----- Original Message ----- 
From: "Rich" <rjseagraves at gmail.com>
To: <squeak-dev at lists.squeakfoundation.org>
Sent: Monday, January 16, 2006 7:19 PM
Subject: Some newb questions


1.  Suppose I were to execute the following block in the Workspace

   |L|
    L := (1 to: 5) collect: [:x | (x * x) + 2]

What would be the "scope" of L?  Does workspace executed code get
wrapped into a no-parameter block which is then executed?
Not exactly. A code snipped that is evaluated is compiled as
method #DoIt in UndefinedObject. You can see that when you
evaluate:
  | l |
  self halt.
  l := (1 to: 5) collect: [x | x * x + 2].

 For compilation, L is made a temporary variable of the
method #DoIt. That is: The variable is part of the activation
record of that method and exists until that activation record
becomes eligible for garbage collection.

2.  Is there a short-hand (syntactic) way to say "OrderedCollection
new", the way you can say #() to get a new, empty Array?  Thank you
very much.
You can write  
    #() asOrderedCollection
but that is not really shorter than
    OrderedCollection new.
Writing forms exist only for
 * the logical values true and false
 * the UndefinedObject (nil, this constant represents also what is called
     nil pointer in some other languages)
 *  Integers and floats
 * characters
 * strings
 * symbols
 * arrays.
All other Objects have to be constructed. I should mention that
the class protocol of Collection has some convenient constructors
that simplify the task of object construction. Try for example

  OrderedCollection with: (Array with: 'Hello' with: 123)
                               with: (Set with: 2 with: 15 with: 2)


Try also these two examples:

 | a b |
  a := #().
  b := #().
 a == b  " two equal signs in sequence: identity check"


 | a b |
  a :=OrderedCollection new.
  b := OrderedCollection new.
 a == b "two equal signs in sequence: identity check "


The first example answers true (within one method, all
occurrences of  the denotation #()  are mapped to the same
empty array)
The second example answers false (Every call of new creates
a new instance. Therefore the identity check says no)



Boris