[lambda.git] / topics / _week5_system_F.mdwn

# System F and recursive types

In the simply-typed lambda calculus, we write types like <code>&sigma;
-> &tau;</code>.  This looks like logical implication.  We'll take
that resemblance seriously when we discuss the Curry-Howard
correspondence.  In the meantime, note that types respect modus
ponens: 

<pre>
Expression    Type      Implication
-----------------------------------
fn            &alpha; -> &beta;    &alpha; &sup; &beta;
arg           &alpha;         &alpha;
------        ------    --------
(fn arg)      &beta;         &beta;
</pre>

The implication in the right-hand column is modus ponens, of course.

System F was discovered by Girard (the same guy who invented Linear
Logic), but it was independently proposed around the same time by
Reynolds, who called his version the *polymorphic lambda calculus*.
(Reynolds was also an early player in the development of
continuations.)  

System F enhances the simply-typed lambda calculus with abstraction
over types.  In order to state System F, we'll need to adopt the
notational convention that "<code>x:&alpha;</code>" represents an
expression `x` whose type is <code>&alpha;</code>.

Then System F can be specified as follows (choosing notation that will
match up with usage in O'Caml, whose type system is based on System F):

        System F:
        ---------
        types τ ::= c | 'a | τ1 -> τ2 | ∀'a. τ
        expressions e ::= x | λx:τ. e | e1 e2 | Λ'a. e | e [τ]

In the definition of the types, "`c`" is a type constant (e.g., `e` or
`t`, or in arithmetic contexts, `N` or `Int`).  "`'a`" is a type
variable (the tick mark just indicates that the variable ranges over
types rather than over values).  "`τ1 -> τ2`" is the type of a
function from expressions of type `τ1` to expressions of type `τ2`.
And "`∀'a. τ`" is called a universal type, since it universally
quantifies over the type variable `'a`.  (You can expect that in
`∀'a. τ`, the type `τ` will usually have at least one free occurrence
of `'a` somewhere inside of it.)

In the definition of the expressions, we have variables "`x`" as usual.
Abstracts "`λx:τ. e`" are similar to abstracts in the simply-typed lambda
calculus, except that they have their shrug variable annotated with a
type.  Applications "`e1 e2`" are just like in the simply-typed lambda calculus.

In addition to variables, abstracts, and applications, we have two
additional ways of forming expressions: "`Λ'a. e`" is called a *type
abstraction*, and "`e [τ]`" is called a *type application*.  The idea
is that <code>&Lambda;</code> is a capital <code>&lambda;</code>: just
like the lower-case <code>&lambda;</code>, <code>&Lambda;</code> binds
variables in its body, except that unlike <code>&lambda;</code>,
<code>&Lambda;</code> binds type variables instead of expression
variables.  So in the expression

<code>&Lambda; 'a (&lambda; x:'a . x)</code>

the <code>&Lambda;</code> binds the type variable `'a` that occurs in
the <code>&lambda;</code> abstract.  This expression is a polymorphic
version of the identity function.  It defines one general identity
function that can be adapted for use with expressions of any type. In order
to get it ready to apply to, say, a variable of type boolean, just do
this:

<code>(&Lambda; 'a (&lambda; x:'a . x)) [t]</code>    

This type application (where `t` is a type constant for Boolean truth
values) specifies the value of the type variable &alpha;, which is
the type of the variable bound in the &lambda; expression.  Not
surprisingly, the type of this type application is a function from
Booleans to Booleans:

<code>((&Lambda; 'a (&lambda; x:'a . x)) [t]): (b -> b)</code>    

Likewise, if we had instantiated the type variable as an entity (base
type `e`), the resulting identity function would have been a function
of type `e -> e`:

<code>((&Lambda; 'a (&lambda; x:'a . x)) [e]): (e -> e)</code>    

Clearly, for any choice of a type `'a`, the identity function can be
instantiated as a function from expresions of type `'a` to expressions
of type `'a`.  In general, then, the type of the unapplied
(polymorphic) identity function is

<code>(&Lambda; 'a (&lambda; x:'a . x)): (&forall; 'a . 'a -> 'a)</code>

Pred in System F
----------------

We saw that the predecessor function couldn't be expressed in the
simply-typed lambda calculus.  It can be expressed in System F, however.

