Question

I have Ruby code:

def test_111(hash)
  n = nil
  3.times do |c|
      if n
       n[c] = c
      else
       n = hash
      end
  end
end

a = {}
test_111(a)
p a

Why it print {1=>1, 2=>2}, not the {} ??

In the test_111 method, the hash and the a use the same memory?

How can the a value be changed in the test_111 method?

I can’t understand

Answer 1

Hashes are passed by reference. So, when you change a method parameter (which is a Hash), you change the original hash.

To avoid this, you should clone the hash.

test_111(a.dup)

This will create a shallow copy (that is, it will not clone child hashes that you may have).

A little illustration of what shallow copy is:

def mutate hash
  hash[:new] = 1
  hash[:existing][:value] = 2
  hash
end

h = {existing: {value: 1}}

mutate h # => {:existing=>{:value=>2}, :new=>1}
# new member added, existing member changed
h # => {:existing=>{:value=>2}, :new=>1}



h = {existing: {value: 1}}

mutate h.dup # => {:existing=>{:value=>2}, :new=>1}
# existing member changed, no new members
h # => {:existing=>{:value=>2}}