From: Pierre Weis <Pierre.Weis@inria.fr>
Subject: Re: Weak types ?
In-Reply-To: <199802031546.QAA21882@scrasmeustache.labri.u-bordeaux.fr> from Pierre CASTERAN at "Feb 3, 98 04:46:22 pm"
To: Pierre.Casteran@labri.u-bordeaux.fr (Pierre CASTERAN)
Date: Mon, 9 Feb 1998 17:00:34 +0100 (MET)
> I suppose the following problem concerns "weak" types, but i don't see
> the reason (I have no assignment at all in this definitions !)
This is an excerpt from the ``expert'' part of the Caml FAQ
that explains the problem, its solution in Caml, and why it is considered a
You could read it in french as well in the FAQ, at URL:
How to introduce polymorphism ?
Polymorphism appears when we introduce type schemes, that is types
with universally quantified type variables. These type scheme
parameters are denoted by the prefix ' (for instance 'a). All type
scheme parameters are universally quantified at the head of the
schema: so 'a -> 'a means: For all types 'a, 'a -> 'a. This
quantification of type parameters is implicit in Caml:
- : ('a -> 'b) -> 'a list -> 'b list = <fun>
The type (schema) of map means For all types 'a, 'b, ('a -> 'b) -> 'a
list -> 'b list.
The polymorphism is only introduced by the definition of names during
a let construct (either local, or global).
The problem and its solution:
The original rule for the local definition
let name = expr1 in expr2
states that you have to type check expr1 first, then give name a type
scheme (a polymorphic type) according to the type obtained for expr1,
then type-check expr2, taking a copy of the type scheme of name for
each occurrence of name. The schema of name is obtained by finding
type parameters, that is all the type variables that have been
introduced during the type-checking of expr1, and that remain
unconstrained at the end of the type-checking of expr1. For instance:
let id = function x -> x in ...
id gets the type scheme: For all 'a. 'a -> 'a, and can be used with
several incompatible types in the rest of the program, for instance
with types int -> int and bool -> bool, if the in part is id 1; id
Unfortunately this simple rule is not correct in the presence of
mutable values (such as references, arrays, or records with mutable
let bad_ref = ref  in ...
With the original let rule, bad_ref has the type schema ``For all
'a. 'a list ref''. So the reference can be used with different
incompatible types in the rest of the program, which is erroneous
(lists must be homogeneous in Caml). For instance, if you use bad_ref
with types int list ref and bool list ref, then you can write:
let bad_ref = ref  in
bad_ref := ;
if hd (!bad_ref) then ...
(bad_ref :=  is correctly type-checked, since bad_ref can be used
with type int list ref, hd (!bad_ref) is correctly type-checked, since
bad_ref can be used with type type bool list ref. But this program
leads to a fatal error at run-time, since after the assignment,
bad_ref contains an integer list not a boolean list.
The new rule to define names:
In short, a secure type system for Caml must forbid the existence of
polymorphic mutable values at run-time. This has been extensively
studied, and many complex systems have been proposed. It appears now
that a very simple modification of the original rule is almost
completely satisfactory. When type-checking
let name = expr1 ...
The type of expr1 is generalized when expr1 is a function, an
identifier or a constant. Otherwise the identifier name is not
polymorphic (type variables are not generalized).
Which programs are now monomorphic ?
The new rule implies that if expr1 is a function application, then the
identifier name is monomorphic. For instance
let bad_ref = ref  in ...
bad_ref is not polymorphic, and that's what we want. In contrast:
let map_id = map (function x -> x) in ...
Since map_id is defined by an application, its type is not generalized
and map_id cannot be used with several different types. So:
#let map_id = map (function x -> x) in
map_id ; map_id [true];;
>map_id ; map_id [true];;
This expression has type bool list,
but is used with type int list.
is ill-typed (the first use of map_id, that is map_id  implies that
map_id has type int list -> int list, and map_id [true] is
ill-typed). The next section explains how to get a polymorphic
definition for map_id, and how to get a polymorphic result when a
polymorphic function is partially applied.
My function is not polymorphic (enough) ?
How to allow polymorphic definitions obtained via partial application ?
The more common case to get a ``not polymorphic enough'' definition is
when defining a function via partial application of a general
polymorphic function. In this case, you recover a fully polymorphic
definition by clearly exhibiting the functionality to the
type-checker: define the function with an explicit functional
abstraction, that is, add a function construct or an extra parameter
(this rewriting is known as eta-expansion):
let map_id l = map (function x -> x) l in ...
let map_id = function l -> map (function x -> x) l in ...
The new definition is semantically equivalent and can be assigned a
polymorphic type scheme, since it is no more a function application.
A type variable starts with _ ?
When an identifier is defined by an expression which cannot be
generalized, the identifier can have a type with unknowns (that is
type variables which cannot be generalized) which will be determined
(that is assigned a type value) by the usage of the identifier in the
rest of the program. These unknown types are special type variables,
prefixed by an underscore (e.g. '_a). Beware, these variables stand
for unknown types to be determined by the type checking process:
that's why these variables may disappear, precisely because their type
value has been discovered.
let map_id = map (function x -> x) in
After its definition map_id has type '_a list -> '_a list, where '_a
means some type a. When typing map_id  this unknown type gets bound
to int (and map_id has type int list -> int list for the rest of the
This phenomenon appears even more clearly if we break the let in
construct in two phrases:
#let map_id = map (function x -> x);;
map_id : '_a list -> '_a list = <fun>
- : int list = 
Now the unknown '_a in the type of map_id is determined, and map_id
has a type without unknowns:
- : int list -> int list = <fun>
The following section
My program is not polymorphic (enough)?
Semantically, the definition of an identifier by a let construct is
equivalent to a function application. More exactly, let ident = e1 in
e2 is equivalent to (function ident -> e2) e1. This means that the two
programs produce the same results and the same effects.
However, the two programs are not completely equivalent as far as type
checking is concerned: in effect, the let version may introduce some
polymorphism, while the function application does not, and forces
ident to be completely monomorphic when typing the expression
#let id = function x -> x in id id;;
- : '_a -> '_a = <fun>
#(function id -> id id) (function x -> x);;
>(function id -> id id) (function x -> x);;
This expression has type 'a -> 'b,
but is used with type 'a.
INRIA, Projet Cristal, Pierre.Weis@inria.fr, http://pauillac.inria.fr/~weis/
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