Codeacademy teaches that you can chain multiple methods together as such:
user_input.method1.method2.method3
However, in a later lesson they display some methods like this:
user_input = gets.chomp
user_input.downcase!
I combined them:
user_input = gets.chomp.downcase!
When I use it this way:
user_input = gets.chomp.downcase!
if user_input.include? "s"
...
I receive an error "undefined method `include?'". If I change it to the following, it works fine:
user_input = gets.chomp
user_input.downcase!
if user_input.include? "s"
...
I'm at a loss. I'm concerned whether or not this is a quirk with their console or if this is just how I should be doing it in Ruby. If someone could tell me which way is right, I'd appreciate it. If both are right, that's OK too.
Firstly, in case you do not yet fully understand, assignment of values to variables are done through =, and that you could inspect what variable type it is by appending .class to anything.
Consider the following:
name = 'John'
puts name
# => John
puts name.class
# => String
Now, secondly, it should be noted that the return values of ALL methods are ALL different. But all of them can be identified into two types:
Methods that:
return self
return anything other than self
Example for 1.
-- methods that return self, which you could say methods that return the same type of object which in our specific case, a String
name = 'John'
puts name
# => 'John'
puts name.class
# => String
downcased_name = name.downcase
puts downcased_name
# => john
puts downcased_name.class
# => String
deleted_downcased_name = downcased_name.delete('h')
puts deleted_downcased_name
# => jon
puts deleted_downcased_name.class
# => String
# All of the above can be just simplified into:
deleted_downcased_name2 = 'John'.downcase.delete('h')
puts deleted_downcased_name2
# => jon
puts deleted_downcased_name2.class
# => String
Notice that deleted_downcased_name and deleted_downcased_name2 are the same, because you could treat the chained methods as if you are chaining the return values which is 'John' -> 'john' -> 'jon'.
Example for 2
-- methods that return anything but self, which you could say methods that return a different type.
In our specific case, String's downcase! returns either a String or NilClass (reference here)
returning String if the string changes, or
returning nil if string is already downcased to begin with (no change).
or another String's method: start_with? (reference here)
returning true or false
This is where chaining of methods will not work (raises an error), when you try to use a String method as a chain to nil value.
Consider the following
name = 'mary'
puts name
# => 'mary'
puts name.class
# => String
downcased_name = name.downcase!
puts downcased_name
# => nil
puts downcased_name.class
# => NilClass
downcased_name.delete('h')
# => This will raise the following error
# NoMethodError: undefined method `delete' for nil:NilClass
The error above is because downcased_name is a type of NilClass where you are expecting it to be a type of String. Therefore you cannot chain any string method on it anymore. You can only chain String methods on a String type of value. Similarly, you can only chain Number methods on a Number type of value.
Whenever in doubt, you could always check the documentation to check what a method does, and what its return value and type.
The problem you are encountering is with the bang method downcase!.
This is basically saying "mutate the original string so that it is downcase".
The important part is that this returns nil. As such you are actually calling include? on nil.
If you use the non bang method downcase instead, it is saying "downcase the previously chained thing but do not mutate the original". The key difference is that it returns the result rather than nil.
Here is an example:
str = "ABCD"
str.downcase!
=> nil
str
=> "abcd"
str = "ABCD"
str.downcase
=> "abcd"
str
=> "ABCD" # Note str is still unchanged unless you set str = str.downcase
Welcome to Ruby! While your apprenticeship at Codeacademy may be limited, you'll continue to refer to language API documentation throughout your career. API documentation is a description of what the language (or a library) does for you. In this case, you're using downcase! which, as one commenter points out, does not always return a String. When it takes no action, it returns nil. Nil is an Object in Ruby (like everything else), but the 'include?' method isn't defined for nil, which explains your error. (It's one of the most common errors in Ruby, learn its meaning.)
So, in fact, what's breaking here isn't your method chain. It's that one of the intermediate methods isn't returning a value of the type you expect (nil instead of some kind of String).
