## Wednesday, January 2, 2013

If you've used Haskell before, chances are that you've heard the term "monad" a good bit. But, you might not know what they are, or what they are really used for. It is my goal in this blog post to shed some light on what monads are, introduce and define few simple ones, and touch on their usefulness.

I will assume basic familiarity with Haskell syntax in this blog post. A working knowledge of monoids and functors will also be beneficial, although not strictly necessary.

We can see a Monad has two operations, return and >>= (this is commonly pronounced "bind"). You may suspect that you know what return does from other programming languages, but be warned: Haskell's return is very different! Haskell's return only operates on Monads, and essentially acts as a "wrapping" function. That is, if you call return on a value, it will turn it into a monadic value of that type. We will look at an example of this shortly. The second operation, >>=, takes two arguments: an a wrapped in a Monad m (I will refer to this as m a from now on), and a function that converts an a to b wrapped in the same type of Monadm. It produces a value of type b, wrapped in a Monad m (I will call this m b from now on). This may sound complicated, but I'll do my best to explain it after showing the most basic Monad type, namely, the Identity Monad:

The Identity data declaration defines only one type: Identity a. Basically, this is just a wrapper for a value, and nothing more. return is simply defined as Identity. We can, for example, call return 3 and get an Identity 3. Turning values into Identitys is as simple as that. The bind function may look a little bit more obscure, though. Let's look closely: We first use pattern matching to be able to operate on the type that the Identity Monad wraps (Identity a), bind it (>>=) to a function (f). Since f converts normal values into monadic ones (look at the type declaration), we can simply apply f to a, and we've produced a new Identity. I've provided a couple of examples of how this works. addThree adds 3 to an Identity Int (m1) and getLength turns an Identity [a]  (m2) into an Identity Int. Both of the bind examples produce the value Identity 6. Can you see why?

I want to take a little sidebar here and try to explain how I understand Monads before moving on. Basically the real tough part of understanding Monads is just understanding >>=. A little bit of dissection can explain a lot, but it took me months to really grasp the concept. On the surface, you're morphing an m a into an m b. At first, I thought that the m's in the type signature were allowed to different types of Monads. For instance, I thought that Monads made it possible to bind Identity Ints to functions that produced [String]s (we'll look at the list Monad in a minute). This is wrong, and for good reason! It was incredibly confusing to think about a function that could generalize to this degree and it is, to my knowledge, not possible to do so. The Monad (wrapper) type is retained, but the wrapped type can be changed by means of the function you bind to.

The second thing that I want to stress is what >>= really does. What we're doing with the >>= function is turning an m a into an m b, but it's not so direct as you might want to think. The function argument in >>=  ( (a -> m b) ) was really confusing to me to begin with, so I'm going to try to explain it. The function you're binding your m a to must take an a and return an m b. This means that you're acting on the wrapped part of the Monad in order to produce an entirely new Monad. It took me a while to realize what exactly was happening, that I wasn't supposed to be directly converting m a to m b by means of my argument function (that's actually what >>= is for). Just remember that the function you bind to should operate on an a to produce an m b.

With that knowledge, let's take a look at some more examples of Monads:

If you're a Haskell user, you're likely already familiar with the Maybe data type:

Defined in the Prelude

The Maybe Monad is hardly an extension of the aforementioned Identity Monad. All we are really adding is the functionality to fail -- that is, to produce a Nothing value if necessary. In fact, the Just a type is exactly the same as Identity a, except that a Nothing value anywhere in the computation will produce a Nothing as a result. Next, we'll look at the Monad instance for lists:

Defined in the Prelude

The list Monad is a little more complicated, but it's really nothing too new. For return, we just have to wrap our value inside a list. Now, let's look at >>=concatMap actually is the >>= instance for lists. Why? Well, concatMap is, simply, (concat . map). We know that map takes a list of as and maps a function over them, and concat is a function that flattens lists, that is, turns [[a]]s into [a]s. When we compose these two functions (or use concatMap), we're making a function that must produce [[a]]s out of [a]s, and then flatten them back to [a]s. That is exactly how >>=  works for lists! Now, let's take a look at a slightly more complicated Monad:

This one requires some explanation, because it's a bit of a leap from the last three, since we're defining a Monad instance for functions. I have added a couple of comments in the above code in order to further explain how everything is working in the (->) r Monad.

Well, first things's first. We define return as const, which takes some arbitrary argument and produces a function that produces a constant value (what we pass in to return). This is exactly what we want; a minimal context for functions.

(>>=) is a bit more complex. First, let's take a look at the type signature, specific to this Monad. We're binding a function (r -> a) to a function (a -> (r -> b)) to produce a function of type (r -> b). So in the declaration of the function, f is of type (r -> a) and g is of type (a -> (r -> b)). I've conveniently named the argument our function is going to take r, for clarity. This is what the user will pass into the function that (>>=) produces.

f takes an r, so we apply the value from the lambda into it to produce an a. We then use that a to produce an (r -> b), using g. This produces our function, but it needs to be fed some value, so we finally apply it to r in order to complete everything. If any of this sounds confusing, try to work through it on paper and read the comments in my code -- after a bit of staring, it should start to make sense.

I've also implemented the mean function here, using the (->) r Monad with and without do-notation. I think it's a pretty elegant way to express it.

We've only seen the tip of the iceberg when it comes to Monads, but this should serve as a good basic introduction. Once you get used to seeing Monads, they become a bit less scary. The more time you spend with them, the more comfortable you'll feel with them.

Before I end this post, I want to leave a question for the reader. Given the general Tree data type represented below, can you intuitively make a Monad instance?

I'll explain why this is difficult when we talk about a similar mathematical type in the next post.

Ben

For more introductory monad material, I suggest the following resources:

1. Hello Ben,

Both bindExample1 and bindExample2 don't compile with the definitions given above for addThree and getLength.

Their definitions should be changed as follows:

addThree x = return (x + 3)

getLength :: Monad m => [a] -> m Int
getLength x = return (length x)

Best wishes for the New Year.

JH

1. You're right! I'll correct that now. Thanks for catching that!

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