Question

I’m trying to figure out why the following codes doesn’t return the same result:

CODE 1

p0 = "hello"
a = []
b = p0
1.upto(5) do |i|
  b.insert(2,"B")
  a.push b
end

a => ["heBBBBBllo", "heBBBBBllo", "heBBBBBllo", "heBBBBBllo", "heBBBBBllo"]

CODE 2

p0 = "hello"
a = []
b = p0
1.upto(5) do |i|
  b.insert(2,"B")
  a.push b.inspect
end

a => ["\"heBllo\"", "\"heBBllo\"", "\"heBBBllo\"", "\"heBBBBllo\"", "\"heBBBBBllo\""]

What I need is the Code 2’s result, but I don’t need the escaped char like the inspect method does.

Honestly, I really don’t understand why with the inspect method works, and why in the code 1 doesn’t.
It seems like that in code 1, “b” is used as a pointer, and every time it’s updated, all the “linked”-b are updated.

Any clue??

thank you in advance.

Answer 1

It seems like that in code 1, “b” is used as a pointer, and every time
it’s updated, all the “linked”-b are updated.

Correct! In the first case, you’re pushing b into the array five times, so the result is that a contains five pointers to b. In the second case you’re pushing b.inspect – which is a different object from b.

The easiest way to fix your first example would be to call a.push b.dup instead, which creates a duplicate of b which won’t be affected by future changes to b.