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The Monadic Way

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==An evaluation of Philip Wadler's "Monads for functional programming"==
+
'''Note''' Since the size of the previous file was getting too big for a wiki, the tutorial has been divided into two parts: [[The Monadic Way Part I]] and [[The Monadic Way Part II]]. See below for some introductory remarks.
   
This tutorial is a "translation" of Philip Welder's "Monads for
+
'''Contents'''
functional programming".
 
(avail. from [http://homepages.inf.ed.ac.uk/wadler/topics/monads.html here])
 
   
I'm a Haskell newbie trying to grasp such a difficult concept as the
+
;[[Meet Bob The Monadic Lover]]
ones of Monad and monadic computation.
+
:A (supposed-to-be) funny and short introduction to Monads, with code but without any reference to category theory: what monads look like and what they are useful for, from the perspective of a ... lover. It could be an introduction to "The Monadic Way" tutorial.
While [http://www.cs.utah.edu/~hal/htut/ "Yet Another Haskell
 
Tutorial"] gave me a good understanding of the type system when it
 
comes to monads I find it almost unreadable.
 
But I had also Welder's paper, and started reading it. Well, just
 
wonderful! It explains how to ''create'' a monad!
 
So I decided to "translate it", in order to clarify to myself the
 
topic. And I'm now sharing this traslation (not completed yet) with
 
the hope it will be useful to someone else.
 
Moreover, that's a wiki, so please improve it. And, specifically,
 
correct my poor English. I'm Italian, afterall.
 
   
==A Simple Evaluator==
+
;[[The Monadic Way/Part I]]
  +
:In the first part of the tutorial we will start from a very simple evaluator that will be transformed into a monadic evaluator with an increasing number of features: output, exceptions, and state: a very simple counter for tracking the number of recursions of the evaluation precess.
   
Let's start with something simple: suppose we want to implement a new
+
;[[The Monadic Way/Part II]]
programming language. We just finished with
+
:In the second part of the tutorial we will see how to take complexity out of our monadic evaluator with the use of monadic transformers, and specifically StateT. This part is just a skeleton, since, for the time being, it contains only the code.
[http://swiss.csail.mit.edu/classes/6.001/abelson-sussman-lectures/
 
Abelson and Sussman's Structure and Interpretation of Computer
 
Programs] and we want to test what we have learned.
 
   
Our programming language will be very simple: it will just compute the
 
sum operation.
 
So we have just one primitive operation (Add) that takes two constants
 
and calculates their sum
 
For instance, something like:
 
   
(Add (Con 5) (Con 6))
+
==Preliminary remarks==
   
should yeld:
+
When I started writing this tutorial I though that the only way to
  +
explain monads to a newcomer was to show them from the inside, with
  +
the use of lambda abstraction. Not only because this is the way Philip
  +
Wedler's
  +
[http://homepages.inf.ed.ac.uk/wadler/papers/marktoberdorf/baastad.pdf paper]
  +
adopts, but also because I believed, and still believe, that the
  +
only way to understand what bind (>>=) does is to explain it as a
  +
function that takes a monad and an anonymous function.
   
11
+
I had this feeling because I am a newcomer, and this is the way I came to understand monads.
   
We will implement our language with the help of a data type
+
I did not received very much feedback for this tutorial, and I must
constructor such as:
+
admit that I would like to. But one person, on the haskell-cafe
  +
mailing list, [http://www.haskell.org/pipermail/haskell-cafe/2006-September/017740.html told me]
  +
that:
  +
<pre>
  +
imho your tutorial makes the error that is a very typical: when you
  +
write your tutorial you already know what are monads and what the
  +
program you will construct at the end. but your reader don't know all these!
  +
for such fresh reader this looks as you made some strange steps, write
  +
some ugly code and he doesn't have chances to understand that this ugly
  +
code is written just to show that this can be simplified by using monads.
  +
</pre>
   
<haskell>
+
I believe that Bulat is right. In this tutorial you go through some
  +
code that is '''really''' ugly and then, with some kind of magic, it
  +
turns out in the (redundant but) clean evaluator of the end of [[The Monadic Way/Part II]].
   
module MyMonads where
+
I took that mail as a challenge and
data Term = Con Int
+
[http://www.haskell.org/pipermail/haskell-cafe/2006-September/017778.html I responded]
| Add Term Term
+
by writing [[Meet Bob The Monadic Lover]].
deriving (Show)
 
   
</haskell>
+
In "Meet Bob" the code is clean, variable names are very descriptive,
  +
and you see that I can create a monad without any use of lambda
  +
abstraction.
   
