[Newbie] Question about instance method scoping/parsing (?)

[Chris Gernon]

Can anyone explain to me what's going on here?

-------------------------------------
#!/usr/bin/env ruby -w

class Thingy
attr_accessor :name

def test
puts "self.name: #{self.name}"
puts "name: #{name}"

name ||= 'default value'

puts "self.name: #{self.name}"
puts "name: #{name}"
end
end

thing1 = Thingy.new
thing1.name = 'custom value'
thing1.test
-------------------------------------
$ ruby dditest.rb
self.name: custom value
name: custom value
self.name: custom value
name: default value
-------------------------------------

I thought calling a method without a receiver (i.e. name, name=) within
an instance method would always call that method on the current object.
Is this not true for setters (i.e. Thingy#name=)?

Additionally, why is the first instance of name interpreted as a method
call and the last instance interpreted as (I think) a local variable?

Thanks,
Chris

 

[dblack]

Hi --

I thought calling a method without a receiver (i.e. name, name=) within
an instance method would always call that method on the current object.
Is this not true for setters (i.e. Thingy#name=)?

It is not true for setters. The way it works is that when the parser
sees:

bareword = ...

it defines bareword as a local variable. Therefore, you always need
an explicit receiver for that kind of method, so Ruby will know it's a
method.

There are some interesting side-effects from this. For example:

if false
a = 1
end

puts a # nil

Even though "a = 1" is never executed, it's picked up by the parser,
and a gets defined. That's why "puts a" doesn't raise an "unknown
method or local variable" error.

Additionally, why is the first instance of name interpreted as a method
call and the last instance interpreted as (I think) a local variable?

If both are defined, the local variable takes precedence. You can
force method interpretation with parentheses:

name()

but of course in practice it's better not to reuse the names.


David

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