Lexical vs Dynamic Scope

ruby-talk
1

Forgive this very basic question, but Googling has not answered my
question, and I'm sure it's a simple one for the gurus here.

I've been using Ruby for years, and I've always had questions about
how it handles scope. Usually Ruby just does what I would expect it
to.

In reading about lexical vs. dynamic scope on various places on the
Web, I read that Ruby has lexical (static) scope. But I cannot prove
it to myself with code. For example, the following produces one (1) --
not zero (0) as I would expect it to if Ruby was truly statically
scoped:

x = 0
f = Proc.new { x }
g = Proc.new {
  x = 1
  f.call
}
puts g.call
# => 1

(I purposely used Procs instead of regular methods here since Ruby
methods cannot see the top-level "x" variable at all, which is a whole
other issue.)

Is Ruby really dynamically scoped?

2

The x in g is the same as the x outside of g. There is only one single x in
your program, so this doesn't show much. Maybe this instead?

def show_error(f)
  f.call
rescue => e
  e
end

f = lambda { x }
show_error f # => #<NameError: undefined local variable or method `x' for
main:Object>

x = 1
show_error f # => #<NameError: undefined local variable or method `x' for
main:Object>

g = lambda { x }
show_error g # => 1

···

On Sat, Jan 22, 2011 at 12:20 AM, Tim Morgan <tim@timmorgan.org> wrote:

Forgive this very basic question, but Googling has not answered my
question, and I'm sure it's a simple one for the gurus here.

I've been using Ruby for years, and I've always had questions about
how it handles scope. Usually Ruby just does what I would expect it
to.

In reading about lexical vs. dynamic scope on various places on the
Web, I read that Ruby has lexical (static) scope. But I cannot prove
it to myself with code. For example, the following produces one (1) --
not zero (0) as I would expect it to if Ruby was truly statically
scoped:

x = 0
f = Proc.new { x }
g = Proc.new {
x = 1
f.call
}
puts g.call
# => 1

(I purposely used Procs instead of regular methods here since Ruby
methods cannot see the top-level "x" variable at all, which is a whole
other issue.)

Is Ruby really dynamically scoped?

3

Ahhh. I see. Thanks!

···

On Jan 22, 1:12 am, Josh Cheek <josh.ch...@gmail.com> wrote:

The x in g is the same as the x outside of g. There is only one single x in
your program, so this doesn't show much. Maybe this instead?

4

After doing the above, try this.

def x
  "Anything returned by the (Receiver).x method"
end

show_error f # => "Anything returned by the (Receiver).x method"
show_error g # => 1

At compile time...
f = lambda { x }
x is not defined, and is thought to be a method!

Look...

Abinoam Jr.

···

On Sat, Jan 22, 2011 at 4:12 AM, Josh Cheek <josh.cheek@gmail.com> wrote:

def show_error(f)
f.call
rescue => e
e
end

f = lambda { x }
show_error f # => #<NameError: undefined local variable or method `x' for
main:Object>

x = 1
show_error f # => #<NameError: undefined local variable or method `x' for
main:Object>

g = lambda { x }
show_error g # => 1