In the previous article I gave the conceptual background necessary to understand what monads are. Now I'm going to get into more of the details.

## The two fundamental monadic operations

Remember when I said above that monads generalize function composition and function application? We'll work through that here. Have patience: it'll take a while.

By this point, I hope you have at least a vague sense of what monads "are" and what they are used for. However, as I said before, one of the keys to functional programming is the ability to compose functions to create new functions. Functional programmers talk about "composability" all the time, with the implication that if some aspect of a programming language isn't composable, it's probably not worth much. So if our newly-minted monadic functions were not composable, they wouldn't be nearly as useful as they would be if they were. But as we'll see, they aren't composable using the standard Haskell function composition operator. Something more will be needed, and this will lead us to derive the two fundamental monadic operations (or at least their types).

Let's say that we have two monadic functions:

```
f :: a -> m b
g :: b -> m c
```

for some monad `m`

. If you want a more specific example, you can imagine
that `f`

and `g`

are in the `IO`

monad, so we'd have

```
f :: a -> IO b
g :: b -> IO c
```

but the same argument will apply for all monads. Remember that (for the `IO`

case) the function `f`

takes a value of type `a`

and outputs a value of type
`b`

, possibly doing some (file or terminal) I/O along the way. Similarly,
`g`

takes a value of type `b`

and outputs a value of type `c`

, possibly doing
some I/O along the way. Therefore, if we wanted to compose them, we'd
hopefully end up with a function like this:

```
h :: a -> IO c
```

*i.e.* a function that takes a value of type `a`

, outputs a value of type
`c`

, and possibly does some I/O along the way (with the I/O somehow being the
combination of the I/O activity for functions `f`

and `g`

). We can write
this out as follows:

```
compose:
(f :: a -> IO b)
with:
(g :: b -> IO c)
to get:
(h :: a -> IO c)
```

However, our normal Haskell function composition operators won't work for
this purpose, because they don't want the `IO`

in the types. Let's compare
with similarly-typed pure functions `p`

, `q`

, and `r`

that don't do I/O:

```
p :: a -> b
q :: b -> c
r :: a -> c
```

Then we could compose them using either the `(.)`

or the `(>.>)`

operator as
described above:

```
(.) :: (b -> c) -> (a -> b) -> (a -> c)
r = q . p
(>.>) :: (a -> b) -> (b -> c) -> (a -> c)
r = p >.> q
```

Neither `(.)`

or `(>.>)`

will work with our monadic functions:

```
f :: a -> IO b
g :: b -> IO c
h :: a -> IO c
g . f --> type error! mismatch between IO b and b
f >.> g --> type error! mismatch between IO b and b
```

The point is, you can't use a monadic value of type `IO b`

when a type of `b`

is needed. (This is a very common bug when writing monadic Haskell
programs.) What we want is a special monadic composition function which I'll
call `mcompose`

(standing for "monadic compose") which has the following type
signature:

```
mcompose :: (a -> m b) -> (b -> m c) -> (a -> m c)
```

This will work for any monad `m`

, including the `IO`

monad. Specialized to
the `IO`

monad, it will have the following type signature:

```
mcompose :: (a -> IO b) -> (b -> IO c) -> (a -> IO c)
```

Then we could use it like this:

```
f :: a -> IO b
g :: b -> IO c
h :: a -> IO c
h = f `mcompose` g
```

and `h`

would have the correct type signature. (We're using a spiffy
syntactic feature of Haskell here, whereby any two-argument function can be
turned into an infix operator by putting backquotes around it. Remember,
operators in Haskell are just functions which happen to be placed between
their operands.) Somehow, through (currently) mysterious means, the
`mcompose`

function (or operator, if you like) is able to

- take the original input value of type
`a`

- apply
`f`

to it (this is just normal function application) to get a result of type`IO b`

- take the value of type
`IO b`

output from`f`

and extract the value of type`b`

(this is what we couldn't do before) - take the value of type
`b`

and apply`g`

to it (again, this is just normal function application) to get the value of type`IO c`

, which is the result.

The only thing we can't already do is step (3), extracting a value of type `b`

from a value of type `IO b`

. Now, we *could* do this if we had a function
called `extract`

with this type:

```
extract :: IO b -> b
```

or more generally for arbitrary monads,

```
extract :: m b -> b
```

It turns out that such a function, if it existed, would destroy all the advantages of monads and pure functional programming! One of the reasons we wanted monads in the first place was to keep these special notions of computation (monadic functions) separate from normal (pure) functions, because otherwise there would be no way to guarantee that pure functions were in fact pure. This is an important point, so I'm going to spend a little bit of time on it, after which we'll return to monadic composition.