[See Benjamin C. Pierce. 2002. *Types and Programming Languages*, MIT
Press, pp. 350--353, for `tail` for lists in System F.]


Types in OCaml
--------------

OCaml has type inference: the system can often infer what the type of
an expression must be, based on the type of other known expressions.

For instance, if we type

    # let f x = x + 3;;

The system replies with

    val f : int -> int = <fun>

Since `+` is only defined on integers, it has type

     # (+);;
     - : int -> int -> int = <fun>

The parentheses are there to turn off the trick that allows the two
arguments of `+` to surround it in infix (for linguists, SOV) argument
order. That is,

    # 3 + 4 = (+) 3 4;;
    - : bool = true

In general, tuples with one element are identical to their one
element:

    # (3) = 3;;
    - : bool = true

though OCaml, like many systems, refuses to try to prove whether two
functional objects may be identical:

    # (f) = f;;
    Exception: Invalid_argument "equal: functional value".

Oh well.

[Note: There is a limited way you can compare functions, using the
`==` operator instead of the `=` operator. Later when we discuss mutation,
we'll discuss the difference between these two equality operations.
Scheme has a similar pair, which they name `eq?` and `equal?`. In Python,
these are `is` and `==` respectively. It's unfortunate that OCaml uses `==` for the opposite operation that Python and many other languages use it for. In any case, OCaml will accept `(f) == f` even though it doesn't accept
`(f) = f`. However, don't expect it to figure out in general when two functions
are equivalent. (That question is not Turing computable.)

        # (f) == (fun x -> x + 3);;
        - : bool = false

Here OCaml says (correctly) that the two functions don't stand in the `==` relation, which basically means they're not represented in the same chunk of memory. However as the programmer can see, the functions are extensionally equivalent. The meaning of `==` is rather weird.]



Booleans in OCaml, and simple pattern matching
----------------------------------------------

Where we would write `true 1 2` in our pure lambda calculus and expect
it to evaluate to `1`, in OCaml boolean types are not functions
(equivalently, they're functions that take zero arguments). Instead, selection is
accomplished as follows:

    # if true then 1 else 2;;
    - : int = 1

The types of the `then` clause and of the `else` clause must be the
same.

The `if` construction can be re-expressed by means of the following
pattern-matching expression:

    match <bool expression> with true -> <expression1> | false -> <expression2>

That is,

    # match true with true -> 1 | false -> 2;;
    - : int = 1

Compare with

    # match 3 with 1 -> 1 | 2 -> 4 | 3 -> 9;;
    - : int = 9

Unit and thunks
---------------

All functions in OCaml take exactly one argument.  Even this one:

    # let f x y = x + y;;
    # f 2 3;;
    - : int = 5

Here's how to tell that `f` has been curry'd:

    # f 2;;
    - : int -> int = <fun>

After we've given our `f` one argument, it returns a function that is
still waiting for another argument.

There is a special type in OCaml called `unit`.  There is exactly one
object in this type, written `()`.  So

    # ();;
    - : unit = ()

Just as you can define functions that take constants for arguments

    # let f 2 = 3;;
    # f 2;;
    - : int = 3;;

you can also define functions that take the unit as its argument, thus

    # let f () = 3;;
    val f : unit -> int = <fun>

Then the only argument you can possibly apply `f` to that is of the
correct type is the unit:

    # f ();;
    - : int = 3

Now why would that be useful?

Let's have some fun: think of `rec` as our `Y` combinator.  Then

    # let rec f n = if (0 = n) then 1 else (n * (f (n - 1)));;
    val f : int -> int = <fun>
    # f 5;;
    - : int = 120

We can't define a function that is exactly analogous to our &omega;.
We could try `let rec omega x = x x;;` what happens?

