[英]Order of evaluation for multiple operators in infix form
Given this: 鉴于这种:
data Base = Base {
key1 :: Text,
key2 :: Text,
key3 :: Text
} deriving (Show)
instance FromJSON Base where
parseJSON (Object v) = Base <$>
((v .: "base123") >>= (.: "key1")) <*> -- 1
((v .: "base123") >>= (.: "key2")) <*> -- 2
((v .: "base123") >>= (.: "key3")) -- 3
parseJSON _ = mzero
What's the order of in which the infix operators <$>
, <*>
and <*>
are applied? 前缀运算符<$>
, <*>
和<*>
的应用顺序是什么? In other words, if I rewrite it in prefix form: 换句话说,如果我以前缀形式重写它:
instance FromJSON Base where
parseJSON (Object v) = Base <$> ((<*>) ((v .: "base123") >>= (.: "key1")) $ (<*>) ((v .: "base123") >>= (.: "key2")) ((v .: "base123") >>= (.: "key3")))
parseJSON _ = mzero
(notice $
operator), will the right part of the second <*>
be evaluated first because only in this case it makes sense because the first <*>
requires 2 arguments? (注意$
运算符),将首先评估第二个<*>
的右侧部分,因为仅在这种情况下才有意义,因为第一个<*>
需要2个参数? And since it requires 2 arguments, we have to use $
also. 并且由于它需要2个参数,因此我们也必须使用$
。
I might've asked my question so that it was difficult to understand what I meant but I hope you did understand. 我可能会问我的问题,以致难以理解我的意思,但我希望你确实理解。
Actually your prefix form is not quite correct, it should be like this: 实际上您的前缀形式不是很正确,应该是这样的:
parseJSON (Object v) = ((<*>)
((<*>)
((<$>) Base ((v .: "base123") >>= (.: "key1")))
(((v .: "base123") >>= (.: "key2"))))
(((v .: "base123") >>= (.: "key3"))))
The above definition is still not in complete prefix form. 上面的定义仍然不是完整的前缀形式。 You have to take >>=
and .:
to the left to make them completely prefix. 您必须在左边加上>>=
和.:
使其完全前缀。 That being said, to find the exact order of evaluation of multiple operators in infix form I would suggest you to play up in ghci to get more insights into types. 话虽这么说,要找到以infix形式表示的多个运算符的确切评估顺序,建议您使用ghci进行操作,以获取更多有关类型的见解。 As an initial step, check the associativity and the precedence order for all the operators: 首先,请检查所有运算符的关联性和优先顺序:
λ> :i (<$>)
(<$>) :: Functor f => (a -> b) -> f a -> f b
infixl 4 <$>
λ> :i (<*>)
(<*>) :: f (a -> b) -> f a -> f b
infixl 4 <*>
So, they both are left associative and have the same precedence. 因此,它们都保持关联并且具有相同的优先级。 The infix form of the definition is quite clear on how the evaluation will take place: they start from left and initially <$>
is applied over Base
and then followed by application of two <*>
functions. 定义的中缀形式非常清楚如何进行评估:它们从左开始,首先在Base
上应用<$>
,然后再应用两个<*>
函数。 The type Base
is initially applied to <$>
: 类型Base
最初应用于<$>
λ> :t Base
Base :: Text -> Text -> Text -> Base
λ> :t (Base <$>)
(Base <$>) :: Functor f => f Text -> f (Text -> Text -> Base)
Now, ((v .: "base123") >>= (.: "key1"))
is applied to the resultant of the above type: 现在,将((v .: "base123") >>= (.: "key1"))
应用于上述类型的结果:
λ> let (Object v) = undefined :: Value
λ> :t (Base <$> ((v .: "base123") >>= (.: "key1")))
(Base <$> ((v .: "base123") >>= (.: "key1"))) :: Parser (Text -> Text -> Base)
You can see that it returns a function wrapped in Parser
type. 您会看到它返回包装在Parser
类型中的函数。 And to extract the underlying function out of the Parser
type, you have to use <*>
: Parser
类型中提取基础函数,必须使用<*>
:
λ> :t (<*>)
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
λ> :t (Base <$> ((v .: "base123") >>= (.: "key1")) <*>)
(Base <$> ((v .: "base123") >>= (.: "key1")) <*>) :: Parser Text -> Parser (Text -> Base)
You can follow the similar steps to see how it is applied to the other parts of the function definition. 您可以按照类似的步骤查看如何将其应用于函数定义的其他部分。 At the end, you will get a type of Parser Base
. 最后,您将获得一种Parser Base
。
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