Side note: In fact, some monadsdohave the equivalent of an`extract`

function, and for most of those monads it doesn't cause problems. All I'm saying is that a generic`extract`

function that works forallmonads is not allowed.

What we would like is to ensure that functions that have non-monadic type
signatures are pure functions. Now, in a sense, even our monadic functions are
pure functions, because they are implemented in Haskell as pure functions that
return monadic values. But what we want to guarantee is that non-monadic (pure)
functions don't even do that *i.e.* don't even return monadic values. If that's
the case, they are certainly going to be pure functions. So a pure function
`hh`

of type

```
hh :: a -> c
```

should never do (file or terminal) input/output, for instance, because if it did it would be required by the type system to have the type

```
hh :: a -> IO c
```

instead. Guarantees like this, enforced by the type system, are one of the major strengths of Haskell. They allow us to glance at the type of a function and be 100% sure that that function doesn't do input/output, for instance.

However, if we had the `extract`

function, we could compose the supposedly pure
function `hh`

it out of monadic functions that do I/O like this:

```
ff :: a -> IO b
gg :: b -> c
hh = ff >.> extract >.> gg -- or equivalently: hh = gg . extract . ff
```

So even though `hh`

is never supposed to be doing I/O, if there was an
`extract`

function then you could build an `hh`

function using normal
function composition, it would have the type signature of a pure function,
and yet it would do I/O. So much for separating I/O (and other monadic
computations) from pure computations (recall that this was one of the main
reasons for wanting monads in the first place). Note, by the way, that this
is exactly the situation in most conventional programming languages, which is
why the type systems of those languages can offer no guarantees that a
function is pure. In Haskell we like pure functions and we use the type
system to give us guarantees that pure functions are actually pure —
and that means no `extract`

function.

There's one slight problem with what I just said: technically, it's a lie.
There is a function called `unsafePerformIO`

that has the type ```
IO a ->
a
```

*i.e.* it's an `extract`

function for the `IO`

monad only. The word
"unsafe" is a clue that tells you that you should avoid using it unless you know
exactly what you're doing and are prepared for weird failures. I myself have
never needed to use `unsafePerformIO`

, but there are legitimate uses for it (for
instance, deep down in the implementation of Haskell compilers). Just forget I
even brought this up, OK? It's embarrassing. Excuse me while I go wash my
hands.

OK, I'm back. So far, we've established that (a) we want to be able to compose
monadic functions, (b) we can't do that with normal function composition in
Haskell because we can't convert monadic types into regular types, and (c) we
can't define an `extract`

function to do that conversion, because that would
screw up the purity of the rest of the language. So what do we do?

Well, first of all, note that we can get by with something simpler than an
`mcompose`

function. Let's say we had an `mapply`

(monadic apply) function
that had this type signature:

```
mapply :: m b -> (b -> m c) -> m c
```

or, more specifically for the `IO`

monad:

```
mapply :: IO b -> (b -> IO c) -> IO c
```

It's called `mapply`

because it's very similar to the regular function
application operators. For instance, recall the `>$>`

operator we defined
previously, which had this type signature (using `b`

and `c`

instead of `a`

and `b`

for type variables):

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

This is the same as `mapply`

except that the `m`

s are gone (the types are not
monadic types). With `mapply`

, we could trivially define `mcompose`

as follows:

```
mcompose :: (a -> m b) -> (b -> m c) -> (a -> m c)
mcompose f g x = (f x) `mapply` g -- or: mapply (f x) g
```

Note that since the `->`

associates to the right in type signatures, the type
signature of `mcompose`

can be written without the final set of parentheses
as:

