* (no subject)
@ 2002-10-16 14:07 climb
2002-10-17 0:58 ` [Caml-list] Re: how to start Alessandro Baretta
0 siblings, 1 reply; 2+ messages in thread
From: climb @ 2002-10-16 14:07 UTC (permalink / raw)
To: caml-list
Dear caml-list
i am a new commer of Caml language. and start with online mannul
i am confused at this sentence in Chapter 1
# type idref={ mutable id : 'a. 'a -> 'a};;
what does "." mean?
why i can not write
# type intidref= { mutable id : int. int->int};;
thanks
climb
onlyclimb@163.com
2002-10-16
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* [Caml-list] Re: how to start
2002-10-16 14:07 climb
@ 2002-10-17 0:58 ` Alessandro Baretta
0 siblings, 0 replies; 2+ messages in thread
From: Alessandro Baretta @ 2002-10-17 0:58 UTC (permalink / raw)
To: climb; +Cc: caml-list
climb wrote:
> Dear caml-list
>
> i am a new commer of Caml language.
Welcome on board!
> and start with online mannul
> i am confused at this sentence in Chapter 1
> # type idref={ mutable id : 'a. 'a -> 'a};;
> what does "." mean?
'a. 'a -> 'a
This is read as: "For every type 'a, a function from 'a to
'a". In this context 'a is a type variable. The type
variable preceding the "." are "universally quantified":
this is where the "For every type ..." comes into the game.
This basically states that the type of id is polymorphic: 'a
-> 'a is a type schema as opposed to a specific type. Many
different types fit this schema:
int -> int
string -> string
(int -> int) -> (int -> int), and so on.
The type schema 'a. 'a -> 'a is entirely equivalent to the
axiom schema of propositional calculus A => A: "If A then
A." A stands for *any* proposition, not for any one specific
proposition.
> why i can not write
> # type intidref= { mutable id : int. int->int};;
This does not make sense because "int" is a type value and
not a type variable.
Alex
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