Variable changing value, ruby

Question

I am not sure how this variable called origString is changing value in my loop

def scramble_string(string, positions)
  i = 0
  origString = string
  puts origString
  newString = string
  while i < string.length
    newString[i] = origString[positions[i]]
    i = i + 1
  end
  puts origString
  return newString
end

for example if I run scramble_string("abcd", [3, 1, 2, 0]) origString changes from "abcd" in the first "puts" to "dbcd" in the second one. How am I changing the value of origString if I am only declaring it once?

Ruby strongly recommends using names like orig_string, no upper-case letters, as case has significant meaning in Ruby. — tadman
– tadman, Commented Sep 3, 2017 at 23:56
The variable isn't changing its value. The value is changing its content, because you tell it to on line 7. — Jörg W Mittag
– Jörg W Mittag, Commented Sep 4, 2017 at 13:11
@JörgWMittag can you elaborate? how it works " value is changing its content". — Aparichith
– Aparichith, Commented Sep 5, 2017 at 10:11
@Dharshan: If an object supports changing its state, you can change its state by sending it a message telling it to change its state. For example, you can change the state of a string by sending it the []= message or the upcase! message. — Jörg W Mittag
– Jörg W Mittag, Commented Sep 5, 2017 at 10:38
@JörgWMittag based on tadman's answer when we say x = y both object_ids are same, If we change entire value of x = 'pest', object_ids will change completely. If we check before changing value object_ids of x[0] and y[0] are different. . But changing a single value in x (i,e x[0]) reflects in original string y, HOW? — Aparichith
– Aparichith, Commented Sep 5, 2017 at 11:05

tadman · Accepted Answer · 2017-09-04 00:07:53Z

5

When you say x = y in Ruby that creates a variable with a reference to exactly the same object. Any modifications to x will apply to y and vice-versa:

y = "test"
x = y

x[0] = "b"

x
# => "best"
y
# => "best"

You can tell because of this:

x.object_id == y.object_id
# => true

They're identical objects. What you want is to make a copy first:

x = y.dup
x[0] = "b"
x
# => "best"
y
# => "test"

This results in two independent objects:

x.object_id == y.object_id
# => false

So in your case what you need is to change it like:

orig_string = string.dup

Now that being said, often the best way to process things in Ruby is by using functions that return copies, not manipulating things in place. A better solution is this:

def scramble_string(string, positions)
  (0...string.length).map do |p|
    string[positions[p]]
  end.join
end

scramble_string("abcd", [3, 1, 2, 0])
"dbca"

Note that's a lot more succinct than the version with string manipulation.

edited Sep 4, 2017 at 0:07

answered Sep 3, 2017 at 23:59

tadman

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3 Comments

Stefan Over a year ago

Or string.chars.values_at(*positions).join

Aparichith Over a year ago

@tadman need some explanation on your statements. when we say x = y both object_ids are same, but object_ids of x[0] and y[0] are different. If we change entire value of x = 'pest' object_id will completely change. But changing a single value in x reflects in original string y, HOW? Need to know little bit more about this.

tadman Over a year ago

x[0] returns a single-character substring. Since every substring operation can produce new, unique string, even x[0].object_id == x[0].object_id is generally not true. There's only a few things that are object_id equivalent by design: Symbols, nil, true, false, and some numbers.

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