```
mcompose :: (a -> m b) -> (b -> m c) -> a -> m c
mcompose f g x = (f x) `mapply` g
```

This may be easier to understand than the previous version, but they are
equivalent. Note that `x`

has type `a`

and the result has type `m c`

. So what
we're doing here is applying `f`

to `x`

to get a value of type `m b`

, and using
`mapply`

on the `m b`

value and the `g`

function to get a value of type `m c`

.
So the upshot is, we don't need `mcompose`

to be defined for us if we have
`mapply`

, because we can use `mapply`

to define `mcompose`

ourselves. And, in
fact, `mapply`

is one of the two fundamental monadic operations. It's normally
called "bind" and is written as an infix operator with the symbol `>>=`

as
follows:

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

Note that I did a switch in the type signature, using `a`

in place of `b`

and
`b`

in place of `c`

. It doesn't matter since `a`

, `b`

, and `c`

are type
variables — they work for any types.

I'd just like to point out here that `>>=`

has an incredibly abstract type. Its
first argument is a value of type `m a`

, where `a`

can be any type at all and
`m`

is any monadic type constructor whatsoever. The second argument is a
function of type `a -> m b`

, where `a`

and `b`

can be any types at all and `m`

is again any monadic type constructor. The return value has type `m b`

, where
again `b`

can be any type and `m`

is any monadic type constructor. When you
program in Haskell for long enough, this kind of type signature becomes second
nature, but it can be intimidating to new Haskell programmers. If you
specialize it to the `IO`

monad, you get:

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

which, of course, is the type signature of an `IO`

-specific monadic apply
operator. We'll see below that Haskell's type class mechanism allows us to use
the same operator name `>>=`

for all the different specializations of this
operator to different monads (how cool is that?).

Assuming we have the `>>=`

operator, we can now compose `f`

and `g`

to get
`h`

as follows:

```
-- assume we have:
f :: a -> m b
g :: b -> m c
-- definition of h:
h :: a -> m c
h x = f x >>= g
```

We can also write `h`

directly as:

```
h = \x -> f x >>= g
```

where the `\x -> ...`

is, as I mentioned above, Haskell's notation for an
anonymous function (in this case with a single argument `x`

); both versions of
`h`

mean the same thing. Using `mcompose`

we can write this as:

```
h = f `mcompose` g = mcompose f g = \x -> (f x >>= g)
```

Our definition of `mcompose`

is thus just:

```
mcompose f g = \x -> (f x >>= g)
```

and in fact, Haskell has a standard operator for monadic composition called
`>=>`

:

```
f >=> g = \x -> (f x >>= g) -- same as (f `mcompose` g) but more concise
```

So, assuming we have this monadic apply operator `>>=`

, we can easily define the
monadic composition operator `>=>`

. So the monadic apply operator (the bind
operator) is the important concept here. As we'll see, each individual monad
has to define its own specific version of this operator, which will be different
from every other monad's version. That's where Haskell's type classes will come
in very handy. Incidentally, in the ghc Haskell
compiler, the `>=>`

operator is defined in the `Control.Monad`

module.

Now remember that we could write the normal apply operator in two ways:

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

and

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

depending on what order we wanted the arguments to be in. (Of course, it's also
fine to define both operators and use whichever one is most convenient in any
given situation.) Similarly, we can write the monadic apply operator in two
ways. The first way is as the bind operator `>>=`

with type

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

which is analogous to the non-monadic `>$>`

apply operator. We can also
trivially define a monadic apply operator that takes its operands in the reverse
order:

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

You can also use the `flip`

function, which takes a function of two arguments
and returns a new function which is the same as the old one except that it
takes the arguments in the reverse order:

```
flip :: (a -> b -> c) -> (b -> a -> c)
flip f = \x y -> f y x
```

Then we can define `=<<`

as follows:

```
(=<<) = flip (>>=)
```

You get extra points for functional coolness if you write concise definitions like this.

Similarly again, we can define a monadic composition operator that takes its operands in the reverse order:

```
(>=>) :: (a -> m b) -> (b -> m c) -> (a -> m c) -- already defined
(<=<) :: (b -> m c) -> (a -> m b) -> (a -> m c)
(<=<) = flip (>=>)
```

So just as was the case for the regular (non-monadic) apply and compose
operators, we can define monadic apply and compose operators which take their
operands in whichever order we want. In practice, though, the monadic operator
Haskell programmers seem to use the most is the `>>=`

operator (or at least,
it's the one *I* use the most).

If you've understood everything so far, congratulations! It's all downhill from here. Or so I hope.

There is one more fundamental monadic operation I need to talk about. To motivate it, consider this scenario. You want to compose a monadic function with a non-monadic function. In other words, you have these functions

```
f :: a -> m b -- monadic
g :: b -> c -- non-monadic
```

The problem is this: you can't use regular function composition to compose
`f`

and `g`

, because `m b`

is not the same type as `b`

. And you can't use
monadic function composition either, because `g`

doesn't have the type ```
b ->
m c
```

, which is what monadic composition would require. So what can you do?