[Note: if you want to learn more OCaml, you might come back here someday and try:

        # let id x = x;;
        val id : 'a -> 'a = <fun>
        # let unwrap (`Wrap a) = a;;
        val unwrap : [< `Wrap of 'a ] -> 'a = <fun>
        # let omega ((`Wrap x) as y) = x y;;
        val omega : [< `Wrap of [> `Wrap of 'a ] -> 'b as 'a ] -> 'b = <fun>
        # unwrap (omega (`Wrap id)) == id;;
        - : bool = true
        # unwrap (omega (`Wrap omega));;
    <Infinite loop, need to control-c to interrupt>

But we won't try to explain this now.]


Even if we can't (easily) express omega in OCaml, we can do this:

    # let rec blackhole x = blackhole x;;

By the way, what's the type of this function?

If you then apply this `blackhole` function to an argument,

    # blackhole 3;;

the interpreter goes into an infinite loop, and you have to type control-c
to break the loop.

Oh, one more thing: lambda expressions look like this:

    # (fun x -> x);;
    - : 'a -> 'a = <fun>
    # (fun x -> x) true;;
    - : bool = true

(But `(fun x -> x x)` still won't work.)

You may also see this:

        # (function x -> x);;
        - : 'a -> 'a = <fun>

This works the same as `fun` in simple cases like this, and slightly differently in more complex cases. If you learn more OCaml, you'll read about the difference between them.

We can try our usual tricks:

    # (fun x -> true) blackhole;;
    - : bool = true

OCaml declined to try to fully reduce the argument before applying the
lambda function. Question: Why is that? Didn't we say that OCaml is a call-by-value/eager language?

Remember that `blackhole` is a function too, so we can
reverse the order of the arguments:

    # blackhole (fun x -> true);;

Infinite loop.

Now consider the following variations in behavior:

    # let test = blackhole blackhole;;
    <Infinite loop, need to control-c to interrupt>

    # let test () = blackhole blackhole;;
    val test : unit -> 'a = <fun>

    # test;;
    - : unit -> 'a = <fun>

    # test ();;
    <Infinite loop, need to control-c to interrupt>

We can use functions that take arguments of type `unit` to control
execution.  In Scheme parlance, functions on the `unit` type are called
*thunks* (which I've always assumed was a blend of "think" and "chunk").

Question: why do thunks work? We know that `blackhole ()` doesn't terminate, so why do expressions like:

        let f = fun () -> blackhole ()
        in true

terminate?

Bottom type, divergence
-----------------------

Expressions that don't terminate all belong to the **bottom type**. This is a subtype of every other type. That is, anything of bottom type belongs to every other type as well. More advanced type systems have more examples of subtyping: for example, they might make `int` a subtype of `real`. But the core type system of OCaml doesn't have any general subtyping relations. (Neither does System F.) Just this one: that expressions of the bottom type also belong to every other type. It's as if every type definition in OCaml, even the built in ones, had an implicit extra clause:

        type 'a option = None | Some of 'a;;
        type 'a option = None | Some of 'a | bottom;;

Here are some exercises that may help better understand this. Figure out what is the type of each of the following:

        fun x y -> y;;

        fun x (y:int) -> y;;

        fun x y : int -> y;;

        let rec blackhole x = blackhole x in blackhole;;

        let rec blackhole x = blackhole x in blackhole 1;;

        let rec blackhole x = blackhole x in fun (y:int) -> blackhole y y y;;

        let rec blackhole x = blackhole x in (blackhole 1) + 2;;

        let rec blackhole x = blackhole x in (blackhole 1) || false;;

        let rec blackhole x = blackhole x in 2 :: (blackhole 1);;

By the way, what's the type of this:

        let rec blackhole (x:'a) : 'a = blackhole x in blackhole


Back to thunks: the reason you'd want to control evaluation with
thunks is to manipulate when "effects" happen. In a strongly
normalizing system, like the simply-typed lambda calculus or System F,
there are no "effects." In Scheme and OCaml, on the other hand, we can
write programs that have effects. One sort of effect is printing.
Another sort of effect is mutation, which we'll be looking at soon.
Continuations are yet another sort of effect. None of these are yet on
the table though. The only sort of effect we've got so far is
*divergence* or non-termination. So the only thing thunks are useful
for yet is controlling whether an expression that would diverge if we
tried to fully evaluate it does diverge. As we consider richer
languages, thunks will become more useful.