Related
#user.update_languages(params[:language][:language1],
params[:language][:language2],
params[:language][:language3])
lang_errors = #user.errors
logger.debug "--------------------LANG_ERRORS----------101-------------"
+ lang_errors.full_messages.inspect
if params[:user]
#user.state = params[:user][:state]
success = success & #user.save
end
logger.debug "--------------------LANG_ERRORS-------------102----------"
+ lang_errors.full_messages.inspect
if lang_errors.full_messages.empty?
#user object adds errors to the lang_errors variable in the update_lanugages method.
when I perform a save on the #user object I lose the errors that were initially stored in the lang_errors variable.
Though what I am attempting to do would be more of a hack (which does not seem to be working). I would like to understand why the variable values are washed out. I understand pass by reference so I would like to know how the value can be held in that variable without being washed out.
The other answerers are all correct, but a friend asked me to explain this to him and what it really boils down to is how Ruby handles variables, so I thought I would share some simple pictures / explanations I wrote for him (apologies for the length and probably some oversimplification):
Q1: What happens when you assign a new variable str to a value of 'foo'?
str = 'foo'
str.object_id # => 2000
A: A label called str is created that points at the object 'foo', which for the state of this Ruby interpreter happens to be at memory location 2000.
Q2: What happens when you assign the existing variable str to a new object using =?
str = 'bar'.tap{|b| puts "bar: #{b.object_id}"} # bar: 2002
str.object_id # => 2002
A: The label str now points to a different object.
Q3: What happens when you assign a new variable = to str?
str2 = str
str2.object_id # => 2002
A: A new label called str2 is created that points at the same object as str.
Q4: What happens if the object referenced by str and str2 gets changed?
str2.replace 'baz'
str2 # => 'baz'
str # => 'baz'
str.object_id # => 2002
str2.object_id # => 2002
A: Both labels still point at the same object, but that object itself has mutated (its contents have changed to be something else).
How does this relate to the original question?
It's basically the same as what happens in Q3/Q4; the method gets its own private copy of the variable / label (str2) that gets passed in to it (str). It can't change which object the label str points to, but it can change the contents of the object that they both reference to contain else:
str = 'foo'
def mutate(str2)
puts "str2: #{str2.object_id}"
str2.replace 'bar'
str2 = 'baz'
puts "str2: #{str2.object_id}"
end
str.object_id # => 2004
mutate(str) # str2: 2004, str2: 2006
str # => "bar"
str.object_id # => 2004
In traditional terminology, Ruby is strictly pass-by-value. But that's not really what you're asking here.
Ruby doesn't have any concept of a pure, non-reference value, so you certainly can't pass one to a method. Variables are always references to objects. In order to get an object that won't change out from under you, you need to dup or clone the object you're passed, thus giving an object that nobody else has a reference to. (Even this isn't bulletproof, though — both of the standard cloning methods do a shallow copy, so the instance variables of the clone still point to the same objects that the originals did. If the objects referenced by the ivars mutate, that will still show up in the copy, since it's referencing the same objects.)
Ruby uses "pass by object reference"
(Using Python's terminology.)
To say Ruby uses "pass by value" or "pass by reference" isn't really descriptive enough to be helpful. I think as most people know it these days, that terminology ("value" vs "reference") comes from C++.
In C++, "pass by value" means the function gets a copy of the variable and any changes to the copy don't change the original. That's true for objects too. If you pass an object variable by value then the whole object (including all of its members) get copied and any changes to the members don't change those members on the original object. (It's different if you pass a pointer by value but Ruby doesn't have pointers anyway, AFAIK.)
class A {
public:
int x;
};
void inc(A arg) {
arg.x++;
printf("in inc: %d\n", arg.x); // => 6
}
void inc(A* arg) {
arg->x++;
printf("in inc: %d\n", arg->x); // => 1
}
int main() {
A a;
a.x = 5;
inc(a);
printf("in main: %d\n", a.x); // => 5
A* b = new A;
b->x = 0;
inc(b);
printf("in main: %d\n", b->x); // => 1
return 0;
}
Output:
in inc: 6
in main: 5
in inc: 1
in main: 1
In C++, "pass by reference" means the function gets access to the original variable. It can assign a whole new literal integer and the original variable will then have that value too.
void replace(A &arg) {
A newA;
newA.x = 10;
arg = newA;
printf("in replace: %d\n", arg.x);
}
int main() {
A a;
a.x = 5;
replace(a);
printf("in main: %d\n", a.x);
return 0;
}
Output:
in replace: 10
in main: 10
Ruby uses pass by value (in the C++ sense) if the argument is not an object. But in Ruby everything is an object, so there really is no pass by value in the C++ sense in Ruby.
