# Staring at ($), (<$>), (<*>) and (>>=)

Recently I’ve spent some time staring at type signatures. The goal was to develop a better intuition by absorbing their wisdom. Last week it was Monad’s `bind`

. This time I’ve decided to compare the following four:

```
($) :: (a -> b) -> a -> b
(<$>) :: (a -> b) -> f a -> f b
(<*>) :: f (a -> b) -> f a -> f b
(>>=) :: (a -> m b) -> m a -> m b
```

## Function Application or `($)`

`($) :: (a -> b) -> a -> b`

It takes a funtion from a value of type `a`

to a value of type `b`

, an `a`

and returns `b`

. There’s only one possible way to implement `($)`

which is to apply the funtion to the value of type `a`

.

## Functor’s `fmap`

or `(<$>)`

`(<$>) :: (a -> b) -> f a -> f b`

The only difference from the previous is that `a`

and `b`

exist in a context `f`

. For example, we could have an `Int`

in a `List`

context (i.e. `[Int]`

), which means we went from one `Int`

to any number of `Int`

s. Or we could have an `Int`

in a `Maybe`

context (i.e. `Maybe Int`

), in other words there could be either no `Int`

s or just one `Int`

. And so on and so forth depending on the semantics of each functor.

Again, it’s easy to see how the value of type `a`

must feed the function from `a`

to `b`

to generate the output. The only difference from `($)`

is that depending on the semantics of the context `f`

, the function will be applied in a different way.

## Applicative Functor’s sequential application or `(<*>)`

`(<*>) :: f (a -> b) -> f a -> f b`

In this instance, the function from `a`

to `b`

has a context `f`

too. Therefore, the way the output is calculated depends on both the first and the second `f`

(which must be the same `f`

).

## Monad’s `bind`

or `(>>=)`

`(>>=) :: (a -> m b) -> m a -> m b`

This time, the way the function is applied depends only on the second `m`

. This is the same situation as for `(<$>)`

. But there’s one important change: the previous functions could only transform an `a`

into a `b`

. In the case of `bind`

, the funtion decides not only on the `b`

but also on the `m`

, which must be the same `m`

for both.

## Concretely

Let’s see the above in action in the context of `Either`

which has an instance for Functor, Applicative Functor and Monad. Notice that the instances are defined for `Either e`

because the context they provide is around one type, not two. For example, given an `Int`

we can provide it an `Either String`

context by doing `Either String Int`

.

```
show 1
--> "1"
-- ($)
show $ 1
--> "1"
-- (<$>)
show <$> Right 1
--> Right "1"
show <$> Left "string"
--> Left "string"
-- Either maps the function only when the value is a `Right`.
-- (<*>)
Right show <*> Right 1
--> Right "1"
Right show <*> Left "string"
--> Left "string"
Left show <*> Right 1
--> Left show
Left show <*> Left "string"
--> Type error: the type on the left should be the same for both `Either`s.
-- Either applies the function only when both values are `Right`.
-- (>>=)
Right 1 >>= (\x -> Right (show x))
--> Right "1"
Left "string" >>= (\x -> Right (show x))
--> Left "string"
Right 1 >>= (\x -> Left (show x))
--> Left "string"
Left "string" >>= (\x -> Left (show x))
--> Left "string"
-- Either binds the function only when the value before `>>=` is a `Right`.
-- Contrarily to the previous cases, `>>=` can decide to return `Left` or `Right`.
```

**I need your help** to make the blog awesome.
Here's a survey. Since there are no required questions, **you decide how much time to spend on it**.