# Monad is just an ADT. Period.

The amount of nonsense repeated about monads is unimaginable. The famous meme *Idiots, idiots everywhere..* comes to mind.

Monad is a *parametrized type*. This means that there is some *type-constructor* that takes a *value of type a* and returns a

*value of type*.

`m a`

The technical term "type-constructor" is misleading, because what it returns is *data*, not a *type*, which is mere an abstraction. So it should be called *data-constructor*.

Logically (and technically), it is equivalent to attaching of *an additional type-tag* to a value of some "primitive" type (already tagged).

Just having a new type tag attached is not enough. The tags (or similar "marking" mechanism) are used by *type-checker*.

To be a Monadic, an ADT must implement at least two procedures. So-called *bind* which is traditionally denoted in Haskell as `>>=`

and `return`

.

`return`

is simple - it is just a data-constructor. It what they call it *lifts a value in a world of Monads*. Logically it just attaches an additional type-tag to a given value.

`>>=`

is messy. Its type in Haskell is `m a -> (a -> m b) -> m b`

. It is a *binary operator*
which supposed to be written `m >>= f`

which in Haskell means a *curried procedure*.

The function `f`

must be of the type `a -> m b`

which means, it must apply the `return`

procedure to the result of computation as its last *expression*. This would be in turn a return value of `>>=`

procedure.

This is an *interface* an ADT should implement to be *Monadic*.

Makes any sense? NO. Here is why.

In a *lazy* "pure-functional" languages such as Haskell (in contrast with, say, Erlang, which is also "pure-functional" but *eager*), the order of evaluation of an expression is undefined. This means that sub-expressions of a compound expression could be evaluated (reduced) in *any* order - not just left-to-right or right-to-left (sequentially) but *out-of-order*, even (in parallel).

What `>>=`

does is "takes a value of type `m a`

, strips Monad type-tag from a value, takes a procedure `f`

to be applied to a value of type `a`

, wraps an application of a procedure to the value into a *thunk* and returns it. (Technically, thunk is created implicitly by language runtime). This thunk, when evaluated, would produce a value of type `m b`

.

Does it make sense? Little.

Here comes the realization. **Monads makes no sense in strict languages**. In these languages we don't have to create an implicit *thunk* (which is just a *lambda expression*) to ensure a *sequential* order of evaluation. There are special forms for that, like `begin`

or `progn`

or `for-each`

.

So what is a Monad type really? It is an awkward way of what they call it "to abstracting a sequential computation" in a language without well-defined evaluation order. All these idioms and silly function names are mere *accidents* from some obscure "scientific" paper.

It is here not because it is some fundamental thing (why, it is just an awkward ADT), but because *idiots everywhere* copy-pasted pieces of code from each other without deep understanding what it really does and *what it is for*.

Now look, given that `ma`

and `mb`

are *thunks* of type `m a`

and `m b`

respectively (*types are for a type-checker*), the expression `ma >> mb`

is nothing but `(begin (force ma) mb)`

and `ma >>= f`

is `(f (force ma))`

where `f`

must apply `(delay ...)`

as its last expression.

So, `ma >> mb`

is *force ma, return mb as is (unevaluated)*. This is for side-effects.

and `ma >>= f`

is *force ma, then make a new "thunk", which when evaluated, will apply f to a*. This is for "serialization" in a

*non-strict language*.

Note that `>>=`

"forces" `ma`

into `a`

. Or we could say "strips `m`

form `m a`

".

Note that `f`

must use `return`

.

This "generator-like" `do`

-notation

do a <- ma b <- mb return (a+b)

is just a *syntactic sugar* for *nested lambdas*

ma >>= (\a -> mb >>= (\b -> return (a + b)))

For a nice visualization we could think of `ma >> mb`

as a rough equivalent of `$ cp xxx yyy ; sync`

and `ma >>= f`

is a-la `$ cat xxx | grep yyy`

or a *pipe*. The crucial difference that only the first (leftmost) parameter is fully evaluated, and a *promise* (a *thunk* to be "forced" later) is returned.

Notice that these expressions are made of nested *lambdas* and a "binding" procedure `>>=`

, so `x`

and `y`

are *lexical scoped bindings*, used in `return`

expression.

getChar >>= \x -> getChar >>= \y -> return (x,y)

which is rather an awkward way *to pass a value* from "effectful", "non-pure", "unsafe" `(getChar)`

to a nested `(lambda (x) ...`

(>>= (getChar) (lambda (x) (>>= (getChar) (lambda (y) (return (list x y))))))

Remember that `>>=`

does "strict evaluation" of its first argument - a monad (in Scheme it would be `(force m)`

) and "binds" (or *captures*) the value to a given procedure (creates a *closure*).

(define (>>= m f) (f (force m)))

In Haskell everything *type-checks* (ensures correct types of each binding and a valid structure of whole expressions. *It does NOT guarantee correctness of execution or any kind of "safety").
*

Actually, in Haskell `x`

and `y`

are *patterns*. In this case they are unbound variables, so they match any value.

do x <- getChar y <- getChar return (x,y)

**a Monad is mere an ADT** - a construct to explicitly pass a value from in a non-strict language. To ensure the order of evaluation they are using "nested lambda expressions" or `do`

notation as a *syntactic sugar*.

Why do we have this `>>=`

procedure instead just nesting? To make a type-checker "happy", to annotate "passing of a value" from `m a`

to a procedure that takes `a`

and produces a new Monad `m b`

.

That's all.

*In a strict languages Monad is just a meaningless ADT*:

constructor (add a type-tag)

(define (make-monad x) (cons 'monad x))

selectror (strip a type-tag)

(define (monad-get m) (cdr m))

predicate

(define (monad? x) (eqv? 'monad (car x)))

to be used in `a -> m a`

procedures

(define return make-monad)

procedure application, `f`

must use `return`

(define (>>= m f) (f (monad-get m)))

That is it. This is what your Monadic IO really does. `>>=`

strips type information and applies a given procedure `f`

to a value of type `a`

. The `f`

procedure itself must call `return`

with the value it produces to stick the type-tag back. The result of the last expression will be the result of a whole procedure `f`

.

eval a, return b

(define (>> a b) (>>= a (lambda (_) b)))

Show me, please, where the "safety" of Monadic IO comes from..

For each new type we have to `>>=`

and `return`

to be implemented.

For lists it would be

(define return (lambda (x) (cons x '())))

(define (>>= xs f) (map f xs))

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