After that we build our interpreter:
+
Bind (in "Meet Bob" is askLover) now takes a monad and a partial
  +
application, not an anonymous function.
   
<haskell>
+
Obviously you can see an anonymous function as a partial application.
   
eval :: Term -> Int
+
The problem, I think, is that, in "Meet Bob", you cannot understand
eval (Con a) = a
+
the strict relation between what I did before and what I do when I
eval (Add a b) = eval a + eval b
+
start using the "do-notation". You see that the same functions are
  +
being used ("tellMyself" and "newLove"), but "andrea <- tellMyself 1"
  +
is not explained. It's just magic.
   
</haskell>
+
The fact is that you cannot understand "andrea <- tellMyself 1"
+
without the use of lambda abstraction.
That's it. Just an example:
 
 
*MyMonads> eval (Add (Con 5) (Con 6))
 
11
 
*MyMonads>
 
 
Very very simple. The evaluator checks if its argument is a Cons: if
 
it is it just returns it.
 
If it's not a Cons, but it is a Term, it evaluates the right one and
 
sums the result with the result of the evaluation of the second term.
 
 
== Some Output, Please!==
 
 
Now, that's fine, but we'd like to add some features, like providing
 
some output, to show how the computation was carried out.
 
Well, but Haskell is a pure functional language, with no side effects,
 
we were told.
 
Now we seem to be wanting to create a side effect of the computation,
 
its output, and be able to stare at it...
 
If we had some global variable to store the out that would be
 
simple...
 
But we can create the output and carry it along the computation,
 
concatenating it with the old one, and present it at the end of the
 
evaluation together with the evaluation of the expression!
 
Simple and neat!
 
   
  +
I should have written an intermediate step, something like this:
 
<haskell>
 
<haskell>
+
drunk = do newLove "Paula " >>
type MOut a = (a, Output)
+
(tellLover 1 10) >>= \paula ->
type Output = String
+
(tellMyself 3) >>= \lorena -> tellMyself (paula + lorena)
 
formatLine :: Term -> Int -> Output
 
formatLine t a = "eval (" ++ show t ++ ") <= " ++ show a ++ " - "
 
 
evalO :: Term -> MOut Int
 
evalO (Con a) = (a, formatLine (Con a) a)
 
evalO (Add t u) = ((a + b),(x ++ y ++ formatLine (Add t u) (a + b)))
 
where (a, x) = evalO t
 
(b, y) = evalO u
 
   
 
</haskell>
 
</haskell>
   
Now we have what we want. But we had to change our evaluator quite a
+
With this approach I think you can understand '''why and how''' you
bit. First we added a function, that takes a Term (of the expression
+
come to write something like this:
to be evaluated), an Int (the result of the evaluation) and gives back
 
an output of type Output (that is a synonymous of String).
 
 
The evaluator changed quite a lot! Now it has a different type: it
 
takes a Term data type and produces a new type, we called MOut, that
 
is actually a pair of a variable type a (an Int in our evaluator) and
 
a type Output, a string.
 
So our evaluator, now, will take a Term (the type of the expressions
 
in our new programming language) and will produce a pair, composed of
 
the result of the evaluation (an Int) and the Output, a string.
 
 
So far so good. But what's happening inside the evaluator?
 
The first part will just return a pair with the number evaluated and
 
the output formatted by formatLine.
 
The second part does something more complicated: it returns a pair
 
composed by
 
1. the result of the evaluation of the right Term summed to the result
 
of the evaluation of the second Term
 
2. the output: the concatenation of the output produced by the
 
evaluation of the right Term, the output produced by the evaluation of
 
the left Term (each this evaluation returns a pair with the number and
 
the output), and the formatted output of the evaluation.
 
