18 Aug 2015

Practical Haskell - Using Monads

This is part of a tutorial series intended to introduce Haskell by coding things that work. In this article we will teach you how to use Monads without panicking.

  1. Getting Started with Stack
  2. Importing Code
  3. Using Monads
  4. Build a JSON API

Before you Begin

This article assumes a basic understanding of Haskell Types. You should know how to add a type declaration to a simple function, and understand the difference between String, Maybe String, and [String]. Consider reading Chapter 8 of Learn You a Haskell: Making Our Own Types and Typeclasses. You’ll probably need to have read Chapters 2 and 3 first.

Please work through Getting Started so you know how to run code.

Don’t Panic

Monads are not scary. You’ve been using them since the beginning of this tutorial. Whenever we make an IO function, and use do, we’re using monads. Let’s take a look at a simple example:

main = do
  name <- getLine
  putStrLn ("Hello, " ++ name)

You probably have a pretty good feel for what this is doing. Trust that feeling! There will be plenty of time to get the theory down. Today we’re just going to learn how to use them practically.

Functions that return Actions

putStrLn is a function that prints out a message to the console, right? Not quite. Let’s look at its type. What does it return?

putStrLn :: String -> IO ()

It’s a function that accepts a string, and returns an action to print something out to the console. That’s an important distinction. putStrLn doesn’t actually print anything, it returns an instruction to print something. Let’s call those instructions actions.

What good are actions?

What can we do with actions? They are values, like String or Bool, just fancier. Think of them as Todo list items. We can mix them together, and combine them into larger actions. When you use do syntax, that’s exactly what you’re doing:

main :: IO ()
main = do
  putStrLn "Go shopping"
  putStrLn "Take a nap"
  putStrLn "Learn Haskell"

Look at the type: IO () means that main is one IO action. That do block is combining 3 actions into one.

Since the whole point of do is to combine actions, you can leave it out if there’s only one. These are both equivalent:

main = do
  putStrLn "Hello"

main = putStrLn "Hello"

Calling the function doesn’t perform the action

What will it print out if I do this?

main = do
  let action = putStrLn "Hello World"
  return ()

Calling putStrLn does not print anything, it returns a value of type IO () which can be performed later. In the above, we are never executing it.

Anything at the same indentation level (except let) in a do-block is expected to be an action of the same type, and will be executed when the parent is.

-- what will this do?
main = do
  let action = putStrLn "Hello World"
  action
  action
  action
  return ()

We can see this from the type. When we call putStrLn with one argument, it returns IO (), or an IO action, all ready to be used.

putStrLn :: String -> IO ()

action :: IO ()
action = putStrLn "Hello World"

Actions can have results

Actions can yield something when they are performed. Let’s look at the type of getLine. What does it return?

getLine :: IO String

It doesn’t return a String, it returns an action, right? Then what does that String mean? It means that when the action is performed, it will yield a String. Let’s call that the result of the action.

You can use result from an action with the <- operator.

main = do
  name <- getLine
  putStrLn ("Hello " ++ name)

name is no longer an action. Its type is just String, not IO String like getLine was. It’s the difference between a plan to go to the grocery store (getLine) and the groceries you picked by actually going (name)

A do-block always yields whatever its last action does. If we put return last, we can make our do-block yield any simple value.

sayHello :: IO String
sayHello = do
  name <- getLine
  putStrLn ("Hello " ++ name)
  return name

Different types of Actions

Within a given do-block, all actions must be the same type. So for an IO block like the one in main, every line has to have the type IO a. You can’t put regular values there (without let), because they aren’t actions.

main :: IO ()
main = do
  -- good. putStrLn "Hello" is of type IO ()
  putStrLn "Hello"

  -- error! String is not an IO a!
  "whatever"

Remember that functions like putStrLn aren’t IO actions, they return IO actions. So this won’t work:

main = do

  -- error! putStrLn is a function with type String -> IO ()
  putStrLn

  -- good. If we give it one argument, it's now an IO action
  putStrLn "Hello"

Let’s look at some examples of different kinds of monads, each with their own actions. To use the Maybe monad, each step must be of type Maybe a. You can’t mix different kinds of actions in the same do block.

doesNotWork :: Maybe Int
doesNotWork = do

  -- these work, because they are all Maybe a
  Just 6
  Nothing

  -- error! this is type IO (), not Maybe a!
  putStrLn "Hello"

  return 5

main :: IO ()
main = do
  -- error! this is type Maybe Int, not IO a
  Just 6

Monads decide how to combine actions

Each type of Monad combines actions differently. We already looked at IO. It performs the actions in order with IO side effects. Maybe, on the other hand, specifies that if any step is Nothing, the whole thing stops and exits with a Nothing.

beCareful :: Maybe Int
beCareful = do
  Just 6
  Nothing
  return 5

The do-block here always yields Nothing, because of that Nothing in the second to last step.

Time to code

Add beCareful and sayHello to src/Main.hs. Then head over to GHCI and try them out.

$ stack ghci
Prelude> :load Main

*Main> beCareful
Nothing

*Main> sayHello
woot
Hello woot
"woot"

Now try mixing them up. Add an IO action into beCareful.

beCareful :: Maybe Int
beCareful = do
  Just 6
  putStrLn "oops"
  Nothing
  return 5

What happens when we reload GHCI?

*Main> :r
[1 of 1] Compiling Main             ( src/Main.hs, interpreted )

src/Main.hs:11:3:
    Couldn't match expected type ‘Maybe a0’ with actual type ‘IO ()’
    In a stmt of a 'do' block: putStrLn "oops"
    In the expression:
      do { Just 6;
          putStrLn "oops";
          Nothing;
          return 5 }
Failed, modules loaded: none.

Hopefully that error message makes sense to you now.

Other monads in the wild

The Parsec libary uses Parser actions to define a grammer for parsing. Here’s one to parse an IPv4 address. It combines actions by matching them against an input. If the input doesn’t match any of the steps, it exits and tells its parent it didn’t match.

data IP = IP Word8 Word8 Word8 Word8 deriving Show

parseIP :: Parser IP
parseIP = do
  d1 <- decimal
  char '.'
  d2 <- decimal
  char '.'
  d3 <- decimal
  char '.'
  d4 <- decimal
  return $ IP d1 d2 d3 d4

Can you guess what the type of decimal and char are?

char :: Char -> Parser Char
decimal :: Parser Word8

The Scotty library uses the ScottyM monad to define a web API. Here each action is combined to create an API description which is later matched against incoming HTTP requests. The first ones are given priority when matching.

routes :: ScottyM ()
routes = do
  get "/"      (text "homepage!")
  get "/hello" (text "hello")

Don’t sweat the little stuff

So there you go. Don’t worry about it. You can use do notation, monads, and write programs. You don’t have to understand exactly what monads are doing under the hood to have an intuition for how to use them. Later when you learn the theory, it’ll be much easier if you’ve been using them.

Assignment

Build a program that asks the user for a message and a number on the command line, and print out that message N times. Use the control functions from Control.Monad, like mapM, or replicateM

Read Input and Output from Learn You a Haskell. Implement the unix cat command, which takes a command-line argument as a path to a file and prints out its contents. Refer to Getting Started to see how to use stack exec.

Answers

What’s next

In the next tutorial, Build a JSON API, we code a real webserver that responds to JSON requests over HTTP.

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