In Ruby, "pass by object reference" (to use Python's terminology) is used:
Inside the function, any of the object's members can have new values assigned to them and these changes will persist after the function returns.*
Inside the function, assigning a whole new object to the variable causes the variable to stop referencing the old object. But after the function returns, the original variable will still reference the old object.
Therefore Ruby does not use "pass by reference" in the C++ sense. If it did, then assigning a new object to a variable inside a function would cause the old object to be forgotten after the function returned.
class A
attr_accessor :x
end
def inc(arg)
arg.x += 1
puts arg.x
end
def replace(arg)
arg = A.new
arg.x = 3
puts arg.x
end
a = A.new
a.x = 1
puts a.x # 1
inc a # 2
puts a.x # 2
replace a # 3
puts a.x # 2
puts ''
def inc_var(arg)
arg += 1
puts arg
end
b = 1 # Even integers are objects in Ruby
puts b # 1
inc_var b # 2
puts b # 1
Output:
1
2
2
3
2
1
2
1
* This is why, in Ruby, if you want to modify an object inside a function but forget those changes when the function returns, then you must explicitly make a copy of the object before making your temporary changes to the copy.
Is Ruby pass by reference or by value?
Ruby is pass-by-value. Always. No exceptions. No ifs. No buts.
Here is a simple program which demonstrates that fact:
def foo(bar)
bar = 'reference'
end
baz = 'value'
foo(baz)
puts "Ruby is pass-by-#{baz}"
# Ruby is pass-by-value
Ruby is pass-by-value in a strict sense, BUT the values are references.
This could be called "pass-reference-by-value". This article has the best explanation I have read: http://robertheaton.com/2014/07/22/is-ruby-pass-by-reference-or-pass-by-value/
Pass-reference-by-value could briefly be explained as follows:
A function receives a reference to (and will access) the same object in memory as used by the caller. However, it does not receive the box that the caller is storing this object in; as in pass-value-by-value, the function provides its own box and creates a new variable for itself.
The resulting behavior is actually a combination of the classical definitions of pass-by-reference and pass-by-value.
There are already some great answers, but I want to post the definition of a pair of authorities on the subject, but also hoping someone might explain what said authorities Matz (creator of Ruby) and David Flanagan meant in their excellent O'Reilly book, The Ruby Programming Language.
[from 3.8.1: Object References]
When you pass an object to a method in Ruby, it is an object reference that is passed to the method. It is not the object itself, and it is not a reference to the reference to the object. Another way to say this is that method arguments are passed by value rather than by reference, but that the values passed are object references.
Because object references are passed to methods, methods can use those references to modify the underlying object. These modifications are then visible when the method returns.
This all makes sense to me until that last paragraph, and especially that last sentence. This is at best misleading, and at worse confounding. How, in any way, could modifications to that passed-by-value reference change the underlying object?
Is Ruby pass by reference or by value?
Ruby is pass-by-reference. Always. No exceptions. No ifs. No buts.
Here is a simple program which demonstrates that fact:
def foo(bar)
bar.object_id
end
baz = 'value'
puts "#{baz.object_id} Ruby is pass-by-reference #{foo(baz)} because object_id's (memory addresses) are always the same ;)"
=> 2279146940 Ruby is pass-by-reference 2279146940 because object_id's (memory addresses) are always the same ;)
def bar(babar)
babar.replace("reference")
end
bar(baz)
puts "some people don't realize it's reference because local assignment can take precedence, but it's clearly pass-by-#{baz}"
=> some people don't realize it's reference because local assignment can take precedence, but it's clearly pass-by-reference
Parameters are a copy of the original reference. So, you can change values, but cannot change the original reference.
