>> a = 5
=> 5
>> b = a
=> 5
>> b = 4
=> 4
>> a
=> 5
how can I set 'b' to actually be 'a' so that in the example, the variable a will become four as well. thanks.
class Ref
def initialize val
#val = val
end
attr_accessor :val
def to_s
#val.to_s
end
end
a = Ref.new(4)
b = a
puts a #=> 4
puts b #=> 4
a.val = 5
puts a #=> 5
puts b #=> 5
When you do b = a, b points to the same object as a (they have the same object_id).
When you do a = some_other_thing, a will point to another object, while b remains unchanged.
For Fixnum, nil, true and false, you cannot change the value without changing the object_id. However, you can change other objects (strings, arrays, hashes, etc.) without changing object_id, since you don't use the assignment (=).
Example with strings:
a = 'abcd'
b = a
puts a #=> abcd
puts b #=> abcd
a.upcase! # changing a
puts a #=> ABCD
puts b #=> ABCD
a = a.downcase # assigning a
puts a #=> abcd
puts b #=> ABCD
Example with arrays:
a = [1]
b = a
p a #=> [1]
p b #=> [1]
a << 2 # changing a
p a #=> [1, 2]
p b #=> [1, 2]
a += [3] # assigning a
p a #=> [1, 2, 3]
p b #=> [1, 2]
You can't. Variables hold references to values, not references to other variables.
Here's what your example code is doing:
a = 5 # Assign the value 5 to the variable named "a".
b = a # Assign the value in the variable "a" (5) to the variable "b".
b = 4 # Assign the value 4 to the variable named "b".
a # Retrieve the value stored in the variable named "a" (5).
See this article for a more in-depth discussion of the topic: pass by reference or pass by value.
As has been noted the syntax you are using can not be done. Just throwing this out there though you could make a wrapper class it depends what you actually want to do
ruby-1.8.7-p334 :007 > class Wrapper
ruby-1.8.7-p334 :008?> attr_accessor :number
ruby-1.8.7-p334 :009?> def initialize(number)
ruby-1.8.7-p334 :010?> #number = number
ruby-1.8.7-p334 :011?> end
ruby-1.8.7-p334 :012?> end
=> nil
ruby-1.8.7-p334 :013 > a = Wrapper.new(4)
=> #<Wrapper:0x100336db8 #number=4>
ruby-1.8.7-p334 :014 > b = a
=> #<Wrapper:0x100336db8 #number=4>
ruby-1.8.7-p334 :015 > a.number = 6
=> 6
ruby-1.8.7-p334 :016 > a
=> #<Wrapper:0x100336db8 #number=6>
ruby-1.8.7-p334 :017 > b
=> #<Wrapper:0x100336db8 #number=6>
You can use arrays:
a = [5]
b = a
b[0] = 4
puts a[0] #=> 4
This idea is based on this answer.
Just for the sake of reference.
>> a = 5
=> 5
>> a.object_id
=> 11
>> b = a
=> 5
>> b.object_id
=> 11
>> b = 4
=> 4
>> b.object_id
=> 9
>> a.object_id
=> 11
# We did change the Fixnum b Object.
>> Fixnum.superclass
=> Integer
>> Integer.superclass
=> Numeric
>> Numeric.superclass
=> Object
>> Object.superclass
=> BasicObject
>> BasicObject.superclass
=> nil
I hope this gives us all a little better understanding about objects in Ruby.
One option in cases where you feel you would like to have direct pointer operations is to use the replace method of Hashes, Arrays & Strings.
this is useful for when you would like to have a method return a variable that a proc the method sets up will change at a later date, and don't want the annoyance of using a wrapper object.
example:
def hash_that_will_change_later
params = {}
some_resource.on_change do
params.replace {i: 'got changed'}
end
params
end
a = hash_that_will_change_later
=> {}
some_resource.trigger_change!
a
{i: 'got changed'}
It's probably better generally to use explicit object wrappers for such cases, but this pattern is useful for building specs/tests of asynchronous stuff.
