The State Monad: a tutorial for the confused?
I’ve written this brief tutorial on haskell’s State
monad to
help bridge some of the elusive gaps that I encountered in other explanations
I’ve read, and to try to cut through all the sticky abstraction. This is
written for someone who has a good understanding of the Maybe
and List
monads, but has gotten stuck trying to understand State. I hope it’s helpful!
The Data Declaration:
To understand a monad you look at it’s datatype and then at the definition for
bind (>>=
). Most monad tutorials start by showing you the data declaration
of a State s a
in passing, as if it needed no explanation:
newtype State s a = State { runState :: s > (a, s) }
But this does need explanation! This is crazy stuff and nothing like what we’ve seen before in the list monad or the Maybe monad:

The constructor
State
holds a function, not just a simple value like Maybe’sJust
. This looks weird. 
Furthermore there is an accessor function
runState
with a weirdly imperativesounding name. 
Finally, there are two free variables on the left side, not just one.
Yikes! Let’s try to get our head on straight and figure this out:
First of all the State monad is just an abstraction for a function that
takes a state and returns an intermediate value and some new state value. To
formalize this abstraction in haskell, we wrap the function in the newtype
State
allowing us to define a Monad
class instance.
Stepping back from the abstract and conceptual, what we have is the State
constructor acting as a container for a function :: s > (a,s)
, while the
definition for bind just provides a mechanism for “composing” a function
state > (val,state)
within the State
wrapper.
Just as you can chain together functions using (.)
as in (+1) . (*3) . head :: (Num a) => [a] > a
, the state monad gives you (>>=)
to chain together
functions that look essentially like :: a > s > (a,s)
into a single
function :: s > (a,s)
.
Let’s bring the discussion back to actual code and try to make sure we understand those three points of weirdness outlined above. Here’s a stupid example of a function that can be “contained” in our state type:
 look at our counter and return "foo" or "bar"
 along with the incremented counter:
fromStoAandS :: Int > (String,Int)
fromStoAandS c  c `mod` 5 == 0 = ("foo",c+1)
 otherwise = ("bar",c+1)
If we just wrap that in a State
constructor, we’re in the State monad:
stateIntString :: State Int String
stateIntString = State fromStoAandS
But what about runState
? All that does of course is give us the “contents”
of our State constructor: i.e. a single function :: s > (a,s)
. It could
have been named stateFunction
but someone thought it would be really clever
to be able to write things like:
runState stateIntString 1
See, all we’ve done there is used runState
to take our function
(fromStoAandS
) out of the State wrapper; it is then applied to its
initial state (1
). We would do this runState
business after building up
our composed function with (>>=)
, mapM
, etc.
That leaves point 3 unanswered. Let’s start exploring the instance declaration for State.
The Instance Declaration
We’ll start with the first line:
instance Monad (State s) where
We create a Monad instance for (State s) not State. You can think of this as a partiallyapplied type, which is equivalent to a partiallyapplied function:
(State) <==> (+)
(State s) <==> (1+)
(State s a) <==> (1+2)
So (State s
) is the m
in our m a
. This means the type of our state will remain the same as we compose our function with (>>=)
, whereas the intermediate values (the a
s) may well change type as they move through the chain.
Before we move on to the meat of the instance declaration, I’d like to get
your mind calibrated to look at the definitions for return
and (>>=)
:
Whenever you see m a
, as in
return :: (Monad m) => a > m a
…remember that m a
is actually
State s a
…and when you remember (State s a)
, think
(s > (s,a))
So in your mind, m a
becomes function :: s > (a,s)
everywhere you see it.
Just forget about the silly State
wrapper
(the compiler does)!
The definition of return and Bind
Let’s wet our feet with the definition for return
:
return a = State $ \s > (a, s)
All return does is take some value a and make a function that takes a state
value and returns (value, state value). If we ignore the whole State wrapping
business, then return is just (,) :: a > b > (a, b)
Now recall the definition of bind:
(>>=) :: (Monad m) => m a > (a > m b) > m b
Which in our case is:
(>>=) :: State s a >
(a > State s b) >
State s b
And which is just a silly abstraction for the super special function composition that’s going on, which looks like:
(>>=) :: (s > (a,s)) >
(a > s > (b,s)) >
(s > (b,s))
So on the left hand side of (>>=
) is a function that takes some initial
state and produces a (value,new_state)
. On the right hand side is a function
that takes that value and that new_state and generates it’s own (new_value, newer_state)
. The job of bind is simply to combine those two functions into
one bigger function from the initial state to (new_value,newer_state)
,
just like the simple function composition operator
(.) :: (b > c) > (a > b) > a > c
At this point, we can show you bind’s definition:
m >>= k = State $ \s > let (a, s') = runState m s
in runState (k a) s'
You can work through that on your own, keeping in mind that we’re doing
function composition here. The main thing to remember is that the s
at the
top, right after State
, won’t actually be bound to a value until we unwrap
the function with runState
and pass it the initial state value, at which
point we can evaluate the entire chain.