Try this:--
1.object_id
#=> 3
2.object_id
#=> 5
a = 1
#=> 1
a.object_id
#=> 3
b = 2
#=> 2
b.object_id
#=> 5
identifier a contains object_id 3 for value object 1 and identifier b contains object_id 5 for value object 2.
Now do this:--
a.object_id = 5
#=> error
a = b
#value(object_id) at b copies itself as value(object_id) at a. value object 2 has object_id 5
#=> 2
a.object_id
#=> 5
Now, a and b both contain same object_id 5 which refers to value object 2.
So, Ruby variable contains object_ids to refer to value objects.
Doing following also gives error:--
c
#=> error
but doing this won't give error:--
5.object_id
#=> 11
c = 5
#=> value object 5 provides return type for variable c and saves 5.object_id i.e. 11 at c
#=> 5
c.object_id
#=> 11
a = c.object_id
#=> object_id of c as a value object changes value at a
#=> 11
11.object_id
#=> 23
a.object_id == 11.object_id
#=> true
a
#=> Value at a
#=> 11
Here identifier a returns value object 11 whose object id is 23 i.e. object_id 23 is at identifier a, Now we see an example by using method.
def foo(arg)
p arg
p arg.object_id
end
#=> nil
11.object_id
#=> 23
x = 11
#=> 11
x.object_id
#=> 23
foo(x)
#=> 11
#=> 23
arg in foo is assigned with return value of x.
It clearly shows that argument is passed by value 11, and value 11 being itself an object has unique object id 23.
Now see this also:--
def foo(arg)
p arg
p arg.object_id
arg = 12
p arg
p arg.object_id
end
#=> nil
11.object_id
#=> 23
x = 11
#=> 11
x.object_id
#=> 23
foo(x)
#=> 11
#=> 23
#=> 12
#=> 25
x
#=> 11
x.object_id
#=> 23
Here, identifier arg first contains object_id 23 to refer 11 and after internal assignment with value object 12, it contains object_id 25. But it does not change value referenced by identifier x used in calling method.
Hence, Ruby is pass by value and Ruby variables do not contain values but do contain reference to value object.
It should be noted that you do not have to even use the "replace" method to change the value original value. If you assign one of the hash values for a hash, you are changing the original value.
def my_foo(a_hash)
a_hash["test"]="reference"
end;
hash = {"test"=>"value"}
my_foo(hash)
puts "Ruby is pass-by-#{hash["test"]}"
Two references refer to same object as long as there is no reassignment.
Any updates in the same object won't make the references to new memory since it still is in same memory.
Here are few examples :
a = "first string"
b = a
b.upcase!
=> FIRST STRING
a
=> FIRST STRING
b = "second string"
a
=> FIRST STRING
hash = {first_sub_hash: {first_key: "first_value"}}
first_sub_hash = hash[:first_sub_hash]
first_sub_hash[:second_key] = "second_value"
hash
=> {first_sub_hash: {first_key: "first_value", second_key: "second_value"}}
def change(first_sub_hash)
first_sub_hash[:third_key] = "third_value"
end
change(first_sub_hash)
hash
=> {first_sub_hash: {first_key: "first_value", second_key: "second_value", third_key: "third_value"}}
Ruby is interpreted. Variables are references to data, but not the data itself. This facilitates using the same variable for data of different types.
Assignment of lhs = rhs then copies the reference on the rhs, not the data. This differs in other languages, such as C, where assignment does a data copy to lhs from rhs.
So for the function call, the variable passed, say x, is indeed copied into a local variable in the function, but x is a reference. There will then be two copies of the reference, both referencing the same data. One will be in the caller, one in the function.
Assignment in the function would then copy a new reference to the function's version of x. After this the caller's version of x remains unchanged. It is still a reference to the original data.
In contrast, using the .replace method on x will cause ruby to do a data copy. If replace is used before any new assignments then indeed the caller will see the data change in its version also.
Similarly, as long as the original reference is in tact for the passed in variable, the instance variables will be the same that the caller sees. Within the framework of an object, the instance variables always have the most up to date reference values, whether those are provided by the caller or set in the function the class was passed in to.
