Ruby classes in the erlang world

those whom interested in both ruby and erlang,
I’d like to showing my small work.

I’ve been implemented a bit of ruby classes (String, Array, Range,
Regexp ,) using by erlang.
so could do something like

A = ruby:new([1,2,3]),
Stack = ruby:new([]),
[1,2,3] = (A(each))(fun(X) -> Stack({push,X+1}) end),
[2,3,4] = Stack(value),

I sent this to the erlang mailing list on a couple of weeks ago,
and got no replys.
well, I felt that erlang programmers don’t like this :slight_smile:
http://www.erlang.org/pipermail/erlang-questions/2007-June/027592.html

so what I really want to do is that
you know, erlang’s powerful concurrency.
check this working between two erlang nodes.

$ erl -sname b
([email protected])1> A=ruby:new([]).
#Fun<ruby.1.26108348>
([email protected])2> A({register, a}).
true

$ erl -sname c
([email protected])1> N=‘Node’:‘Node’({a,
[email protected]’}).
#Fun<ruby.1.26108348>
([email protected])2> N({push,1}).
[1]

([email protected])3> A(value).
[1]
([email protected])4> A({push,2}).

([email protected])3> N(value).
[1,2]

you could get the code from

svn co http://wookay.googlecode.com/svn/trunk/erlang/ruby/
cd ruby
make test

thank you.
have a good one.

On 7/6/07, Woo-Kyoung N. [email protected] wrote:

[2,3,4] = Stack(value),
check this working between two erlang nodes.
#Fun<ruby.1.26108348>
[1,2]

thank you.
have a good one.

I’m sure Erlang has fascilities for doing what you’re doing without
faking Ruby. What’s the value in using a crippled library nicked from
another language? Perhaps the users of the erlang mailing list simply
didn’t understand. I don’t think I understand either. What do you hope
to achieve by embedding Ruby standard classes in Erlang?

If you’re trying to implement Ruby on the Erlang platform, I bet a lot
of people would be interested, but simply being able to fake Array,
String, Range etc. just isn’t very interesting IMHO.

My guess you didn’t get a reply is because it’s like reinventing the
wheel
and add warts to it.
Compare with list comprehension
A = [1,2,3],
S = [X+1 || X <- A].

or even
A = [1,2,3],
S = lists:map(fun(X) -> X+1 end, A).

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