If we had the `extract`

function I described above, you could compose the two
functions the way I showed there:

```
h :: a -> c
h = f >.> extract >.> g
```

but as I mentioned, you're not allowed to do this. In other words, you're not
allowed to compose a monadic function with a non-monadic function to get a
non-monadic function (because that would screw up the functional purity of the
language). What you *are* allowed to do is to compose a monadic function with a
non-monadic function to get a monadic function, which would work like this:

```
h :: a -> m c
h = f [somehow composed with] g
```

Now, we know we can't use our monadic composition operator for this, because
`g`

doesn't have the right type (which would be `b -> m c`

). But we *could*
use the monadic composition operator if we had some way to convert a
non-monadic function to a monadic one. In other words, if we had a function
`functionToMonadicFunction`

with this type signature:

```
functionToMonadicFunction :: (b -> c) -> (b -> m c)
```

then we could define `h`

as:

```
h :: a -> m c
h = f >=> (functionToMonadicFunction g)
```

It turns out that all you need in order to define `functionToMonadicFunction`

is an even simpler thing, which is a monadic function with the (possibly
confusing) name of `return`

. It has the following type signature:

```
return :: a -> m a
```

for any type `a`

and any monadic type constructor `m`

. What `return`

does is
convert a regular value into the corresponding monadic value for a given monad
`m`

. We'll see the specifics of this below.

If you have `return`

, then `functionToMonadicFunction`

can trivially be defined
as:

```
functionToMonadicFunction :: (a -> b) -> (a -> m b)
functionToMonadicFunction f = \x -> return (f x)
```

or, if I wanted to be cool and use function composition, as:

```
functionToMonadicFunction :: (a -> b) -> (a -> m b)
functionToMonadicFunction f = return . f
```

or even as:

```
functionToMonadicFunction :: (a -> b) -> (a -> m b)
functionToMonadicFunction = (return .)
```

using a cool syntactic feature of Haskell called operator sections. All three functions are equivalent.

Note that I once again switched `b`

for `a`

and `c`

for `b`

in the type
signature of `functionToMonadicFunction`

; again, it doesn't matter. The point
is, with this `return`

function, we can now compose monadic functions with
non-monadic ones to create new monadic functions. And `return`

is the second
fundamental monadic operation.

Side note: If you've done a lot of imperative programming, you will probably find the name`return`

more than a little annoying at first. Just remember that it isnota keyword in Haskell and it hasnothingto do with returning from a function. So try to keep those ideas out of your head when dealing with`return`

.

The name "return" actually comes from the characterization of monadic values
as "actions"; seen in that light, the function `return`

takes a value and
produces a monadic value which is an "action" which "does something" and then
"returns" the original value unchanged. Note also that `return`

is in fact a
monadic function. Putting these two ideas together, you can see (or at least
guess) that `return`

is the monadic version of the identity function (the
function that maps a value to itself, *i.e.* `\x -> x`

). We'll see this again
later when I talk about monad laws.

Let's put `return`

to work composing our monadic function `f`

with the
non-monadic function `g`

to get the monadic function `h`

. Here's the
definition:

```
h = f >=> (return . g)
```

because, as we saw above, `return . g`

will convert `g`

into a monadic function.

After all this, you might wonder how many more monadic operations we're going
to have to plow through before we're done defining them all. As Professor
Farnsworth would say: Good news, everybody! There are only two! There are
also a couple of non-critical operations that we will eventually want to
define for convenience, but `>>=`

and `return`

are the only ones that
absolutely have to be there.

There is one rather peculiar aspect of `return`

. We say that `return`

has
the type `a -> m a`

, but when we say *e.g.* `return 10`

, what is the type of
the output? It could have the type `IO Int`

, or `Maybe Int`

or some other
monadic type involving `Int`

. How do we know which of the many possibilities
is the correct one? Note, by the way, that the monadic value of type ```
IO
Int
```

is a completely different value than the monadic value of type ```
Maybe
Int
```

, so it's not just about getting the right type — it's not even
obvious what kind of value `return 10`

*is*!