A Final Note About The State Monad with do
Notation
State
is often used like this:
stateFunction :: State [a] ()
stateFunction = do x < pop
pop
push x
Remember that the functions above are desugaring to m >>= \a> f...
or if
there is no left arrow on the previous line: m >>= \_> f...
That a
in
there is an intermediate value, the fst
in the tuple. The push function
might look like:
push :: State [a] ()
push a = State $ \as > (() , a:as)
The function doesn’t return any meaningful a
value, so we don’t bind it by
using the <
. For more work with do notation and some fine pictures, see
Bonus’s post on something awful.
Getting more general: StateT
and MonadState
added by request 2/16/2012
So now you’re an expert on the State monad. Unfortunately (actually it’s a good
thing) the State
type I describe above isn’t in any of the standard
libraries. Instead State
is defined in terms of the StateT
monad transformer
here.
type State s = StateT s Identity
If you haven’t seen Monad transformers before, see if you can figure out how
StateT s Identity
is equivalent to State s
as I defined it above. Just
follow the links on hackage.
You might also have noticed a typeclass called MonadState
, also in the mtl package, and be wondering how that fits in.
Here’s what it looks like; I’ll explain the oddlooking bits in a moment:
class Monad m => MonadState s m  m > s where
get :: m s
put :: s > m ()
Whereas above we were discussing State
, a concrete data type, MonadState
is
a new typeclass for types that are monads and for which we can define get
and put
operations.
The class allows for a variety of Monad Transformer “stacks” that use statepassing to share a common interface for the basic state operations.
Small aside: since I neglected doing this above, this is how we would define get
and
put
as regular functions (not methods of MonadState
) on our State
type:
 return the state value being passed around:
get :: State s s
get = State $ \s > (s,s)
 replace the current state value with 's':
put :: s > State s ()
put s = State $ \_ > (s,())
Quite useful. See if you can understand how those work now that you’ve got a
better grasp of State
. Excercise: define modify :: (s>s) > State s ()
.
Back to MonadState
: If this class and its instances look confusing to
you, you need to know about two extensions to haskell98 (both of which are
very common and sticking around): MultiParameter Type Classes, and Functional
Dependencies.
In GHC you can enable both these extensions in your source by putting this at the top:
{# LANGUAGE MultiParamTypeClasses, FunctionalDependencies #}
Multiparameter Type Classes
Ignoring the functional dependencies (a.k.a “fundeps”) in MonadState
you get:
class Monad m => MonadState s m where
In which…
MonadState
is the name of the classs
is the first class type variable (in this case the type of our state)m
is the second type variable (the type of our statelike monad, e.g.State s
)
Multiparameter type classes are tricky in that they define a relationship between multiple types with associated operations. You can read more about them in the GHC docs.
Functional dependencies
With multiple parameters in a single class, you can often end up with instances that are disallowed or difficult to use because they are ambiguous. Functional Dependencies help resolve ambiguity by allowing a way to specify that certain type parameters can be determined by knowledge of one or more of the other parameters.
In the case of MonadState
, the part of the class declaration that looks like:
 m > s
says that the type of m
(say StateT Int IO
) uniquely determines the type
of s
(Int). Again please see the linked GHC docs for details; the Collects
example used is very similar to MonadState
.
Putting it all together, the class declaration might read in english as:
The relation of types
m
ands
wherem
uniquely determiness
is in theMonadState
class.
A MonadState
instance for State
Finally here’s what the MonadState
instance for the State
type we’ve been
discussing would look like (but again, it’s not because mtl
builds
around the more general StateT
):
instance MonadState s (State s) where
get = State $ \s > (s,s)
put s = State $ \_ > (s,())