[Caml-list] p4 (newbie) question

[Caml-list] p4 (newbie) question

[Ian Zimmerman]

Looking at the documentation for the MLast quotations in camlp4
reference manual, the very first node is

<:expr< $e1$ . $e2$ >> : access in records an in modules

Now I understand that camlp4 is all syntax, but even so, aren't these
two fundamentally different things?  In other words, shouldn't there
be 2 distinct nodes like this

<:expr< $e1$ . $e2$ >> : access in records
<:expr< $me1$ . $e2$ >> : access in modules

This is a practical issue when constructing my grammar --- I have
rules that derive module_expr, so how do I put them together with the
core expr rules to derive possibly qualified identifiers?

-- 
Ian Zimmerman, Oakland, California, U.S.A.
GPG: 433BA087  9C0F 194F 203A 63F7 B1B8  6E5A 8CA3 27DB 433B A087
In his own soul a man bears the source
from which he draws all his sorrows and his joys.
Sophocles.
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Re: [Caml-list] p4 (newbie) question

[Daniel de Rauglaudre]

Hi,

On Mon, Jan 21, 2002 at 10:22:23AM -0800, Ian Zimmerman wrote:

> shouldn't there be 2 distinct nodes like this
> 
> <:expr< $e1$ . $e2$ >> : access in records
> <:expr< $me1$ . $e2$ >> : access in modules

No: in these two quotations, e1, e2, me1, e2 are not some kind of
"keywords" like you seem to believe, but variables. These quotations
are (resp.) equivalent to:
       MLast.ExAcc (loc, e1, e2)
       MLast.ExAcc (loc, me1, e2)
which cannot discriminate according to the cases record/modules as
you see.

The difference is done by the first parameter when it represents (or
not) an uppercase identifier:
       MLast.ExAcc (loc, MLast.ExUid loc s, e2)
which can be written with quotations as:
       <:expr< $uid:s$ . $e2$ >>

That is (above) the code of contructing a module access if s is a
string holding the module name and e2 is an expression. If you already
know the name of your module, and if it is e.g. Foo, you can write it:
       <:expr< Foo . $e2$ >>
which is equivalent to:
       MLast.ExAcc (loc, MLast.ExUid loc "Foo", e2)

Note that you don't need to know the form with MLast.ExAcc, MLast.ExUid
and so on, but it is just to explain you what the quotations represent.

-- 
Daniel de RAUGLAUDRE
daniel.de_rauglaudre@inria.fr
http://cristal.inria.fr/~ddr/
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Re: [Caml-list] p4 (newbie) question

[Ian Zimmerman]

itz> shouldn't there be 2 distinct nodes like this
itz> 
itz> <:expr< $e1$ . $e2$ >> : access in records
itz> <:expr< $me1$ . $e2$ >> : access in modules

Daniel> No: in these two quotations, e1, e2, me1, e2 are not some kind
Daniel> of "keywords" like you seem to believe, but variables. These
Daniel> quotations are (resp.) equivalent to:

Daniel>        MLast.ExAcc (loc, e1, e2)
Daniel>        MLast.ExAcc (loc, me1, e2)

Daniel> which cannot discriminate according to the cases
Daniel> record/modules as you see.

Well, I guess I can rephrase my question then:  shouldn't there be both

MLast.ExRecAcc (loc, e1, e2)  and
MLast.ExModAcc (loc, me1, e2)

?

After all, doesn't p4 have to pass distinct trees to the compiler
proper in these two cases?

Daniel> The difference is done by the first parameter when it
Daniel> represents (or not) an uppercase identifier:

Daniel>        MLast.ExAcc (loc, MLast.ExUid loc s, e2)

Daniel> which can be written with quotations as:

Daniel>        <:expr< $uid:s$ . $e2$ >>

Daniel> That is (above) the code of contructing a module access if s
Daniel> is a string holding the module name and e2 is an
Daniel> expression. If you already know the name of your module, and
Daniel> if it is e.g. Foo, you can write it:

Daniel>        <:expr< Foo . $e2$ >>

But I don't know the name like that, it is not fixed; I am trying to
parse that as well.  So my me1 can be any module access path in
general (OK, no functor applications for now, but arbitrary depth).

It would be a different story if the dot were right-associative; then,
indeed, I could write something like

let barexp = <:expr< Bar . $exp$ >> in
<:expr< Foo . $barexp$ >>

But this is ungrammatical; normal ocaml grammar says that

Foo.Bar.exp

is to be parsed as

(Foo.Bar).exp


Antoher question, and one I am afraid I know the answer to:
where/what is the quotation for applicative record update 
{foo with bar = expr} ?

-- 
Ian Zimmerman, Oakland, California, U.S.A.
GPG: 433BA087  9C0F 194F 203A 63F7 B1B8  6E5A 8CA3 27DB 433B A087
In his own soul a man bears the source
from which he draws all his sorrows and his joys.
Sophocles.
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Re: [Caml-list] p4 (newbie) question

[Daniel de Rauglaudre]

Hi,

On Tue, Jan 22, 2002 at 06:46:32AM -0800, Ian Zimmerman wrote:

> Well, I guess I can rephrase my question then:  shouldn't there be both
> MLast.ExRecAcc (loc, e1, e2)  and
> MLast.ExModAcc (loc, me1, e2)

I prefer being far from semantics. I just consider syntax as
syntax. Same thing for "types" and "types declarations" which are
different in OCaml syntax tree, but not in Camlp4 syntax tree where
there are only "types".

> So my me1 can be any module access path in general (OK, no functor
> applications for now, but arbitrary depth).

Then use <:expr< $uid:s$ . $...$ >> in an iterator to create your
Camlp4 syntax tree.

> Antoher question, and one I am afraid I know the answer to:
> where/what is the quotation for applicative record update 
> {foo with bar = expr} ?

   <:expr< {(foo) with bar = expr} >>

-- 
Daniel de RAUGLAUDRE
daniel.de_rauglaudre@inria.fr
http://cristal.inria.fr/~ddr/
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