 
Let's try it:
 
*MyMonads> evalO (Add (Con 5) (Con 6))
 
(11,"eval (Con 5) <= 5 - eval (Con 6) <= 6 - eval (Add (Con 5) (Con 6)) <= 11 - ")
 
*MyMonads>
 
 
It works! Let's put the output this way:
 
eval (Con 5) <= 5 -
 
eval (Con 6) <= 6 -
 
eval (Add (Con 5) (Con 6)) <= 11 -
 
 
Great! We are able to produce a side effect of our evaluation and
 
present it at the end of the computation, after all.
 
 
Let's have a closer look at this expression:
 
 
<haskell>
 
<haskell>
+
drunk = do newLove "Paula "
evalO (Add t u) = ((a + b),(x ++ y ++ formatLine (Add t u) (a + b)))
+
paula <- (tellLover 1 10)
where (a, x) = evalO t
+
lorena <- tellMyself 3
(b, y) = evalO u
+
tellMyself (paula + lorena)
 
 
</haskell>
 
</haskell>
   
Why all that? The problem is that we need "a" and "b" to calculate their
+
That is to say, in this way you can see a do block as a series of
sum, together with the output coming from their calculation (to be
+
nested anonymous functions whose arguments are bound to some value by
concatenated by the expression x ++ y ++ formatLine ...).
+
the application of >>=. Anonymous functions that bind to some value
So we need to separate the pairs produced by "evalO t" and "eval u"
+
the variables appearing after the "->" ("paula" and "lorena").
(remember: eval now produces a value of type M Int, i.e. a pair of an
 
Int and a String!).
 
   
== Let's Go Monadic==
+
To summarize, I think that even if you can start using monads without
  +
understanding that what happens inside a "do" block is strictly
  +
related with lambda calculus, I don't think you can claim you
  +
understand monads just because you are using them.
   
Is there a more general way of doing so?
+
And I'm quite sure that if a problem arises somewhere you can have a
  +
very difficult time in trying to find out what the problem is. This is
  +
even more true when you start doing monadic transformations.
   
Let's analyze the evaluator from another perspective. From the type
+
==How to explain monads to newcomers?==
perspective.
 
We solved our problem by creating a new type, a pair of an Int (the
 
result of the evaluation) and a String (the output of the process of
 
evaluation).
 
   
The first part of the evaluator does nothing else but creating, from
+
Monads are not an impossible-to-understand-unless-you-have-a-phd
a value of type Int, an object of type M Int (Int,Output). It does so
+
topic. I don't know if I can claim I'm a living proof of this
by creating a pair with that Int and some text.
+
assumption, since I have a PhD. But please take into account that I
  +
have a PhD in Comparative Private Law! Nothing related to computer
  +
science. And I claim I understand monads.
   
The second part evaluates the two Term(s) and "stores" the values thus
+
Moreover, since I have come to understand them, I think I can try to
produced in some variables to be use later to compute the output.
+
explain them to newcomers like me. This is why I started writing this
  +
tutorial.
   
Let's focus on the "stores" action. The correct term should be
+
Still, in order to understand them I studied, and I studied hard. I
"binds".
+
wanted to understand, it was difficult, and I kept studying. Going
Take a function:
+
back to the foundation when needed.
<haskell>
 
f x = x + x
 
</haskell>
 
"x" appears on both sides of the expression. We say that on the right
 
side "x" is bound to the value of x given on the left side.
 
So
 
<haskell>
 
f 3
 
</haskell>
 
binds x to 3 for the evaluation of the expression "x + x".
 
Our evaluator binds "a" and "x" / "b" and "y" with the evaluation of
 
"eval t" and "eval u" respectively. "a","b","x" and "y" will be then
 
used in the evaluation of ((a+)(x ++ ...).
 
We know that there is an ad hoc operator for binding variables to a
 
value: lambda, or \.
 
Indeed f x = x + x is syntactic sugar for:
 
<haskell>
 
f = \x -> x + x
 
</haskell>
 
When we write f 3 we are actually binding "x" to 3 within what's next
 
"->", that will be used (substituted) for evaluating f 3.
 
