身份单片作为免费的单子
身份单元的函数可以定义为：
data Identity a = Identity a
因为这个单子是免费的，所以另一种定义是：
data Term f a = Pure a | Impure (f (Term f a))
data Zero a
type IdentityF a = Term Zero a
由于这是通过两种方式定义的单一单元，所以它们却不 互相转换。也就是说，一个应该能够定义两个 函数f ::身份a - ＆gt; sidentityf a和g :: identityf a - ＆gt;身份a使其构图f。 g和g。 f是 身份。函数f易于定义：
f :: Identity a -> IdentityF a
f (Identity a) = Pure a
但是功能g呢？
g :: IdentityF a -> Identity a
g (Pure a) = Identity a
g (Impure x) = ??????
g（不纯x）的值应该是什么？我可以尝试作弊 是未定义的，但是f。 g不会是身份函数， Identity和Identityf不会是同构。
The functor of the identity monad can be defined as:
data Identity a = Identity a
Because this monad is free, an alternative definition is the following:
data Term f a = Pure a | Impure (f (Term f a))
data Zero a
type IdentityF a = Term Zero a
Since this is the same monad defined in two ways, they shoud be
convertible into each other. That is to say that one should be able to define two
functions f :: Identity a -> IdentityF a and g :: IdentityF a -> Identity a such that their compositions f . g and g . f are
identities. The function f is easy to define:
f :: Identity a -> IdentityF a
f (Identity a) = Pure a
But what about the function g?
g :: IdentityF a -> Identity a
g (Pure a) = Identity a
g (Impure x) = ??????
What should be the value of g (Impure x). I could try to cheat and say it
is undefined but then f . g would not be the identity function andIdentity and IdentityF would not be isomorphic.
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一个合适的定义是：
案例中没有分支。这不是错字。在此情况下，分支机构与Zero a中的构造函数一样多。这是一个完整的模式匹配。（您必须打开
emptricageGHC的扩展名才能接受此AS-I。）One suitable definition is:
There are no branches in the
case. This was not a typo. There are exactly as many branches in the case as there are constructors inZero a, as required; this is a complete pattern match.(You must turn on the
EmptyCaseextension for GHC to accept this as-is.)