Can't block exporter with use_ok

[toddr]

I've got a unit test that parses for all my modules in a LARGE project and does a use_ok($module) on them. The problem is that some of the modules (legacy) export the same methods.

What I would like to do is to have use_ok do a "use module ()" this is not possible at present.

What I'm envisioning is this syntax, which should be meaningless in all other contexts: use_ok($module, undef) causes use $module ();

I've attached a patch for this. I'll provide tests if you're willing to accept the patch.

[toddr]

http://github.com/toddr/test-more/tree/use_ok_undef

[schwern]

Thanks for the work. require_ok is equivalent to what you want. use Some::Module (); is equivalent to BEGIN { require Some::Module } but use_ok doesn't happen at BEGIN time so you just want to do a require.

I'm disinclined towards the special undef syntax. It would cause odd behavior should you do use_ok $module, $import and $import is accidentally undefined.

What would be nice is a doc patch to use_ok saying the above, that if you want to test use Module () just do require_ok "Module".

[toddr]

require_ok doesn't call import. doesn't import get called with use $module ()?

[schwern]

No, that's the whole point of the empty parens. From perldoc -f use...

If you do not want to call the package's "import" method (for
instance, to stop your namespace from being altered),
explicitly supply the empty list:

  use Module ();

That is exactly equivalent to

  BEGIN { require Module }

[toddr]

my bad. Stupid patch then. I'll send you a doc patch in thanks for you pointing out my stupidity.

[toddr]

A pull request has been sent. it should just be a cherry-pick of sha 5be859f

[schwern]

Done and in 0.97_01.