(no subject)

[climb]

Dear caml-list
 
 Could you explain this senence for me in the Ocaml mannul in Using coerecions?
" When the domain of a coercion is left implicit and its co-domain is an abbreviation of a known class type, then the class type rather than the object type, is used to derive the coercion functions".

i am not quite clear the defintion of  codomain ,  class type and object type  respectively?


Best regards
            
              Yours
                         climb
                         onlyclimb@163.com
                         2002-10-18

 




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[Caml-list] Re: how to start

[Florian Hars]

climb wrote:
>  Could you explain this senence for me in the Ocaml mannul in Using coerecions?

Well, the correct answer to your last questions is probably:

  Don't try to learn Ocaml from the reference manual.

It will tell you everything about the language, but not necessaryly in the
order you need to know it, and it doesn't tell you which parts of the language
are fundamental (core language, modules), optional (labels, objects) or arcane
(universal quantification, when to omit the first type in a coercion). Try the
tutorial by Jason Hickey first.

Note to INRIA: Describing the manual with "The Objective Caml manual begins
with short tutorials on the core language, objects and classes, and modules."
on the tutorials page is slightly misleading, those tutorials don't really add
up to an introduction to ocaml.

Yours, Florian.

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[Caml-list] Re: how to start

[Alessandro Baretta]

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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