The 'call by value' or 'call by reference' is muddled here because of confusion over '=' In compiled languages '=' is a data copy. Here in this interpreted language '=' is a reference copy. In the example you have the reference passed in followed by a reference copy though '=' that clobbers the original passed in reference, and then people talking about it as though '=' were a data copy.
To be consistent with definitions we must keep with '.replace' as it is a data copy. From the perspective of '.replace' we see that this is indeed pass by reference. Furthermore, if we walk through in the debugger, we see references being passed in, as variables are references.
However if we must keep '=' as a frame of reference, then indeed we do get to see the passed in data up until an assignment, and then we don't get to see it anymore after assignment while the caller's data remains unchanged. At a behavioral level this is pass by value as long as we don't consider the passed in value to be composite - as we won't be able to keep part of it while changing the other part in a single assignment (as that assignment changes the reference and the original goes out of scope). There will also be a wart, in that instance variables in objects will be references, as are all variables. Hence we will be forced to talk about passing 'references by value' and have to use related locutions.
Lots of great answers diving into the theory of how Ruby's "pass-reference-by-value" works. But I learn and understand everything much better by example. Hopefully, this will be helpful.
def foo(bar)
puts "bar (#{bar}) entering foo with object_id #{bar.object_id}"
bar = "reference"
puts "bar (#{bar}) leaving foo with object_id #{bar.object_id}"
end
bar = "value"
puts "bar (#{bar}) before foo with object_id #{bar.object_id}"
foo(bar)
puts "bar (#{bar}) after foo with object_id #{bar.object_id}"
# Output
bar (value) before foo with object_id 60
bar (value) entering foo with object_id 60
bar (reference) leaving foo with object_id 80 # <-----
bar (value) after foo with object_id 60 # <-----
As you can see when we entered the method, our bar was still pointing to the string "value". But then we assigned a string object "reference" to bar, which has a new object_id. In this case bar inside of foo, has a different scope, and whatever we passed inside the method, is no longer accessed by bar as we re-assigned it and point it to a new place in memory that holds String "reference".
Now consider this same method. The only difference is what with do inside the method
def foo(bar)
puts "bar (#{bar}) entering foo with object_id #{bar.object_id}"
bar.replace "reference"
puts "bar (#{bar}) leaving foo with object_id #{bar.object_id}"
end
bar = "value"
puts "bar (#{bar}) before foo with object_id #{bar.object_id}"
foo(bar)
puts "bar (#{bar}) after foo with object_id #{bar.object_id}"
# Output
bar (value) before foo with object_id 60
bar (value) entering foo with object_id 60
bar (reference) leaving foo with object_id 60 # <-----
bar (reference) after foo with object_id 60 # <-----
Notice the difference? What we did here was: we modified the contents of the String object, that variable was pointing to. The scope of bar is still different inside of the method.
So be careful how you treat the variable passed into methods. And if you modify passed-in variables-in-place (gsub!, replace, etc), then indicate so in the name of the method with a bang !, like so "def foo!"
P.S.:
It's important to keep in mind that the "bar"s inside and outside of foo, are "different" "bar". Their scope is different. Inside the method, you could rename "bar" to "club" and the result would be the same.
I often see variables re-used inside and outside of methods, and while it's fine, it takes away from the readability of the code and is a code smell IMHO. I highly recommend not to do what I did in my example above :) and rather do this
def foo(fiz)
puts "fiz (#{fiz}) entering foo with object_id #{fiz.object_id}"
fiz = "reference"
puts "fiz (#{fiz}) leaving foo with object_id #{fiz.object_id}"
end
bar = "value"
puts "bar (#{bar}) before foo with object_id #{bar.object_id}"
foo(bar)
puts "bar (#{bar}) after foo with object_id #{bar.object_id}"
# Output
bar (value) before foo with object_id 60
fiz (value) entering foo with object_id 60
fiz (reference) leaving foo with object_id 80
bar (value) after foo with object_id 60
Yes but ....
Ruby passes a reference to an object and since everything in ruby is an object, then you could say it's pass by reference.
I don't agree with the postings here claiming it's pass by value, that seems like pedantic, symantic games to me.
