On Mar 1, 1:56 pm, luisfe <lftab...@yahoo.es> wrote:
>
> No, the lazy_import object keeps wrapping the original object, but
> when accessing the lazy_import object it imports the real object in
> the namespace. So, if the lazy_import has not been referenced by
> another object then it will be collected as garbage. I do not have a
> running sage right now, but if you experiment you will see that after
> calling the object the lazy_import object disappears.
>
> It should be something like:
>
> {{{
> sage: from sage.misc.lazy_import import lazy_import
> sage: lazy_import('sage.rings.all', 'ZZ')
> sage: a = ZZ
> sage: type(ZZ)
> <class 'sage.misc.lazy_import.LazyImport'>
> sage: ZZ(4.0)
> 4
> sage: type(ZZ)
> <type 'sage.rings.integer_ring.IntegerRing_class'>
> sage: type(a)
> <class 'sage.misc.lazy_import.LazyImport'>
>
> }}}
I just did the above in 4.6.1 and got:
{{{
sage: lazy_import('sage.rings.all','ZZ')
sage: a = ZZ
sage: type(ZZ)
<class 'sage.misc.lazy_import.LazyImport'>
sage: ZZ(4.0)
4
sage: type(ZZ)
<class 'sage.misc.lazy_import.LazyImport'>
}}}
And naturally, the same type for a. It seems nothing I do to the ZZ
object will unwrap it from the LazyImporter; otherwise, what you say
would naturally be true, and that was exactly what I thought would be
neat to do. Even though the wrapping is very thin, if each and every
callable was wrapped as such, it would probably be quite a large
performance-penalty.
Johan
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