   
So we can try to abstract this phenomenon.
+
So, I cannot explain monads to you unless you are willing to study. I can
What we need is a function that takes our composed type MOut Int and a
+
show you the way, but you must take your time and follow that way.
function in order to produce a new MOut Int, concatenating the
 
output of the computation of the first with the output of the
 
computation of the second.
 
This is what bindM does:
 
   
<haskell>
+
==What do I need to know to understand monads?==
   
bindM :: MOut a -> (a -> MOut b) -> MOut b
+
===Functional programming===
bindM m f = (b, x ++ y)
 
where (a, x) = m
 
(b, y) = f a
 
   
</haskell>
+
First you need at least a basic understanding of functional
  +
programming.
   
It takes:
+
This is going to be the easiest step, because there is an invaluable
1. "m": the compound type MOut Int carrying the result of an "eval
+
resource available on line for free.
Term",
 
2. a function "f". This function will take the Int ("a") extracted by the
 
evaluation of "m" ((a,x)=m). This function will produce anew pair: a
 
new Int produced by a new evaluation; some new output.
 
bindM will return the new Int in pair with the concatenated outputs
 
resulting from the evaluation of "m" and "f a".
 
   
So let's write the new version of the evaluator:
+
This is what I did: I spent 20 hours of my life watching the Video
  +
Lectures by Hal Abelson and Gerald Jay Sussman on
  +
[http://swiss.csail.mit.edu/classes/6.001/abelson-sussman-lectures/ Structure and Interpretation of Computer Programs].
   
<haskell>
+
This course, and the annexed text book (I did not read it entirely),
  +
are an amazing introduction to computer science: you start by learning
  +
some basic Scheme and then Abelson and Sussman will bring you, step by
  +
step, to understand evaluation, interpretation and compilation of Lisp
  +
(Scheme) code.
   
evalM_1 :: Term -> MOut Int
+
The course is clear, interesting, funny and will provide you with a
evalM_1 (Con a) = (a, formatLine (Con a) a)
+
basic, but strong, understanding of functional programming.
evalM_1 (Add t u) = bindM (evalM_1 t) (\a ->
 
bindM (evalM_1 u) (\b ->
 
((a + b), formatLine (Add t u) (a + b))
 
)
 
)
 
   
</haskell>
+
Believe me: if you like programming but don't have a computer science
  +
curriculum, you'll find out that spending 20 hours of your life to
  +
watch that course has been the most productive investment of your
  +
learning life.
   
Ugly, isn't it?
+
===My problem with Haskell===
   
Let's start from the outside:
+
I find Haskell an amazingly expressive programming language. It makes
bindM (evalM_1 u) (\b -> ((a + b), formatLine (Add t u) (a + b)))
+
it incredibly easy to perform very difficult tasks, just like monads,
  +
for instance.
   
bindM takes the result of the evaluation "evalM_1 u", a type Mout Int,
+
The problem is that, in doing so, it makes it difficult to understand
and a function. It will extract the Int from that type and use it to
+
Haskell to newcomers.
bind "b".
 
So in bindM (evalM_1 u)... "b" will be bound to a value.
 
   
Then the outer part (bindM (evalM_1 t) (\a...) will bind "a" to the
+
I must confess that tutorials and other learning material are quite
value needed to evaluate "((a+b), formatLine...) and produce our final
+
dissatisfying on this regards too.
MOut Int.
 
   
We can write the evaluator in a more convinient way, now that we know
+
Since Haskell seems to make easy what is not easy, these tutorial take
what it does:
+
the "it's easy" approach.
   
<haskell>
+
I will give you an example. I think that the
  +
[http://www.cs.utah.edu/~hal/htut/ "Yet Another Haskell Tutorial"] is
  +
a really good attempt to explain Haskell, and if you want to
  +
understand this tutorial you need to read it at least for getting a
  +
good understanding of the Haskell type system.
   
evalM_2 :: Term -> MOut Int
+
The tutorial explains monads and explains the "do notation" in terms
evalM_2 (Con a) = (a, formatLine (Con a) a)
+
of lambda abstraction (par. 9.1, p. 120). Still, to the lambda
evalM_2 (Add t u) = evalM_2 t `bindM` \a ->
+
operator ("\") the tutorial dedicates only 17 lines in par. 4.4.1.
evalM_2 u `bindM` \b ->
+
Moreover "\" is explained just as another way of creating functions.
((a + b), (formatLine (Add t u) (a + b)))
 
   
</haskell>
+
This probably makes sense for a functional programmer who is studying
  +
Haskell for the first time. But, on the other side it will make the
  +
newcomer, who's not a functional programmer, believe that explaining
  +
monads in terms of "\" is just ugly, and definitely not Haskell.
   
