Related
I want to be able to find a custom class in my set given just a string. Like so:
require 'set'
Rank = Struct.new(:name, keyword_init: true) {
def hash
name.hash
end
def eql?(other)
hash == other.hash
end
def ==(other)
hash == other.hash
end
}
one = Rank.new(name: "one")
two = Rank.new(name: "two")
set = Set[one, two]
but while one == "one" and one.eql?("one") are both true, set.include?("one") is still false. what am i missing?
thanks!
Set is built upon Hash, and Hash considers two objects the same if:
[...] their hash value is identical and the two objects are eql? to each other.
What you are missing is that eql? isn't necessarily commutative. Making Rank#eql? recognize strings doesn't change the way String#eql? works:
one.eql?('one') #=> true
'one'.eql?(one) #=> false
Therefore it depends on which object is the hash key and which is the argument to include?:
Set['one'].include?(one) #=> true
Set[one].include?('one') #=> false
In order to make two objects a and b interchangeable hash keys, 3 conditions have to be met:
a.hash == b.hash
a.eql?(b) == true
b.eql?(a) == true
But don't try to modify String#eql? – fiddling with Ruby's core classes isn't recommended and monkey-patching probably won't work anyway because Ruby usually calls the C methods directly for performance reasons.
In fact, making both hash and eql? mimic name doesn't seem like a good idea in the first place. It makes the object's identity ambiguous which can lead to very strange behavior and hard to find bugs:
h = { one => 1, 'one' => 1 }
#=> {#<struct Rank name="one">=>1, "one"=>1}
# vs
h = { 'one' => 1, one => 1 }
#=> {"one"=>1}
what am i missing?
What you are missing is that "one" isn't in your set. one is in your set, but "one" isn't.
Therefore, the answer Ruby is giving you is perfectly correct.
All that you have done with your implementation of Rank is that any two ranks with the same name are considered to be the same by a Hash, Set, or Array#uniq. But, a Rank is not the same as a String.
If you want to be able to have a set-like data structure where you can look up things by one of their attributes, you will have to write it yourself.
Something like (untested):
class RankSet < Set
def [](*args)
super(*args.map(&:name))
end
def each
return enum_for(__callee__) unless block_given?
super {|e| yield e.name }
end
end
might get you started.
Or, instead of writing your own set, you can just use the fact that any arbitrary rank with the right name can be used for lookup:
set.include?(Rank.new(name: "one"))
#=> true
# even though it is a *different* `Rank` object
I have the following two methods, which I believe should have the same behaviour disregarding their names:
def a=(*params)
params
end
def b(*params)
params
end
But when in fact I use them:
a=(1) # => 1
b(1) # => [1]
(a=1) == b(1) # => false
while interestingly:
(a=1,2) == b(1,2) # => true
Why isn't their behaviour the same?
Edit: forgot to wrap the above in a class / call with self. which accidentally produces the same behaviour but for a different reason. It has been pointed out in the answers.
It has nothing to do with splat. It's the assignment operator. In ruby, the assignment operator returns the value assigned. The return value from the method is ignored.
So a=1 return 1, not [1].
But, as mentioned by #mudasobwa, you're not even calling the method here. But if you were, that's what would happen (ignoring the return value).
class Foo
def a=(*params)
params
end
end
f = Foo.new
f.a = 1 # => 1
f.a = 1,2 # => [1, 2]
To not ignore the return value, call that setter without using assignment operator.
f.send 'a=', 1 # => [1]
The thing is that
a = 1
sets the local variable and does not call your method at all. Try with
def a=(*param)
puts "I AM HERE"
end
var= methods require an explicit receiver. To call your method, call it with an explicit receiver:
self.a = 1
It still won’t return anything but 1, because assignment methods return the value (the same way as initialize called through MyClass.new returns an instance, no matter what.) But you might check that splat works with:
def a=(*param)
puts param.inspect
end
self.a = 1
# [1]
#⇒ 1
#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 on an application that takes input from a YAML file, parses them into objects, and let's them do their thing. The only problem I'm having now, is that the YAML parser seems to ignore the objects "initialize" method. I was counting on the constructor to fill in any instance variables the YAML file was lacking with defaults, as well as store some things in class variables. Here is an example:
class Test
##counter = 0
def initialize(a,b)
#a = a
#b = b
#a = 29 if #b == 3
##counter += 1
end
def self.how_many
p ##counter
end
attr_accessor :a,:b
end
require 'YAML'
a = Test.new(2,3)
s = a.to_yaml
puts s
b = YAML::load(s)
puts b.a
puts b.b
Test.how_many
puts ""
c = Test.new(4,4)
c.b = 3
t = c.to_yaml
puts t
d = YAML::load(t)
puts d.a
puts d.b
Test.how_many
I would have expected the above to output:
--- !ruby/object:Test
a: 29
b: 3
29
3
2
--- !ruby/object:Test
a: 4
b: 3
29
3
4
Instead I got:
--- !ruby/object:Test
a: 29
b: 3
29
3
1
--- !ruby/object:Test
a: 4
b: 3
4
3
2
I don't understand how it makes these objects without using their defined initialize method. I'm also wondering if there is anyway to force the parser to use the initialize method.
