Changing value of ruby variables/references

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

irb(main):008:0* a = 'abc'
=> "abc"
irb(main):009:0> a.replace('def')
=> "def"
irb(main):010:0> a
=> "def"

I've had to use replace exactly zero times in all the years I've been programming in Ruby. I wonder if there's a better design for your code.

Most string methods which change the receiver are suffixed by ! (sub!, capitalize!, chomp!, etc.) Some that change the receiver are not suffixed by ! (replace is one). If you call a method that changes the receiver, any and all references to that object will see the change. if you call a method that does not change receiver, the method returns a new string.

gsub does not modify the receiver, but instead returns a new instance of String:

a = "foo"
b = a
p a.sub(/o/, '#')     # "f##"
p a                   # => "foo"
p b                   # => "foo"

gsub! does modify the receiver:

a = "foo"
b = a
p a.sub!(/o/, '#')    # => "f#o"
p a                   # => "f#o"
p b                   # => "f#o"

Answer 2

"How can I avoid to use sub! but having the same result?"

def parse_set()
  "my value changed"
end

a = parse_set()

Set it as an instance variable of an object (although this works just in your main script, I recommend making your own class if you want to use the instance variables approach).

@my_var_a = "nothing happend to me"

def parse_set()
  @my_var_a = "my value changed"
end

Answer 3

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.