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

Ruby
1

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

2

Hi –

On Sat, 6 Jan 2007, Chris G. wrote:

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