In Haskell, this is worked out by the context in which `return 10`

is found.
The type checker has to make sure that all the functions get the right type
of input arguments, and so if `return 10`

is the input to a function
expecting a value of type `IO Int`

, the type checker will decide that ```
return
10
```

has the type `IO Int`

(and similarly for other monads). Put differently,
the value computed by `return 10`

depends on the type it has to have
according to its context. If you want to, you can annotate `return 10`

with the
type you want it to have by writing `(return 10 :: IO Int)`

, for instance, but
this is rarely necessary.

To recap this section:

There are two fundamental monadic operations, called "bind" (the

`>>=`

operator) and`return`

.The bind (

`>>=`

) operator is a monadic apply operator. It can be used to define a monadic composition operator, which is written`>=>`

.The

`return`

operator transforms regular values into monadic values. It can be used to define a function to convert regular functions into monadic functions.

## What do monadic application and composition *mean*?

By now, you should have a reasonable understanding of the mechanics of monadic composition and monadic application, but that's not the same thing as understanding what they mean intuitively. So let's expand on this a bit.

I said above that we can't define an `extract`

function which takes a monadic
value and returns a regular value. However, if we want to compose two
monadic functions to get a third using the monadic composition operator, we
need some way to extract values from monadic values.

```
(>=>) :: (a -> m b) -> (b -> m c) -> (a -> m c)
f >=> g = {- whatever -}
```

Here we see that the function `f`

takes in a value of type `a`

and returns a
monadic value of type `m b`

, whereas `g`

takes in a value of type `b`

and
returns a value of type `m c`

. So it certainly *looks* like some "unpacking" of
the monadic value into a regular value must be happening for this to be
possible. As we showed above, we can define the monadic composition operator
`>=>`

in terms of the monadic apply operator `>>=`

, so we can just look at
`>>=`

.

```
(>>=) :: m b -> (b -> m c) -> m c
mv >>= g
```

where `mv`

refers to some monadic value of type `m b`

. But again, without an
`extract`

function, how do we get the value of type `b`

out of the value of
type `m b`

so we can pass it to the function `g`

?

The answer is different for every monad. Every monad has its own way of
"unpacking" a value from a monadic value and passing it on to a monadic
function. Put differently, every monad has to define its own monadic apply
operator `>>=`

, and the details of how a monadic value gets unpacked and passed
to a monadic function are handled in the definition of that operator.
Similarly, every monad has to define its own `return`

function.

From now on, though, when dealing with arbitrary monads, I'll use the following
terminology. The `>>=`

operator takes as its input a monadic value (also known
as an "action"), "unpacks" a regular (non-monadic) value from it (the details of
which are specific to the particular monad), and passes that value as the input
to the monadic function, which produces a monadic value ("action") as its final
result.

Now that we've got that straight, let's talk about the `Monad`

type class.
Later, we will go over the definition of the `>>=`

operator for several
different monads, and you'll see for yourself how the unpacking is done in those
cases.

## The `Monad`

type class

I've mentioned above that `>>=`

is "the" monadic apply operator, and `return`

is
"the" function which converts a value into (any kind of) monadic value. But
this is sloppy terminology, since, as I've also mentioned, every monad has to
define its own version of these operators/functions, and these are completely
different operators/functions for each monad. On the other hand, it would be a
pain to have to use completely different *names* for every version of `>>=`

and
`return`

; we'd need to have nasty stuff like:

```
IO>>= :: IO a -> (a -> IO b) -> IO b
IOreturn :: a -> IO a
Maybe>>= :: Maybe a -> (a -> Maybe b) -> Maybe b
Maybereturn :: a -> Maybe a
```

and these function names aren't even legal Haskell syntax, because you can't mix symbolic and non-symbolic characters in identifiers and function names can't start with capital letters.

Haskell handles this kind of problem elegantly by using what is called a "type
class". (Remember, I said that this tutorial would be easier to understand if
you already knew about type classes. If you don't, be aware that this has
absolutely nothing, zero, zilch to do with the notion of classes in
object-oriented programming; a Haskell class is more like a compile-time
interface.) A type class is a way of saying that a bunch of different types
have different definitions for the same named functions (or operators) which
have particular types. So, for instance, the `Eq`

type class defines an
operator called `==`

which has the type `a -> a -> Bool`

(where each `a`

represents the same type). This says that for type `a`

to be in type class
`Eq`

, it has to define what `==`

(equality comparison) means for that type. The
numeric types `Int`

and `Float`

are both in type class `Eq`

(they are instances
of type class `Eq`

), and thus they have to define what `==`

means for those
types. We write this as follows in Haskell:

