Question

I just stumbled upon something i don’t quite understand. I know that variables in ruby are references. So that awesome stuff is possible. But when i pass a variable to a method, it behaves strangely:

my_var_a = "nothing happend to me"
my_var_b = "nothing happend to me"

def parse_set(my_var_set)
  my_var_set = "my value changed"
end

def parse_sub(my_var_sub)
  my_var_sub.sub! /(.*)/, "my value changed"
end

parse_set(my_var_a)
parse_sub(my_var_b)

my_var_a # => "nothing happend to me"
my_var_b # => "my value changed"

Can you explain to me why it works with sub! and = leaves the object unchanged? How can I avoid to use sub! but having the same result?

Answer 1

my_var_a and my_var_set are different references, but they point at the same object. If you modify the object in my_var_set, the change shows up in my_var_a. However, if you repoint my_var_set at a new object, that doesn’t change what my_var_a points at.

Edit: clarification…

What Ruby does is called passing references by value. When you say

my_var_a = "nothing happend to me"

Ruby saves the string “nothing happend to me” in a memory location (let’s call it 1000), and saves the my_var_a reference in another memory location (let’s say 2000). When your code uses my_var_a, the interpreter looks at location 2000, see that it points to 1000, then gets the actual string value from 1000.

When you call parse_set(my_var_a), Ruby actually creates a new reference named my_var_set and points it to the string that my_var_a was pointing at (memory location 1000). However, my_var_set is a copy of the my_var_a reference — let’s say my_var_set was created at memory location 3000. my_var_a and my_var_set are 2 completely different references in memory, they just happen to point at the same exact memory location which holds the string value.

The statement my_var_set = "my value changed" in parse_set creates a new string in memory and points my_var_set at that new memory location. However, this doesn’t change what the original my_var_a reference points at! Now that my_var_set points at a different memory location, nothing that you do to that variable will affect my_var_a.

The same reference copy happens for parse_sub as well. But the reason that parse_sub changes the string is because you’re calling a method directly on the my_var_sub reference. When you do this, the interpreter gets the object that my_var_sub is pointing at and then modifies it. So that change will show up in the my_var_a reference, because it still points at the same string.