CODEWITHSUNDEEP
haskell
Incerteza
Incerteza

Reputation: 34944

Order of evaluation for multiple operators in infix form

Given this:

 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? In other words, if I rewrite it in prefix form:

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? And since it requires 2 arguments, we have to use $ also.

I might've asked my question so that it was difficult to understand what I meant but I hope you did understand.

Upvotes: 1

Views: 105

Answers (1)

Sibi
Sibi

Reputation: 48766

Actually your prefix form is not quite correct, it should be like this:

parseJSON (Object v) = ((<*>)
                        ((<*>)
                         ((<$>) Base ((v .: "base123") >>= (.: "key1")))
                         (((v .: "base123") >>= (.: "key2"))))
                        (((v .: "base123") >>= (.: "key3"))))

The above definition is still not in complete prefix form. You have to take >>= and .: to the left to make them completely prefix. 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. As an initial step, check the associativity and the precedence order for all the operators:

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. The infix form of the definition is quite clear on how the evaluation will take place: they start from left and initially <$> is applied over Base and then followed by application of two <*> functions. The type Base is initially applied to <$>:

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:

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. And to extract the underlying function out of the Parser type, you have to use <*>:

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. At the end, you will get a type of Parser Base.

Upvotes: 2

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