Now, look at the first part:
+
This is my problem with Haskell: it hides complexity with
  +
constructions like "let...in", "where", "do".
   
<haskell>
+
Still I believe that if you want to use those constructions you must
evalM_2 (Con a) = (a, formatLine (Con a) a)
+
know what they do. And they do ugly stuff like the one you'll be
</haskell>
+
looking at in this tutorial.
   
We could use a more general way of creating some output.
+
I am not saying that this is ''the'' way to learn Haskell. I'm just
  +
saying that this is the way I'm learning Haskell. And this is the way
  +
this tutorial has been written.
   
First we need a method for creating someting of type M a, starting from
+
===Starting from the inside or the outside?===
something of type a. This is what evalM_2 (Con a) is doing, after all.
 
Very simply:
 
   
<haskell>
+
In This tutorial I show monads from their inside. In "Meet Bob" I do
  +
the opposite. As I said I think (and I did it on purpose) that after
  +
reading "Meet Bob" you can have a general idea of what a monad is, but
  +
you need to go inside a monad to see how it works.
   
mkM :: a -> MOut a
+
As a suggestion, I'd invite you to start with "Meet Bob", and then
mkM a = (a, "")
+
procede with the tutorial.
   
</haskell>
+
I hope the these two approaches will give you an overall image of a
  +
monad.
   
We then need to "insert" some text (Output) in our type M:
+
===Prerequisites===
   
<haskell>
+
As I said, in order to understand this tutorial you need:
+
* a basic understanding of functional programming (this is required to understand Haskell after all)
outPut :: Output -> MOut ()
+
* a basic understanding of the type system:
outPut x = ((), x)
+
** how to create new types (with "data" and "newtype")
+
** how to use type constructors to construct and to match types and their internal components
</haskell>
+
** how to use a label field to retrieve a type internal component
+
* a basic understanding of lambda abstraction
Very simple: we have a string "x" (Output) and create a pair with a ()
 
instead of an Int, and the output.
 
 
This way we will be able to define also this firts part in terms of
 
bindM, that will take care of concatenating outputs.
 
 
So we have now a new evaluator:
 
 
<haskell>
 
 
evalM_3 :: Term -> MOut Int
 
evalM_3 (Con a) = outPut (formatLine (Con a) a) `bindM` \_ -> mkM a
 
evalM_3 (Add t u) = evalM_3 t `bindM` \a ->
 
evalM_3 u `bindM` \b ->
 
outPut (formatLine (Add t u) (a + b)) `bindM` \_ -> mkM (a + b)
 
 
</haskell>
 
 
Well, this is fine, definetly better then before, anyway.
 
Still we use `bindM` \_ -> that binds something we do not use (_). We
 
could write something for this case, when we concatenate computations
 
without the need of binding variables. Let's call it `combineM`:
 
 
<haskell>
 
 
combineM :: MOut a -> MOut b -> MOut b
 
combineM m f = m `bindM` \_ -> f
 
 
</haskell>
 
 
So the new evaluator:
 
 
<haskell>
 
 
evalM :: Term -> MOut Int
 
evalM (Con a) = outPut (formatLine (Con a) a) `combineM`
 
mkM a
 
evalM (Add t u) = evalM t `bindM` \a ->
 
evalM u `bindM` \b ->
 
outPut (formatLine (Add t u) (a + b)) `combineM`
 
mkM (a + b)
 
 
</haskell>
 
 
Let's put everything together (and change some names):
 
