为什么“和[]"是真和“或[]"是假的 [英] Why "and []" is True and "or []" is False

问题描述

为什么空列表上的and"返回true，是否暗示空列表为True?抱歉，我无法正确阅读和理解这一点，所以请纠正我.谢谢.

Why "and" on an empty list returns true, does it imply that an empty list holds True? Sorry but I cannot read and comprehend this correctly, so please correct me. Thanks.

Prelude> and []
True
Prelude> or []
False

推荐答案

在数学中，谈论二元运算通常很有用，例如 &&, ||、+、* 等具有身份.标识是一个值 e 使得以下属性适用于某些通用二进制操作 <>

In mathematics, it's often useful to talk about a binary operation, such as &&, ||, +, *, etc as having an identity. The identity is a value e such that the following property holds for some generic binary operation <>

e <> x = x
x <> e = x

对于我上面列出的操作符，它们是可交换的，这意味着 x <>y＝y＜＞x 代表所有的 x 和 y，所以我们只需要检查上述属性之一.对于and，有问题的二元运算符是&&，对于or，二元运算符是||.如果我们为这些操作制作一个Cayley 表，它看起来像

For the operators I listed above, they are commutative, meaning that x <> y = y <> x for all x and y, so we only have to check one of the above properties. For and, the binary operator in question is &&, and for or the binary operator is ||. If we make a Cayley table for these operations, it would look like

&&    | False | True
------+-------+------
False | False | False
True  | False | True


||    | False | True
------+-------+------
False | False | True
True  | True  | True

如你所见，对于 && 如果你有 True &&False 和 True &&对，答案总是&& 的第二个参数.对于 ||，如果你有 False ||False 和 False ||对，答案总是第二个参数，所以每个的第一个参数必须是这些运算符下的标识元素.简单地说:

So as you can see, for && if you have True && False and True && True, the answer is always the second argument to &&. For ||, if you have False || False and False || True, the answer is always the second argument, so the first argument of each must be the identity element under those operators. Put simply:

True && x = x
x && True = x

False || x = x
x || False = x

因此，当没有元素可以执行操作时，首选答案是每个操作的标识元素.

Thus, the preferred answer when there are no elements to perform the operator on is the identity element for each operation.

考虑 + 和 * 的标识元素可能会有所帮助，它们是 0 和 1> 分别:

It might help to also think about the identity elements for + and *, which are 0 and 1 respectively:

x + 0 = x = 0 + x
x * 1 = x = 1 * x

您还可以将其扩展到列表连接(++ 和 [])、a -> 类型函数的函数组合等操作.a((.) 和 id)，以及许多其他的.由于这开始看起来像一种模式，您可能会问这是否已经在 Haskell 中出现，并且确实如此.Data.Monoid 模块定义了抽象这个模式的 Monoid 类型类，它的最小定义是

You can also extend this to operations like list concatenation (++ with []), function composition for functions of type a -> a ((.) with id), along with many others. Since this is starting to look like a pattern, you might ask if this is already a thing in Haskell, and indeed it is. The module Data.Monoid defines the Monoid typeclass that abstracts this pattern, and it's minimal definition is

class Monoid a where
    mempty :: a                   -- The identity
    mappend :: a -> a -> a        -- The binary operator

它甚至将 mappend 别名为 <> 以方便使用(我在上面选择它作为通用二元运算符并不是偶然的).我鼓励您查看该模块并了解其定义.源代码很容易阅读，很有启发性.

And it even aliases mappend as <> for ease of use (it was no accident that I choose it above for a generic binary operator). I encourage you to look at that module and play around with its definitions. The source code is quite easy to read and is enlightening.

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