Deserializing an object from Yaml doesn’t use the initialize method because in general there is no correspondance between the object’s instance variables (which is what the default Yaml serialization stores) and the parameters to initialize.
As an example, consider an object with an initialize that looks like this (with no other instance variables):
def initialize(param_one, param_two)
#a_variable = some_calculation(param_one, param_two)
end
Now when an instance of this is deserialized, the Yaml processor has a value for #a_variable, but the initialize method requires two parameters, so it can’t call it. Even if the number of instance variables matches the number of parameters to initialize it is not necessarily the case that they correspond, and even if they did the processor doesn’t know the order they shoud be passed to initialize.
The default process for serializing and deserializing a Ruby object to Yaml is to write out all instance variables (with their names) during serialization, then when deserializing allocate a new instance of the class and simply set the same instance variables on this new instance.
Of course sometimes you need more control of this process. If you are using the Psych Yaml processor (which is the default in Ruby 1.9.3) then you should implement the encode_with (for serialisation) or or init_with (for deserialization) methods as appropriate.
For serialization, Psych will call the encode_with method of an object if it is present, passing a coder object. This object allows you to specify how the object should be represented in Yaml – normally you just treat it like a hash.
For deserialization, Psych will call the init_with method if it is present on your object instead of using the default procedure described above, again passing a coder object. This time the coder will contain the information about the objects representation in Yaml.
Note you don’t need to provide both methods, you can just provide either one if you want. If you do provide both, the coder object you get passed in init_with will essentially be the same as the one passed to encode_with after that method has run.
As an example, consider an object that has some instance variables that are calculated from others (perhaps as an optimisation to avoid a large calculation), but shouldn’t be serialized to the Yaml.
class Foo
def initialize(first, second)
#first = first
#second = second
#calculated = expensive_calculation(#first, #second)
end
def encode_with(coder)
# #calculated shouldn’t be serialized, so we just add the other two.
# We could provide different names to use in the Yaml here if we
# wanted (as long as the same names are used in init_with).
coder['first'] = #first
coder['second'] = #second
end
def init_with(coder)
# The Yaml only contains values for #first and #second, we need to
# recalculate #calculated so the object is valid.
#first = coder['first']
#second = coder['second']
#calculated = expensive_calculation(#first, #second)
end
# The expensive calculation
def expensive_calculation(a, b)
...
end
end
When you dump an instance of this class to Yaml, it will look something like this, without the calculated value:
--- !ruby/object:Foo
first: 1
second: 2
When you load this Yaml back into Ruby, the created object will have the #calculated instance variable set.