 
<haskell>
 
 
type MO a = (a, Out)
 
type Out = String
 
 
mkMO :: a -> MO a
 
mkMO a = (a, "")
 
 
bindMO :: MO a -> (a -> MO b) -> MO b
 
bindMO m f = (b, x ++ y)
 
where (a, x) = m
 
(b, y) = f a
 
 
combineMO :: MO a -> MO b -> MO b
 
combineMO m f = m `bindM` \_ -> f
 
 
outMO :: Out -> MO ()
 
outMO x = ((), x)
 
 
evalMO :: Term -> MO Int
 
evalMO (Con a) = outMO (formatLine (Con a) a) `combineMO`
 
mkMO a
 
evalMO (Add t u) = evalMO t `bindMO` \a ->
 
evalMO u `bindMO` \b ->
 
outMO (formatLine (Add t u) (a + b)) `combineMO`
 
mkMO (a + b)
 
 
</haskell>
 
 
== Some Sugar, Please!==
 
Now our evaluator has been completely transformed into a monadic
 
evaluator. That's what it is: a monad.
 
We have a function that constructs an object of type MO Int, formed by
 
a pair: the result of the evaluation and the accumulated
 
(concatenated) output.
 
The process of accumulation and the act of parting the MO Int into its
 
component is buried into bindM, that can also preserve some value for
 
later uses.
 
So we have:
 
MO a type constructor for a type carrying a pair composed by an Int
 
and a String;
 
bindMO, that gives a direction to the process of evaluation: it
 
concatenates computations and captures some side effects we created.
 
mkOM lets us create an object of type MO Int starting from an Int.
 
 
As you see this is all we need to create a monad. In other words
 
monads arise from the type system. Everything else is just syntactic
 
sugar.
 
 
So, let's have a look to that sugar: the famous do-notation!
 
 
Now we have to crate a new type: this is necessary in order to use
 
specific monadic notation and have at our disposal the more practical
 
do-notation.
 
 
<haskell>
 
 
newtype Eval a = Eval a
 
deriving (Show)
 
 
</haskell>
 
 
So, our type will be an instance of the monad class. We will have to
 
define the methods of this class (>>= and return), but that will be
 
easy since we already done that while defining bindMO and mkMO.
 
 
<haskell>
 
 
instance Monad Eval where
 
return a = Eval a
 
Eval m >>= f = f m
 
 
</haskell>
 
 
And then we will take the old version of our evaluator and substitute
 
`bindMO` with >>= and `mkMO` with return:
 
 
<haskell>
 
 
evalM_4 :: Term -> Eval Int
 
evalM_4 (Con a) = return a
 
evalM_4 (Add t u) = evalM_4 t >>= \a ->
 
evalM_4 u >>= \b ->
 
return (a + b)
 
 
</haskell>
 
 
which is, in the do-notation:
 
 
<haskell>
 
 
evalM_5 :: Term -> Eval Int
 
evalM_5 (Con a) = return a
 
evalM_5 (Add t u) = do a <- evalM_5 t
 
b <- evalM_5 u
 
return (a + b)
 
 
</haskell>
 
 
Simple: do binds the result of "eval_M5 t" to a, binds the result of
 
"eval_M5 u" to b and then returns the sum. In a very imperative style.
 
 
We can now have an image of our monad: it is out type (Eval) that is
 
made up of a pair: during evaluation the first member of the pair (the
 
Int) will get the results of our computation (i.e.: the procedures to
 
calculate the final result). The second part, the String called
 
Output, will get filled up with the concatenated output of the
 
computation.
 
The sequencing done by bindMO (now >>=) will take care of passing to
 
the next evaluation the needed Int and will do some more side
 
calculation to produce the output (concatenating outputs resulting
 
from computation of the new Int, for instance).
 
 
So we can grasp the basic concept of a monad: it is like a label which
 
we attach to each step of the evaluation (the String attached to the
 
Int).
 
This label is persistent within the process of computation and at each
 
step bindMO can do some manipulation of it.
 
We are creating side-effects and propagating them within our monads.
 
