the only elements of this lattice.)

布尔运算和和或，也可以在看起来的满足的和加入的，分别在这个格。符合认定的最大下界，并加入找到最上限。这意味着 A和＆安培;假=假是同样的事情的话称，下界底部和其他任何的底部， A ||真= TRUE 是同样的事情的话称，上限顶什么是顶部。因此，满足并加入，其中使用布尔值的顺序性，相当于你所熟悉的布尔运算。

The boolean operations and and or can also be looked at as meet and join, respectively, in this lattice. Meet finds the greatest lower bound and join finds the least upper bound. This means that a && False = False is the same thing as saying that the lower bound of bottom and anything else is bottom, and a || True = True is the same thing as saying that the upper bound of top and anything is top. So meet and join, which use the ordering property of the booleans, are equivalent to the boolean operations you are familiar with.

您可以使用分和最大来显示这个在Haskell：

You can use min and max to show this in Haskell:

False `min` True = False -- this is the greatest lower bound
False  &&   True = False -- so is this

False `max` True = True  -- this is the least upper bound
False  ||   True = True  -- so is this

这表明，你可以定义＆放大器;＆安培; 和 || 的只是的从派生奥德实例：

This shows that you can define && and || just from the derived Ord instance:

(&&) = min
(||) = max

请注意，这些定义不在的presence别样的底部的等效的，因为（安培;＆安培;）和（||）的短路（非严格的第二个参数的时候，首先是假或真，分别），而分和最大都没有。

Note that these definitions are not equivalent in the presence of a different kind of bottom because (&&) and (||) are short-circuiting (non-strict in the second argument when the first is False or True, respectively) while min and max are not.

此外，小更正：导出子句不说布尔派生自奥德。它指示GHC派生类型类的一个实例奥德的为的类型布尔。

Also, a small correction: The deriving clause does not say thatBool "derives from" Ord. It instructs GHC to derive an instance of the typeclass Ord for the type Bool.

*更具体地，一补充分配格的。还更具体地，一个布尔代数。

* More specifically, a complemented distributive lattice. More specifically still, a boolean algebra.

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