```
class Eq a where
(==) :: a -> a -> Bool
instance Eq Int where
(==) = intEquals -- intEquals has type (Int -> Int -> Bool)
instance Eq Float where
(==) = floatEquals -- floatEquals has type (Float -> Float -> Bool)
```

assuming that `intEquals`

is the equality function for `Int`

s, and
`floatEquals`

is the equality function for `Float`

s. (There is also an
inequality operator I've left out, but the idea is the same.) What this does
is allow us to use the same operator `==`

when comparing `Int`

s, or `Float`

s,
or any type which is an instance of the type class `Eq`

. This is a great
notational convenience. Note, though, that `==`

requires that both its
arguments have the same types, so it won't allow you to compare *e.g.* an `Int`

and a `Float`

. (Nor should you want to!)

Now, I mentioned above that all monads in Haskell are type constructors, and
we've seen that all monads have to provide independent definitions for the `>>=`

operator and the `return`

function. Putting this all together, we find that in
Haskell, there is a type class called (you guessed it) `Monad`

which looks
basically like this:

```
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
return :: a -> m a
```

This is actually a little more complicated than the `Eq`

type class. There
are two functions/operators, but that's no big deal, since we already knew
that we'd have to define two functions for every monad. The types of the two
functions/operators are familiar from the discussion above.

The weird part is that `Monad`

isn't a type class the way `Eq`

is. `Monad`

is
what's called a "constructor class" which means that its instances (called `m`

here) are not types but type constructors — and we saw above that all
monads have to be type constructors, so this has to be the case. To define an
instance of a constructor class, we'd write this (using `Maybe`

as our example
instance):

```
instance Monad Maybe where
(>>=) = {- the Maybe version of (>>=) -}
return = {- the Maybe version of return -}
```

The fact that the same notation is used for regular classes and constructor classes can be a bit confusing, but it works out well enough in practice. Perhaps Haskell would be more precise if it used this notation:

```
constructorClass Monad m where
(>>=) :: m a -> (a -> m b) -> m b
return :: a -> m a
constructorInstance Monad Maybe where
(>>=) = {- the Maybe version of (>>=) -}
return = {- the Maybe version of return -}
```

but that's pretty verbose and you can usually tell what's going on from context anyway.

Now let's look at a simple example and then continue talking about the
`Monad`

type class.

## An example

One of the simplest examples of a function in the `IO`

monad is a function
which reads a line from the terminal and prints it right back out (with a
newline at the end). The natural way to write this is to use the `getLine`

and `putStrLn`

functions. As we saw above, they have the types:

```
getLine :: IO String
putStrLn :: String -> IO ()
```

We can't use the monadic composition operator `>=>`

to combine these,
because `getLine`

is a monadic value, not a monadic function. But we *can*
use the monadic apply operator `>>=`

as follows:

```
readAndPrintLine = getLine >>= putStrLn
```

All we are doing here is (monadically) applying the `putStrLn`

monadic function
to the `getLine`

monadic value. Using the "action" terminology we discussed
above, we can think of this as follows: `getLine`

is an "action" which reads a
line of text input to the terminal and "returns" it as a monadic value, and the
`>>=`

operator "unpacks" that line of text from the monadic value, passing it to
`putStrLn`

, which prints it back to the terminal and returns nothing (actually
it returns the meaningless `()`

value as a monadic value).

We could have written this out more explicitly as follows:

```
readAndPrintLine = getLine >>= (\s -> putStrLn s)
```

Notice that `(\s -> putStrLn s)`

is exactly the same function as `putStrLn`

,
in the same way that `(\x -> cos x)`

is the same function as just `cos`

. So
we haven't changed anything important here. But writing it this way makes it
clearer that *something* (a text string) is somehow being returned from
`getLine`

, and what the `>>=`

operator is doing is taking that string
(called `s`

here) and printing it.

We'll expand on this simple example with some variations once we've covered
some more features of the `Monad`

type class, which we will do in the
next article of this
series.

Anonymous

September 18 2010, 15:27:37 UTC 11 years ago

"we'd write this (using Maybe as our example instnace):"

"putStrLn "unpacks" that line of text from the monadic value, prints it back to the terminal, "

Doesn't >>= unpacks the value

"what the >>= operator is doing is taking that string (called s here) and printing it."

Isn't putStrLn doing the printing?

-Lakshmi Narasimhan

## Thanks!

mvanierSeptember 19 2010, 04:23:11 UTC 11 years ago

## good tutorial

Anonymous

March 19 2011, 13:24:00 UTC 10 years ago

I think the topic cannot be explained much better, assuming only some programming experience and the tiniest bit of understanding of functional programming.

## Re: good tutorial

mvanierMarch 20 2011, 02:40:17 UTC 10 years ago

## Thanks

Anonymous

June 12 2011, 17:38:31 UTC 10 years ago

## Re: Thanks

mvanierJune 13 2011, 03:42:31 UTC 10 years ago