[ruby-talk:5487] Re: Some newbye question

> |My question was about variable n, the block does not necessarly introduces
> |a fresh variable, instead it (possibly) binds it to an already existing
> |variable. IMHO this treatement of variables occurring as argument of
> |blocks is unrelated to the creation of closures.
> 
> It is necessary.  For example
> 
>   fact = proc{|n|
>     n == 0 ? 1 : fact.call(n-1)*n
>   }
>   p fact.call(4)
> 
> would return 0 if the block does not introduce a new scope.

I agree, this is the right way. But again my question is different (see
original posting). What I do not understand is the following:

a = 1; x =1
f = proc {|a| a += 1}
f.call(x)
a # results 2

Why a in proc is bound to the external a?
Why the brackets do NOT introduce a new scope when a var with same name
exists?