However, in effect it "hides" the behaviour because most of the operations ruby provides "out of the box" - for example string operations, produce a copy of the object:
> astringobject = "lowercase"
> bstringobject = astringobject.upcase
> # bstringobject is a new object created by String.upcase
> puts astringobject
lowercase
> puts bstringobject
LOWERCASE
This means that much of the time, the original object is left unchanged giving the appearance that ruby is "pass by value".
Of course when designing your own classes, an understanding of the details of this behaviour is important for both functional behaviour, memory efficiency and performance.
I have the following code as an example.
a = [2]
b = a
puts a == b
a.each do |num|
a[0] = num-1
end
puts a == b
I want b to refer to a's value, and the value of b not to change when a is changed.(The second puts should return false).
Thank you in advance.
Edited-
The answer posted by user2864740 seems to work for the example I gave. However, I'm working on a sudoku solving program, and it doesn't seem to work there.
#gridbylines = [[1,0,0,9,2,0,0,0,0],
[5,2,4,0,1,0,0,0,0],
[0,0,0,0,0,0,0,7,0],
[0,5,0,0,0,8,1,0,2],
[0,0,0,0,0,0,0,0,0],
[4,0,2,7,0,0,0,9,0],
[0,6,0,0,0,0,0,0,0],
[0,0,0,0,3,0,9,4,5],
[0,0,0,0,7,1,0,0,6]]
save = Array.new(#gridbylines) #or #gridbylines.dup
puts save == #gridbylines #returns true
puts save.equal?(#gridbylines) #returns false
#gridbylines[0][0] = 'foo'
puts save.equal?(#gridbylines) #returns false
puts save == #gridbylines #returns true, but I want "save" not to change when I change "#gridbylines"
Does this have something to do with the fact that I'm using a global variable, or the version of Ruby I'm using, or even because it's a multidimentional array unlike the previous example?
Variables name or "refer to" objects1. In the code above the same object (which has two names, a and b) is being changed.
A simple solution in this case is to make a (shallow) copy of the original Array object, such as b = a.dup or b = Array.new(a). (With a shallow copy, elements in the array are also shared and will exhibit the similar phoneme as the original question unless they to are [recursively] duplicated, etc.2)
a = [2]
b = Array.new(a) # create NEW array object, a shallow-copy of `a`
puts a == b # true (same content)
puts a.equal?(b) # false (different objects)
a.each do |num|
a[0] = num-1 # now changing the object named by `a` does not
# affect the object named by `b` as they are different
end
puts a == b # false (different content)
And an isolated example of this "naming" phenomena (see the different equality forms):
a = []
b = a # assignment does NOT make a copy of the object
a.equals?(b) # true (same object)
c = a.dup # like Array.new, create a new shallow-copy object
a.equals?(c) # false (different object)
1 I find it most uniform to talk about variables being names, as such a concept can be applied across many languages - the key here is that any object can have one or more names, just as a person can have many nicknames. If an object has zero names then it is no longer strongly reachable and, just like a person, is forgotten.
However, another way to view variables (naming objects) is that they hold reference values, where the reference value identifies an object. This leads to phrasing such as
"variable a contains a reference [value] to object x" - or,
"variable a refers to / references object x" - or, as I prefer,
"variable a is a name for object x".
For the case or immutable "primitive" or "immediate" values the underlying mechanics are slightly different but, being immutable the object values cannot be changed and such a lack-of-shared object nature will not manifest itself.
See also:
String assignment by reference/copy? (examines object relationship after assignment)
Strange Feature? of Ruby Arrays (examines same-object mutation)
Is Ruby pass by reference or by value? (assignment works in the same way as argument passing)
2 As per the updated question with nested arrays, this is explained by the previous rules - the variables (well, really expressions) still name shared objects. In any case, one way to "clone an array of arrays" (to two levels, although not recursively) is to use:
b = a.map {|r| r.dup}
This is because Array#map returns a new array with the mapped values which are, in this case, duplicates (shallow clones) of the corresponding nested arrays.
See How to create a deep copy of an object in Ruby? for other "deep[er] copy" approaches - especially if the arrays (or affect mutable objects) were nested to N-levels.
This is what I tried:
a = "Hello world"
a.object_id # => -633222538
b = a
b.object_id # => -633222538
b << " i say" # => "Hello world i say"
a # => "Hello world i say"
Why is it that both the variables b and a have the same object id? Also, when I change b, how did a also change?