I'm no Ruby expert. But for a technically crazy kluge...that would only work if you felt like going through eval every time you worked with a variable:
>> a = 5
=> 5
>> b = :a
=> :a
>> eval "#{b} = 4"
=> 4
>> eval "#{a}"
=> 4
>> eval "#{b}"
=> 4
Note that a direct usage of b will still give you :a and you can't use it in expressions that aren't in eval:
>> b
=> :a
>> b + 1
NoMethodError: undefined method `+' for :a:Symbol
...and there are certainly a ton of caveats. Such as that you'd have to capture the binding and pass it around in more complex scenarios...
'pass parameter by reference' in Ruby?
#Paul.s has an answer for if you can change the point of declaration to be a wrapper object, but if you can only control the point of reference then here's a BasicReference class I tried:
class BasicReference
def initialize(r,b)
#r = r
#b = b
#val = eval "#{#r}", #b
end
def val=(rhs)
#val = eval "#{#r} = #{rhs}", #b
end
def val
#val
end
end
a = 5
puts "Before basic reference"
puts " the value of a is #{a}"
b = BasicReference.new(:a, binding)
b.val = 4
puts "After b.val = 4"
puts " the value of a is #{a}"
puts " the value of b.val is #{b.val}"
This outputs:
Before basic reference
the value of a is 5
After b.val = 4
the value of a is 4
the value of b.val is 4
Related
I am new to Ruby, so please excuse this question if it is obvious.
I am working with a Module with a function signature that I don't understand. How would I call this function?
module Facter
...
def self.[](name)
collection.fact(name)
end
...
In my code I want to reference something that should be in collection.fact, in this Facter module. What syntax to I use to call this function?
Cheers
It works like this:
class MyModule
def self.[](arg)
puts arg
end
end
MyModule["Hello world"] # will print Hello world
Please see official docs:
https://ruby-doc.org/core/doc/syntax/methods_rdoc.html
Additionally, methods for element reference and assignment may be defined: [] and []= respectively. Both can take one or more arguments, and element reference can take none.
class C
def [](a, b)
puts a + b
end
def []=(a, b, c)
puts a * b + c
end
end
obj = C.new
obj[2, 3] # prints "5"
obj[2, 3] = 4 # prints "10"
So about example from docs
# From docs
obj[2, 3]
# It's the same as
obj.[](2, 3)
More interesting example
# From docs
obj[2, 3] = 4
# will print 10
# => 4
# It's the almost as
obj.[]=(2, 3, 4)
# will print 10
# => nil
As you see when you call as obj[2, 3] = 4 Ruby takes the value after = as the last argument of the []= method and return it as method result
And regardless of whether there is return in the method body. For example
class C
def []=(a, b, c)
puts "Before return"
return 12
puts "After return"
end
end
obj = C.new
obj[2, 3] = 4
# will print Before return
# => 4
obj.[]=(2, 3, 4)
# will print Before return
# => 12
It is desirable to define such method with more than one parameter. Technically, you can have only one, but the call will be like this obj[] = 1
In PHP, I can do this:
$a = 1;
$c = 'a';
$$c = 2;
//now $a == 2
Is there any equivalent in ruby? By which I mean, any simple way to have it dereference a variable during execution like this? I'd rather not use eval, because it looks messy - I've already determined that eval can't be called as a method of a string.
It is possible but it's a bit more complicated., and you actually have two possibilities:
Kernel#local_variables
Returns the names of the current local variables.
fred = 1
for i in 1..10
# ...
end
local_variables #=> [:fred, :i]
Binding#local_variable_get/set
Returns a value of local variable symbol.
def foo
a = 1
binding.local_variable_get(:a) #=> 1
binding.local_variable_get(:b) #=> NameError
end
This method is short version of the following code.
binding.eval("#{symbol}")
if you just need this you can do
a = 1
c = 'a'
eval("#{c} = 2")
a == 2 # => true
... but this is moron way to do this
if you need this for instance variables
class Foo
attr_reader :a
def initialize
#a = 1
end
end
foo = Foo.new
foo.instance_variable_get(:a) #=> 1
foo.a #=> 1
foo.instance_variable_set(:"#a", 2)
foo.a #=> 2
...you can also eval instance like this:
# ...
foo.instance_eval do
#a = 'b'
end
foo.a # => 'b'
a = "test"
b = a
b << "test1"
puts a
puts b
b += "test2"
puts a
puts b
Output
------
$ruby main.rb
testtest1
testtest1
testtest1
testtest1test2
I was expecting following output.
test
testtest1
test
testtest1test2
b and a holding the same object through your assignment b = a.Thus while you are modifying b with b << "test1", it is actually the same object which is pointing by a.
a = "test"
b = a
a.object_id # => 71753220
b.object_id # => 71753220
Thus, I can tell you behavior you are seeing is correct.