 
Ok. Let's translate out evaluator in monadic notation:
 
 
<haskell>
 
 
newtype Eval_IO a = Eval_IO (a, O)
 
deriving (Show)
 
type O = String
 
 
instance Monad Eval_IO where
 
return a = Eval_IO (a, "")
 
(>>=) m f = Eval_IO (b, x ++ y)
 
where Eval_IO (a, x) = m
 
Eval_IO (b, y) = f a
 
print_IO :: O -> Eval_IO ()
 
print_IO x = Eval_IO ((), x)
 
 
eval_IO :: Term -> Eval_IO Int
 
eval_IO (Con a) = do print_IO (formatLine (Con a) a)
 
return a
 
eval_IO (Add t u) = do a <- eval_IO t
 
b <- eval_IO u
 
print_IO (formatLine (Add t u) (a + b))
 
return (a + b)
 
 
</haskell>
 
   
  +
I hope I'll be able to write a short introductory summary to these
  +
topics (I will put it below this introductory remarks).
   
That's it. For today...
+
Have fun with Haskell!
   
(TO BE CONTINUED)
+
- [[User:AndreaRossato|Andrea Rossato]]
   
Andrea Rossato
+
[[Category:Tutorials]]
arossato AT istitutocolli.org
+
[[Category:Idioms]]
  +
[[Category:Monad]]

Latest revision as of 22:35, 24 July 2007

Note Since the size of the previous file was getting too big for a wiki, the tutorial has been divided into two parts: The Monadic Way Part I and The Monadic Way Part II. See below for some introductory remarks.

Contents

Meet Bob The Monadic Lover
A (supposed-to-be) funny and short introduction to Monads, with code but without any reference to category theory: what monads look like and what they are useful for, from the perspective of a ... lover. It could be an introduction to "The Monadic Way" tutorial.
The Monadic Way/Part I
In the first part of the tutorial we will start from a very simple evaluator that will be transformed into a monadic evaluator with an increasing number of features: output, exceptions, and state: a very simple counter for tracking the number of recursions of the evaluation precess.
The Monadic Way/Part II
In the second part of the tutorial we will see how to take complexity out of our monadic evaluator with the use of monadic transformers, and specifically StateT. This part is just a skeleton, since, for the time being, it contains only the code.


Contents

[edit] 1 Preliminary remarks

When I started writing this tutorial I though that the only way to explain monads to a newcomer was to show them from the inside, with the use of lambda abstraction. Not only because this is the way Philip Wedler's paper adopts, but also because I believed, and still believe, that the only way to understand what bind (>>=) does is to explain it as a function that takes a monad and an anonymous function.

I had this feeling because I am a newcomer, and this is the way I came to understand monads.

I did not received very much feedback for this tutorial, and I must admit that I would like to. But one person, on the haskell-cafe mailing list, told me that:

imho your tutorial makes the error that is a very typical: when you
write your tutorial you already know what are monads and what the
program you will construct at the end. but your reader don't know all these!
for such fresh reader this looks as you made some strange steps, write
some ugly code and he doesn't have chances to understand that this ugly
code is written just to show that this can be simplified by using monads.

I believe that Bulat is right. In this tutorial you go through some code that is really ugly and then, with some kind of magic, it turns out in the (redundant but) clean evaluator of the end of The Monadic Way/Part II.

I took that mail as a challenge and I responded by writing Meet Bob The Monadic Lover.

In "Meet Bob" the code is clean, variable names are very descriptive, and you see that I can create a monad without any use of lambda abstraction.

Bind (in "Meet Bob" is askLover) now takes a monad and a partial application, not an anonymous function.

Obviously you can see an anonymous function as a partial application.

The problem, I think, is that, in "Meet Bob", you cannot understand the strict relation between what I did before and what I do when I start using the "do-notation". You see that the same functions are being used ("tellMyself" and "newLove"), but "andrea <- tellMyself 1" is not explained. It's just magic.

The fact is that you cannot understand "andrea <- tellMyself 1" without the use of lambda abstraction.

I should have written an intermediate step, something like this:

drunk = do newLove "Paula " >>
                       (tellLover 1 10) >>= \paula ->
                           (tellMyself 3) >>= \lorena -> tellMyself (paula + lorena)

With this approach I think you can understand why and how you come to write something like this:

drunk = do newLove "Paula "
           paula <- (tellLover 1 10)
           lorena <- tellMyself 3
           tellMyself (paula + lorena)

That is to say, in this way you can see a do block as a series of nested anonymous functions whose arguments are bound to some value by the application of >>=. Anonymous functions that bind to some value the variables appearing after the "->" ("paula" and "lorena").

