# Map and Flatmap

This week, I explained to my colleagues what's the difference between a `map`

and a `flatMap`

^{1}. The trick was: focus on the types and leave the burritos for lunch time.

Monads and cheese pic.twitter.com/nT3DJ51qXt

— R. Peres (@RuiAAPeres) July 24, 2015

**Map**

`[a] -> (a -> b) -> [b]`

`[1,2,3].map { $0*2 }`

**FlatMap**

`[a] -> (a -> [b]) -> [b]`

`[1,2,3].map { [$0*2] }`

Assuming that `flatMap`

uses `map`

^{2} internally:

```
func flatMap(f : Int -> [Int]) -> [Int] {
return map(f)
}
```

With that in place, let's see how it would end up looking:

`Int -> (Int -> [Int]) -> [[Int]]`

If you look closely at how `map`

works, you can see that it takes a function:

`a -> b`

And returns a:

`[b]`

The key here, is that you pass a function of type:

`a -> [b]`

So the outcome:

`[[b]]`

Wait a sec, but `flatMap`

returns a `[b]`

and we get back a `[[b]]`

. That's true, so that's why you need to `flat`

it:

```
func flat(v : [[Int]]) -> [Int]
{
return v.reduce([], combine: +)
}
```

As you might have guessed by now, I lied to you. That `flatMap`

's implementation is wrong. A correct one would look like the following:

```
func flatMap(f : Int -> [Int]) -> [Int] {
return flat(map(f))
}
```

So the entire thing internally looks like this:

`[a] -> (a -> [[b]]) -> ([[b]] -> [b]) -> [b]`

And to the consumer:

`[a] -> (a -> [b]) -> [b]`

Unfortunately, Swift doesn't really help understanding these two operators ^{3}:

```
let a : [Int] = [1,2,3,4] . flatMap { [2 * $0] } // <- This works
let b : [Int] = [1,2,3,4] . map { 2 * $0 } // This works as well
let c : [Int] = [1,2,3,4] . flatMap { 2 * $0 } // <- This works, but in theory it shouldn't(?)
```

I will explore why you should use one over the other, in another post.

It was actually about monads, but we need first maps and flatMaps. ↩

I could use generics, but to keep it simple, I will fix the type. ↩

Ben does an awesome job describing these. ↩