From: Florian Gross <flgr@...>
Date: 2004-06-29T04:42:57+09:00
Subject: Re: (oddity) applying objects with #call() implicitly

gabriele renzi wrote:

>>>foo()
> 0
> => #<Proc:0x028a43a0@(irb):7>
>>x = foo; x() # What will this do?
> 01 #=> #<Proc2>

So (foo)() and foo() aren't the same thing any more? It feels a bit 
inconsistent IMHO.

Also  foo  would be the same as  foo()  in your case if foo is a method, 
but it would be something totally different if it is a lambda.

I think that a solution to this problem should make "def x; 5; end" and 
"x = lambda { 5 }" as equivalent as possible.

But of course I am not matz so I'm not the one to judge over this anyway.

Maybe you could do a RCR? :)

Regards,
Florian Gross