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comp.lang.ruby

Confused by scope in ruby

Mischa Fierer

8/28/2008 9:54:00 AM

Hello,

I've been writing in ruby for a bit, but ran into something while
working on a project that struck me as weird, and could not find a ready
explanation.

Sorry if this is something super basic...

Here is a simplified example.

My question is: Why is the hash not empty at the end?

class Hmmm
def why_not_empty?
hash = {}
puts "Start: hash = #{hash.to_s}"
[1, 2, 3, 4, 5].each do |num|
put_in_hash(num, hash)
end
puts "End hash:\n#{hash.to_s}" # i would expect this to be {}
end

def put_in_hash(num, hsh)
hsh[num.to_s] = "hello from hash index #{num}.\n"
return nil
end

def number
num = 2
puts "Start number: #{num}"
change_number?(num)
puts "End number: #{num}"
end


def change_number?(num)
num = 3
end
end

@a = Hmmm.new
@a.why_not_empty?
@a.number



Output:
Start: hash =
End hash:
1hello from hash index 1.
2hello from hash index 2.
3hello from hash index 3.
4hello from hash index 4.
5hello from hash index 5.
Start number: 2
End number: 2

The number, as I would expect, stays at two, but the hash is actually
changed by the method, as if I were passing it a pointer in C or
something...
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4 Answers

Mischa Fierer

8/28/2008 10:07:00 AM

0

Actually, it appears that I was googling wrong:

If anyone else is curious:

http://www.ruby-...to...
http://www.ruby-...t...
http://www.ruby-...to...
http://www.ruby-...t...
http://www.ruby-...to...
http://al2o3-cr.blogspot.com/2008/08/objec...
--
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Sebastian Hungerecker

8/28/2008 10:09:00 AM

0

Mischa Fierer wrote:
> The number, as I would expect, stays at two, but the hash is actually
> changed by the method, as if I were passing it a pointer in C or
> something...

Well, that's exactly what's happening at C level - in both cases. The
difference is that if you do foo[bar] = baz or foo.method_that_changes_foo!
then you change the object that foo points to (so all other variables that
point to the same object will also reflect those changes), but if you do
foo = bar, you're just changing the pointer, i.e. foo now points to wherever
bar points to and the object that it previously pointed to is unchanged.

HTH,
Sebastian
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TPReal

8/28/2008 10:14:00 AM

0

Mischa Fierer wrote:
> Hello,
>
> I've been writing in ruby for a bit, but ran into something while
> working on a project that struck me as weird, and could not find a ready
> explanation.

Assigning a value to a parameter simply assigns a new variable to the
letter, and adding something to a hash modifies the existing object,
referenced by the variable. Similarily if you pass a string to a method
def change(s) and inside the method do:
s="abc" # does not change the original object
s+="abc" # does not change the original object, because it is equivalent
to s=s+"abc"
s<<"abc" # does change the outside object because it is an in-place
modification
s.upcase! # does change
s=s.upcase # does not

Have a look: http://al2o3-cr.blogspot.com/2008/08/objec...

TPR.
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Mischa Fierer

8/28/2008 8:38:00 PM

0

Yes, that is helpful. Thank you both.
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