From: gabriele renzi <surrender_it@...1.vip.ukl.yahoo.com>
Date: 2004-06-29T04:13:24+09:00
Subject: Re: (oddity) applying objects with #call() implicitly

il Mon, 28 Jun 2004 19:41:54 +0200, Florian Gross <flgr@ccan.de> ha
scritto::

>gabriele renzi wrote:

>> "p" may stand for self#p, or a local variable named "p". 
>
>This case can be resolved fairly easily. (If there's an assignment to p 
>then it is a variable)

Yes, and if there is a variable named "foo" in
 foo arg1
then it is calling that one instead of the old 'foo' method.

>However think about this case:
>
>def foo
>   puts 0
>   lambda { puts 1; lambda { puts 2; lambda { puts 3 }}}
>end
>
>foo # What will this do?
0 #=> #<Proc>

>foo() # What will this do?
0 #=> #<Proc>

>x = foo; x # What will this do?
0 #=> #<Proc>

>x = foo; x() # What will this do?
01 #=> #<Proc2>

>x = foo(); x # What will this do?
0 #=> #<Proc>

>x = foo(); x() # What will this do?
01 #=> #<Proc2>

>x = foo; y = x; y # What will this do?
0 #=> #<Proc>

>x = foo(); y = x; y # What will this do?
0 #=> #<Proc>

>x = foo(); y = x(); y # What will this do?
01 #=> #<Proc2>

>x = foo; y = x(); y # What will this do?
01 #=> #<Proc2>

>x = foo; y = x(); y() # What will this do?
012 #=> #<Proc3>

>x = foo; y = x; y() # What will this do?
01 #=> #<Proc2>

>x = foo(); y = x(); y # What will this do?
01 #=> #<Proc2>

>x = foo(); y = x(); y() # What will this do?
012 #=> #<Proc3>

>Etc. -- there are many cases here.

yes, but the rule is quite simple :)

Note that I may have 'computed' the things myself and I may be wrong.
Anyway, sample implementation:

>> B=binding
=> #<Binding:0x28ba600>
>> def method_missing(s,*a,&b)
>>  eval(s.to_s,B).call *a, &b
>> end
=> nil
>> def foo
>>  puts 0
>>  lambda { puts 1; lambda { puts 2; lambda { puts 3 }}}
>> end
=> nil
>> foo
0
=> #<Proc:0x028a43a0@(irb):7>
>> foo()
0
=> #<Proc:0x028a43a0@(irb):7>
>> x=foo(); x
=> #<Proc:0x028a43a0@(irb):7>
>> x=foo(); x()
0
1
=> #<Proc:0x028a4538@(irb):7>
>> x=foo; x()
0
1
=> #<Proc:0x028a4538@(irb):7>
>> x=foo; y=x; y
0
=> #<Proc:0x028a43a0@(irb):7>
>> x=foo; y=x(); y
0
1
=> #<Proc:0x028a4538@(irb):7>
>> x=foo; y=x(); y()
0
1
2
=> #<Proc:0x028a4598@(irb):7>
>> x=foo(); y=x(); y()
0
1
2
=> #<Proc:0x028a4598@(irb):7>
>>


>I think the problem is that you can't have optional ()-arguments and () 
>as a call operator as the same time.

this is what I usually thought. After reading the msg [see first msg]
I changed slightly my mind.

>However we need optional () because of things like this:

<snip>
yes, totally agree.