To summarize, I think that even if you can start using monads without understanding that what happens inside a "do" block is strictly related with lambda calculus, I don't think you can claim you understand monads just because you are using them.

And I'm quite sure that if a problem arises somewhere you can have a very difficult time in trying to find out what the problem is. This is even more true when you start doing monadic transformations.

[edit] 2 How to explain monads to newcomers?

Monads are not an impossible-to-understand-unless-you-have-a-phd topic. I don't know if I can claim I'm a living proof of this assumption, since I have a PhD. But please take into account that I have a PhD in Comparative Private Law! Nothing related to computer science. And I claim I understand monads.

Moreover, since I have come to understand them, I think I can try to explain them to newcomers like me. This is why I started writing this tutorial.

Still, in order to understand them I studied, and I studied hard. I wanted to understand, it was difficult, and I kept studying. Going back to the foundation when needed.

So, I cannot explain monads to you unless you are willing to study. I can show you the way, but you must take your time and follow that way.

[edit] 3 What do I need to know to understand monads?

[edit] 3.1 Functional programming

First you need at least a basic understanding of functional programming.

This is going to be the easiest step, because there is an invaluable resource available on line for free.

This is what I did: I spent 20 hours of my life watching the Video Lectures by Hal Abelson and Gerald Jay Sussman on Structure and Interpretation of Computer Programs.

This course, and the annexed text book (I did not read it entirely), are an amazing introduction to computer science: you start by learning some basic Scheme and then Abelson and Sussman will bring you, step by step, to understand evaluation, interpretation and compilation of Lisp (Scheme) code.

The course is clear, interesting, funny and will provide you with a basic, but strong, understanding of functional programming.

Believe me: if you like programming but don't have a computer science curriculum, you'll find out that spending 20 hours of your life to watch that course has been the most productive investment of your learning life.

[edit] 3.2 My problem with Haskell

I find Haskell an amazingly expressive programming language. It makes it incredibly easy to perform very difficult tasks, just like monads, for instance.

The problem is that, in doing so, it makes it difficult to understand Haskell to newcomers.

I must confess that tutorials and other learning material are quite dissatisfying on this regards too.

Since Haskell seems to make easy what is not easy, these tutorial take the "it's easy" approach.

I will give you an example. I think that the "Yet Another Haskell Tutorial" is a really good attempt to explain Haskell, and if you want to understand this tutorial you need to read it at least for getting a good understanding of the Haskell type system.

The tutorial explains monads and explains the "do notation" in terms of lambda abstraction (par. 9.1, p. 120). Still, to the lambda operator ("\") the tutorial dedicates only 17 lines in par. 4.4.1. Moreover "\" is explained just as another way of creating functions.

This probably makes sense for a functional programmer who is studying Haskell for the first time. But, on the other side it will make the newcomer, who's not a functional programmer, believe that explaining monads in terms of "\" is just ugly, and definitely not Haskell.

This is my problem with Haskell: it hides complexity with constructions like "let...in", "where", "do".

Still I believe that if you want to use those constructions you must know what they do. And they do ugly stuff like the one you'll be looking at in this tutorial.

I am not saying that this is the way to learn Haskell. I'm just saying that this is the way I'm learning Haskell. And this is the way this tutorial has been written.

[edit] 3.3 Starting from the inside or the outside?

In This tutorial I show monads from their inside. In "Meet Bob" I do the opposite. As I said I think (and I did it on purpose) that after reading "Meet Bob" you can have a general idea of what a monad is, but you need to go inside a monad to see how it works.

As a suggestion, I'd invite you to start with "Meet Bob", and then procede with the tutorial.

I hope the these two approaches will give you an overall image of a monad.

[edit] 3.4 Prerequisites

As I said, in order to understand this tutorial you need:

  • a basic understanding of functional programming (this is required to understand Haskell after all)
  • a basic understanding of the type system:
    • how to create new types (with "data" and "newtype")
    • how to use type constructors to construct and to match types and their internal components
    • how to use a label field to retrieve a type internal component
  • a basic understanding of lambda abstraction

I hope I'll be able to write a short introductory summary to these topics (I will put it below this introductory remarks).

Have fun with Haskell!

- Andrea Rossato