"ocaml_beginners"::[] Easy Vector Graphics from OCaml

We just released our vector graphics library with both free and commercial
editions:

http://www.ffconsultancy.com/products/smoke_vector_graphics/?o

This library lets you create stunning interactive vector graphics quickly and
easily from OCaml programs.

The site contains three downloadable demos for Linux (both 32- and 64-bit) and
the library is freely available for non-commercial use in the form of
compiled bytecode for OCaml 3.09.2.

Happy hacking!

--
Dr Jon D Harrop, Flying Frog Consultancy Ltd.
The F#.NET Journal
http://www.ffconsultancy.com/products/fsharp_journal/?e

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I forget what this kind of recursion is called and I cannot seem to get it to
work.

Here is an (failing) example for a function that counts how many times it has
been called with an even argument:

let numeven b =
let rec aux a b =
(a, aux (if b mod 2 = 0 then a + 1 else a))
in aux 0

It errors with: This expression has type int * (int -> int * 'a) but is here
used with type int * 'a

I know I can write a side-effecting version using a hidden ref, but I need a
functional version. Does any one know how do you do this in OCaml and what
is this type of thing is called?

Thanks in advance,

Peng
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On May 31, 2007, at 4:05 PM, Peng Zang wrote:
>
> let numeven b =
> let rec aux a b =
> (a, aux (if b mod 2 = 0 then a + 1 else a))
> in aux 0
>

You're only applying aux to one argument (both times). Did you mean:

let numeven b =
let rec aux a b =
aux a (if b mod 2 = 0 then + 1 else a)
in
aux 0 b

--Jonathan

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.

Hi, not sure what you are trying to do, but if you want to be
side-effects free, you can carry the state with you in each
invocation, like:

# open Printf ;;
# let multiply_and_count state number =
let n_state = if number mod 2 = 0 then state +1 else state in
n_state, number * 2;;
val multiply_and_count : int -> int -> int * int = <fun>;

# let st,r = multiply_and_count 0 3;;
val st : int = 0
val r : int = 6
# let st2, r2 = multiply_and_count st 4;;
val st2 : int = 1
val r2 : int = 8

2007/5/31, Jonathan Bryant < jtbryant%40valdosta.edu">jtbryantvaldosta.edu>:
>
>; On May 31, 2007, at 4:05 PM, Peng Zang wrote:
&gt; >
>; > let numeven b =
> > let rec aux a b =
> > (a, aux (if b mod 2 = 0 then a + 1 else a))
> > in aux 0
> >
>;
> You're only applying aux to one argument (both times). Did you mean:
&gt;
> let numeven b =
> let rec aux a b =
> aux a (if b mod 2 = 0 then + 1 else a)
> in
> aux 0 b
>
> --Jonathan
>
&gt;

--
Pablo Polvorin

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On May 31, 2007, at 4:05 PM, Peng Zang wrote:
&gt;
> Here is an (failing) example for a function that counts how many
> times it has
> been called with an even argument:
>
Sorry, didn't read that part: I jumped ahead to the error message.
You could use some sort of continuation, but a ref would be much easier.

--Jonathan

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Ha! You're brilliant. That's exactly what I wanted. I didn't know you had
to explictly enable recursive types. Why isn't this enabled by default?
Slow compile times?? Safety?

Peng

On Thursday 31 May 2007 17:24, Zheng Li wrote:
&gt; Peng Zang < peng.zang%40gmail.com">peng.zanggmail.com> writes:
> > -----BEGIN PGP SIGNED MESSAGE-----
> > Hash: SHA1
>; >
>; > I forget what this kind of recursion is called and I cannot seem to get
> > it to work.
&gt; >
>; > Here is an (failing) example for a function that counts how many times it
> > has been called with an even argument:
> >
>; > let numeven b =
> > let rec aux a b =
> > (a, aux (if b mod 2 = 0 then a + 1 else a))
> > in aux 0
> >
>; > It errors with: This expression has type int * (int -> int * 'a) but is
> > here used with type int * 'a
> >
>; > I know I can write a side-effecting version using a hidden ref, but I
> > need a functional version. Does any one know how do you do this in OCaml
&gt; > and what is this type of thing is called?
>
> Enable recursive type, if that's what you really want.
&gt;
> $ ocaml -rectypes
> Objective Caml version 3.09.2
&gt;
> # let numeven b =
> let rec aux a b =
> (a, aux (if b mod 2 = 0 then a + 1 else a)) in
> aux 0;;
> val numeven : 'a -> (int -> int * 'b as 'b) = <fun>;
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