Update:
How about when the variable is passed as an argument to a method? Why is the receiving variable having the same reference?
They are referencing the same object:
a = "Hello world" # a now references #-633222538
b = a # b now references #-633222538, too
b << " i say" # this appends " i say" to #-633222538
a # a still references #-633222538
String#<< doesn't assign a new object, it appends to the given string, thus changing the receiver.
I you want a copy, you can use clone or dup:
b = a.clone
a == b #=> true (same string values)
a.equal? b #=> false (different objects)
Regarding integers
There's no difference in referencing:
a = 100
a.object_id #=> 201
b = a
b.object_id #=> 201
Now both, a and b reference the same object. The only difference is that an integer cannot be changed in Ruby, they are fixed.
Passing variables as arguments
Again, the reference is passed:
a = "foo"
p = proc { |x| x << "bar" }
p.call(a)
a
#=> "foobar"
ENTER REFERENCES
The answer is that variables in Ruby (with a few exceptions, most notably variables bound to integers) don’t hold object values. a doesn’t contain "Hello world". Rather, a contains a reference to a string object. It’s the string object that has the characteristic of containing the letters that make up "Hello World".
In an assignment with a variable name on the left and an object on the right, the variable receives a reference to the object. In an assignment from one variable to another (a = b), the variable on the left receives a copy of the reference stored in the variable on the right, with the result that both variables now contain references to the same object.
The fact that variables hold references to objects has implications for operations that change objects. The string-concat operation
b << " i say"
concats the characters of the string to which b is a reference with the text " i say". The variable a contains another reference to the same string object. Even though the replace message goes to b, it causes a change to the object to which the reference in b refers. When you print out a, you see the result: the contents of the string have changed.
Some objects in Ruby are stored in variables as immediate values. These include in- tegers, symbols (which look like :this), and the special objects true, false, and nil. When you assign one of these values to a variable (x = 1), the variable holds the value itself, rather than a reference to it.
Copied and modified from Manning The Well Grounded Rubyist.
Looks like you called a mutable function on a variable which shared the same object with another variable. if you instead did b = b + 'i say' a would be left unchanged.
The variables a and b are references to a String object. When you did the b = a assignment, you copied the reference. It doesn't make a new copy of the object. If you want to copy the string object into a new object, you might do something like this:
a = "abc"
b = "" [or, b = String.new]
b << a
Now a and b will be different, independent string objects with the value "abc".
I learned that in Ruby, variables hold references to objects, not the objects themselves.
For example:
a = "Tim"
b = a
a[0] = 'J'
Then a and b both have value "Jim".
However if I change the 3rd line to
a = "Jim"
Then a == Jim and b == Tim
I assume that means the code I changed created a new reference for a.
So why does changing a letter or changing the entire string make so much difference?
Follow-up question: Does Java work the same way?
Thank you.
The single thing to learn here is the difference between assignment and method call.
a = 'Jim'
is an assignment. You create a new string object (literal 'Jim') and assign it to variable a.
On the other side,
a[0] = 'J'
is a method call on an object already referenced by the variable a. A method call can't replace the object referenced by the variable with another one, it can just change the internal state of the object, and/or return another object.
I find that things like this are easiest to figure out using IRB:
>> a = 'Tim'
=> "Tim"
>> a.object_id
=> 2156046480
>> b = a
=> "Tim"
>> b.object_id
=> 2156046480
>> a.object_id == b.object_id
=> true
As you can see a and b have the same object_id, meaning they reference the same object. So when you change one, you change the other. Now assign something new to a:
>> a = 'Jim'
=> "Jim"
>> a.object_id
=> 2156019520
>> b.object_id
=> 2156046480
>> a.object_id == b.object_id
=> false
You made a point to a new object, while b still kept the old reference. Changing either of them now will not change the other one.
When you do a[0] = 'J', you're asking
Change the first character of the object referenced by a (which happens to be the same as b) to 'J'
While when you do a = "Jim", you're assigning an entirely new object reference (the string "Jim") to a. b is unaffected because you're not changing anything in the original reference.
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?
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.
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"
"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