haskell *> sequencing operator

I have a question arising from the answer to my question at: haskell Either and Validation Applicative

My code is posted there.

It concerns the use of the *> sequencing operator instead of the <*> applicative operator.

Based on the explanation at https://hackage.haskell.org/package/base-4.15.0.0/docs/Control-Applicative.html#v:-42--62-, I understand that *> sequences actions, discarding the value of the first argument. So for my code, I've tried fail6 = fail2 *> success, which works, but it's not supposed to work because the value of the first argument, namely fail2, should be discarded. Why does fail6 work?

The output of fail6 is Failure [MooglesChewedWires,StackOverflow].

Solution

With "discard the result", it means the result of an applicative. So for an Either that means a Right y. (*>) is thus equivalent to:

(*>) :: Applicative f => f a -> f b -> f b
(*>) f g = liftA2 (const id) f g

or in another form:

(*>) :: Applicative f => f a -> f b -> f b
(*>) f g = (const id) <$> f <*> g

It thus runs the two actions and returns the second result, but this in the "Applicative context".

For example for an Either, this is implemented as:

instance Applicative (Either a) where
    pure = Right
    Left e <*> _ = Left e
    _ <*> Left e = Left e
    Right f <*> Right x = Right (f x)

This thus means that (*>) is implemented for an Either as:

-- (*>) for Either
(*>) :: Either a b -> Either a c -> Either a c
Left x *> _ = Left x
_ *> Left x = Left x
Right _ *> Right x = Right x

or equivalent:

-- (*>) for Either
(*>) :: Either a b -> Either a c -> Either a c
Left x *> _ = Left x
_ *> x = x

If the first operand is thus a Right …, it will return the second operand, if the first operand is Left x, it will return Left x.