Chaining non destructive methods like:
string.chomp.downcase...
has the advantage that the code is concise, but is not efficient if you are not interested in the original state of the object, and just want the final result because it creates intermediate objects during the chain.
On the other hand, applying destructive methods sequentially to the same object:
string.chomp!
string.downcase!
...
is more efficient if you do not need to keep the original state of the object, but is not concise.
Combining methods that may return an object of a different class (particularly nil) as:
string = gets.chomp!.downcase!...
is wrong because the result can become nil at some point in the chain.
Applying a potentially nil-returning method at only the last position as you did:
string = gets.chomp.downcase!
is still not useful if you expect string to always be a string, and can easily lead to an error as you did.
If you want to chain these methods in you example, perhaps you can do this:
user_input = gets.tap(&:chomp!).tap(&:downcase!)
if user_input.include?("s")
...
#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 am working though LearnTocodethehardway.com http://ruby.learncodethehardway.org/book/ex25.html
On ex25. In the example there is a module that has a bunch of methods that return or print values. The Print_last_word method when supplied with an array of strings just puts nil. it does this even in his example output. My question would then be why?
To be precise, it doesn't puts nil - it puts the last word and returns nil. Here's the example output:
>> Ex25.print_last_word(words)
wait. # <- this is the output
=> nil # <- this is the return value
puts always returns nil.
UPDATE
There seems to be a bug in print_first_word:
module Ex25
def Ex25.print_first_word(words)
word = words.pop(0)
puts word
end
end
Ex25.print_first_word(["foo", "bar", "baz"])
#=> nil
This is because ["foo", "bar", "baz"].pop(0) returns an empty array and puts [] just returns nil without printing anything.
A working implementation (in the exercise's style) could look like this:
module Ex25
def Ex25.print_first_word(words)
word = words.shift
puts word
end
end
To make it more easy to understand:
"puts" never is used for its return value. It is used for its side effect if you wanted a method that would return a word, you would then do something with that word.
words = ["apple", "cucumber", "apple"]
def give_me_first_word(array_of_words)
array_of_words.first
end
variable_to_be_used_later = give_me_first_word(words)
This function would return "apple"(and in this case the variable would be assigned "apple"), but it would puts nothing to the screen. This value would then be used in another program or function.
If you puts a value, you would then not need to return it to any other program as it's already served its purpose. It's actually intuitive because if puts also returned the value, it would be doing two things. The counterpart to "puts" would be "return" that simply returns the value and does not display it to the screen.
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".
If I execute this ruby code:
def foo
100
end
p defined?(foo), foo
if false
foo = 200
end
p defined?(foo), foo
The output I get is:
"method"
100
"local-variable"
nil
Can someone explain to me why foo is set to nil after not executing the if? Is this expected behavior or a ruby bug?
Names on the left hand side of assignments get set to nil, even if the code can't be reached as in the if false case.
>> foo
NameError: undefined local variable or method `foo' for main:Object
...
>> if false
.. foo = 1
.. end #=> nil
>> foo #=> nil
When Ruby tries to resolve barewords, it first looks for local variables (there's a reference to that in the Pickaxe book, which I can't seem to find at the moment). Since you now have one called foo it displays nil. As Mischa noted, the method still can be called as foo().
This is what my pal and Ruby super-expert Josh Cheek had to say:
When Ruby sees the assignment, it initializes the variable in the current scope and sets it to nil. Since the assignment didn't get run, it didn't update the value of foo.
if statements don't change scope like blocks do. This is also the most important difference between
for x in xs
and
xs.each { |x| }
Here's another example:
a = 123 if a # => nil
a # => nil
We shouldn't be able to say if a because we never set a, but Ruby sees the a = 123 and initializes a, then gets to if a at which point a is nil
I'd consider it a quirk of the interpreter, really. Gary Bernhardt makes fun of it in wat (https://www.destroyallsoftware.com/talks/wat) with a = a
-Josh