I'm writing something which is partially written in ruby, but uses an
extension which adds methods to some of the classes in the ruby part.
Right now I've got a directory which contains both the ruby code and
the so containing the extensions
$ls
mycode.rb mycode_prims.so
in mycode.rb I've got something like:
class Mycode
def foo
end
end
# link in the primitives, mycode_prims includes various calls to
# rb_define_method to add methods to the Mycode class.
require 'mycode_prims'
Now this works if the directory containing both parts is the current
directory, but not otherwise. Actually I guess it would work as long
as the directory was on the load path, but I'm concerned about the
possibility of loading the wrong mycode_prims.so.
Is there a way to get the directory containing the current source
file? __FILE__ only gives the file name, but not the path.
What is the right way to do this? I'd like to know how to approach
this either with the code packaged as a gem or as 'source'.
···
--
Rick DeNatale
My blog on Ruby
http://talklikeaduck.denhaven2.com/
one idea is to do something like this
in foo.rb
require 'rbconfig'
arch = Config::CONFIG["arch"]
dlext = Config::CONFIG["DLEXT"]
this = File.dirname __FILE__
archdir = File.join this, arch
lib = File.join archdir, "foo.#{ dlext }"
require lib
then you layout dirs like so
lib/foo.rb
lib/i686-linux/foo.so
lib/mswin/foo.dll
and you can distribute one gem, one tarball, etc. you also could just dump
the *.so right into the libdir and do a
lib = File.join(File.dirname(__FILE__), "foo.#{ dlext }")
require lib
for example. the key point is to bury the native lib close to the *.rb file
and do a relative require.
hope that made sense...
-a
···
On Tue, 10 Oct 2006, Rick DeNatale wrote:
I'm writing something which is partially written in ruby, but uses an
extension which adds methods to some of the classes in the ruby part.
Right now I've got a directory which contains both the ruby code and
the so containing the extensions
$ls
mycode.rb mycode_prims.so
in mycode.rb I've got something like:
class Mycode
def foo
end
end
# link in the primitives, mycode_prims includes various calls to
# rb_define_method to add methods to the Mycode class.
require 'mycode_prims'
Now this works if the directory containing both parts is the current
directory, but not otherwise. Actually I guess it would work as long
as the directory was on the load path, but I'm concerned about the
possibility of loading the wrong mycode_prims.so.
Is there a way to get the directory containing the current source
file? __FILE__ only gives the file name, but not the path.
What is the right way to do this? I'd like to know how to approach
this either with the code packaged as a gem or as 'source'.
--
in order to be effective truth must penetrate like an arrow - and that is
likely to hurt. -- wei wu wei
The problem is that __FILE__ only has the file name, not the path, so
File.dirname(__FILE__) => '.'
···
On 10/9/06, ara.t.howard@noaa.gov <ara.t.howard@noaa.gov> wrote:
this = File.dirname __FILE__
archdir = File.join this, arch
lib = File.join archdir, "foo.#{ dlext }"
require lib
--
Rick DeNatale
My blog on Ruby
http://talklikeaduck.denhaven2.com/
na. it's always relative to the current directory, so you just need to expand
it (i forgot this part - sorry)
mussel:~ > cat /tmp/b.rb
puts File.dirname((__FILE__))
mussel:~ > ruby /tmp/b.rb
/tmp
mussel:~ > cat /tmp/a.rb
puts File.dirname(File.expand_path(__FILE__))
mussel:~ > ruby /tmp/b.rb
/tmp
mussel:~ > ruby -I/tmp -r a.rb -e nil
/tmp
i use this technique all over the place for relative requires:
http://codeforpeople.com/lib/ruby/rq/rq-2.3.4/lib/rq-2.3.4.rb
sorry for confusion.
cheers.
-a
···
On Tue, 10 Oct 2006, Rick DeNatale wrote:
On 10/9/06, ara.t.howard@noaa.gov <ara.t.howard@noaa.gov> wrote:
this = File.dirname __FILE__
archdir = File.join this, arch
lib = File.join archdir, "foo.#{ dlext }"
require lib
The problem is that __FILE__ only has the file name, not the path, so
File.dirname(__FILE__) => '.'
--
in order to be effective truth must penetrate like an arrow - and that is
likely to hurt. -- wei wu wei