If you wanted you could call initialize from within init_with, but I think it would be better to keep the a clear separation between initializing a new instance of the class, and deserializing an existing instance from Yaml. I would recommend extracting the common logic into methods that can be called from both instead,
If you only want this behavior with pure ruby classes that use #-style instance variables (not those from compiled extensions and not Struct-style), the following should work. YAML seems to call the allocate class method when loading an instance of that class, even if the instance is nested as a member of another object. So we can redefine allocate. Example:
class Foo
attr_accessor :yaml_flag
def self.allocate
super.tap {|o| o.instance_variables.include?(:#yaml_flag) or o.yaml_flag = true }
end
end
class Bar
attr_accessor :foo, :yaml_flag
def self.allocate
super.tap {|o| o.instance_variables.include?(:#yaml_flag) or o.yaml_flag = true }
end
end
>> bar = Bar.new
=> #<Bar:0x007fa40ccda9f8>
>> bar.foo = Foo.new
=> #<Foo:0x007fa40ccdf9f8>
>> [bar.yaml_flag, bar.foo.yaml_flag]
=> [nil, nil]
>> bar_reloaded = YAML.load YAML.dump bar
=> #<Bar:0x007fa40cc7dd48 #foo=#<Foo:0x007fa40cc7db90 #yaml_flag=true>, #yaml_flag=true>
>> [bar_reloaded.yaml_flag, bar_reloaded.foo.yaml_flag]
=> [true, true]
# won't overwrite false
>> bar.foo.yaml_flag = false
=> false
>> bar_reloaded = YAML.load YAML.dump bar
=> #<Bar:0x007fa40ccf3098 #foo=#<Foo:0x007fa40ccf2f08 #yaml_flag=false>, #yaml_flag=true>
>> [bar_reloaded.yaml_flag, bar_reloaded.foo.yaml_flag]
=> [true, false]
# won't overwrite nil
>> bar.foo.yaml_flag = nil
=> nil
>> bar_reloaded = YAML.load YAML.dump bar
=> #<Bar:0x007fa40cd73518 #foo=#<Foo:0x007fa40cd73360 #yaml_flag=nil>, #yaml_flag=true>
>> [bar_reloaded.yaml_flag, bar_reloaded.foo.yaml_flag]
=> [true, nil]
I intentionally avoided a o.nil? check in the tap blocks because nil may actually be a meaningful value that you don't want to overwrite.
One last caveat: allocate may be used by third party libraries (or by your own code), and you may not want to set the members in those cases. If you want to restrict allocation, to just yaml loading, you'll have to do something more fragile and complex like check the caller stack in the allocate method to see if yaml is calling it.
I'm on ruby 1.9.3 (with psych) and the top of the stack looks like this (path prefix removed):
psych/visitors/to_ruby.rb:274:in `revive'",
psych/visitors/to_ruby.rb:219:in `visit_Psych_Nodes_Mapping'",
psych/visitors/visitor.rb:15:in `visit'",
psych/visitors/visitor.rb:5:in `accept'",
psych/visitors/to_ruby.rb:20:in `accept'",
psych/visitors/to_ruby.rb:231:in `visit_Psych_Nodes_Document'",
psych/visitors/visitor.rb:15:in `visit'",
psych/visitors/visitor.rb:5:in `accept'",
psych/visitors/to_ruby.rb:20:in `accept'",
psych/nodes/node.rb:35:in `to_ruby'",
psych.rb:128:in `load'",
from_yaml(input)
Special loader for YAML files. When a Specification object is loaded from a YAML file, it bypasses the normal Ruby object initialization routine (initialize). This method makes up for that and deals with gems of different ages.
input can be anything that YAML.load() accepts: String or IO.
This is the reason that the initialize method was not being run when you executed YAML::Load.
In Ruby, is it possible to pass by reference a parameter with value-type semantics (e.g. a Fixnum)?
I'm looking for something similar to C#'s 'ref' keyword.
Example:
def func(x)
x += 1
end
a = 5
func(a) #this should be something like func(ref a)
puts a #should read '6'
Btw. I know I could just use:
a = func(a)
You can accomplish this by explicitly passing in the current binding:
def func(x, bdg)
eval "#{x} += 1", bdg
end
a = 5
func(:a, binding)
puts a # => 6
Ruby doesn't support "pass by reference" at all. Everything is an object and the references to those objects are always passed by value. Actually, in your example you are passing a copy of the reference to the Fixnum Object by value.
The problem with the your code is, that x += 1 doesn't modify the passed Fixnum Object but instead creates a completely new and independent object.
I think, Java programmers would call Fixnum objects immutable.
In Ruby you can't pass parameters by reference. For your example, you would have to return the new value and assign it to the variable a or create a new class that contains the value and pass an instance of this class around. Example:
class Container
attr_accessor :value
def initialize value
#value = value
end
end
def func(x)
x.value += 1
end
a = Container.new(5)
func(a)
puts a.value
You can try following trick:
def func(x)
x[0] += 1
end
a = [5]
func(a) #this should be something like func(ref a)
puts a[0] #should read '6'
http://ruby-doc.org/core-2.1.5/Fixnum.html
Fixnum objects have immediate value. This means that when they are assigned or
passed as parameters, the actual object is passed, rather than a reference to
that object.
Also Ruby is pass by value.
However, it seems that composite objects, like hashes, are passed by reference:
fp = {}
def changeit(par)
par[:abc] = 'cde'
end
changeit(fp)
p fp
gives
{:abc=>"cde"}