Now b += "test2", here you are assigning a new object to b.
b += "test2"
a.object_id # => 71753220
b.object_id # => 72602390
Better to read
String#+
Concatenation—Returns a new String containing other_str concatenated to str
String#<<
Append—Concatenates the given object to str.
If I have variables as below,
i = 1
k1 = 20
is there any ways to get values of k1 with the interpolation of i?
Something like,
k"#{i}"
=> 20
Thanks in advance.
It depends on whether it's a local variable or a method. send "k#{i}" should do the trick with methods:
class Foo
attr_accessor :i, :k1
def get
send "k#{i}"
end
end
foo = Foo.new
foo.i = 1
foo.k1 = "one"
foo.get
# => "one"
If you really need to, you can access local variables using the current Binding and local_variable_get:
i = 1
k1 = "one"
local_variables
# => [:i, :k1]
binding.local_variable_get("k#{i}")
# => "one"
This is pretty awful though. In this instance you'd be better off using a Hash:
i = 1
k = {1 => "one"}
k[i]
# => "one"
Let's say I have a Ruby class:
class MyClass
def self.property
return "someVal"
end
def self.property=(newVal)
# do something to set "property"
success = true
return success # success is a boolean
end
end
If I try and do MyClass.property=x, the return value of the whole statement is always x. It is a convention in a lot of C-based/inspired languages to return a boolean "success" value - is it possible to do this for a setter using the "equals syntax" in Ruby?
Furthermore - if this isn't possible, why not? Is there any conceivable downside to allowing an "equals setter" operation return a value?
One downside is that you would break the chained assignment semantics:
$ irb
irb(main):001:0> x = y = 3
=> 3
irb(main):002:0> p x
3
=> nil
irb(main):003:0> p y
3
=> nil
irb(main):004:0>
Consider:
x = MyClass.property = 3
Then x would take true if this worked as you had expected (right-associativity). That could be a surprise for people using your interface and used to the typical semantics.
You also got me thinking about parallel assignment, eg:
x, y = 1, 2
Apparently the return value from that expression is implementation specific... I guess I won't be chaining parallel assignments :)
Nice question!
Like Martin says, this would break assignment chaining.
The way ruby assignment methods are defined to work expands MyClass.property = 3 to the equivalent of (lambda { |v| MyClass.send('property=', v); v })[3] (not really, but this shows how chaining works). The return value of the assignment is always the value assigned.
If you want to see the result of your MyClass#property= method, then use #send:
irb> o = Object.new
=> #<Object:0x15270>
irb> def o.x=(y)
irb> #x = y+1
irb> puts "y = #{y}, #x = ##x"
irb> true
irb> end
=> nil
irb> def o.x
irb> puts "#x = ##x"
irb> #x
irb> end
=> nil
irb> o.x = 4
y = 4, #x = 5
=> 4
irb> o.x
#x = 5
=> 5
irb> o.send('x=', 3)
y = 3, #x = 4
=> true
However, the ruby way to do this is with exceptions - if something goes wrong during
the assignment, raise an exception. Then all invokers must handle it if something goes
wrong, unlike a return value, which can be easily ignored:
# continued from above...
irb> def o.x=(y)
irb> unless y.respond_to? :> and (y > 0 rescue false)
irb> raise ArgumentError, 'new value must be > 0', caller
irb> end
irb> #x = y + 1
irb> puts "y = #{y}, #x = ##x"
irb> end
=> nil
irb> o.x = 4
y = 4, #x = 5
=> 4
irb> o.x = 0
ArgumentError: new value must be > 0
from (irb):12
from :0
irb> o.x = "3"
ArgumentError: new value must be > 0
from (irb):13
from :0
irb> o.x
#x = 5
=> 5
I'm not a Ruby expert but I'd say no for that case I'm afraid. A property setter is solely there to set the value of a private field, not to have any side effects like returning result codes.
If you want that functionality then forget the setter and write a new method called TrySetProperty or something which tries to set the property and returns a boolean.