If a block is a closure, why does this code does not work, and how to make it work?
def R(arg)
Class.new do
def foo
puts arg
end
end
end
class A < R("Hello!")
end
A.new.foo #throws undefined local variable or method `arg' for #<A:0x2840538>
Blocks are closures and arg is indeed available inside the Class.new block. It's just not available inside the foo method because def starts a new scope. If you replace def with define_method, which takes a block, you'll see the result you want:
def R(arg)
Class.new do
define_method(:foo) do
puts arg
end
end
end
class A < R("Hello!")
end
A.new.foo # Prints: Hello!
If you define the class dynamically, you can alter it as you like:
def R(arg)
c = Class.new
# Send the block through as a closure, not as an inline method
# definition which interprets variables always as local to the block.
c.send(:define_method, :foo) do
arg
end
c
end
class A < R("Hello!")
end
puts A.new.foo.inspect
# => "Hello!"
Related
When inheriting from Module you can pass a block to super() and it is executed:
class Foo < Module
def initialize
super() do
# do some other stuff...
end
end
end
Why would you use this over just calling super(), without the block, followed by the rest of the code, such as:
class Foo < Module
def initialize
super()
# do some other stuff...
end
end
An less contrived example would be:
class QuestionAttr < Module
def initialize(method_name)
super() do
define_method "#{method_name}?" do
!!public_send(method_name)
end
end
end
end
Fruit = Struct.new(:colour) do
include QuestionAttr.new(:colour)
end
fruit = Fruit.new('red')
fruit.colour? # => true
Why would I define_method inside the block passed to super(), as per the above example, instead of omitting the block passed to super() and define_method afterwards.
There is a pretty good documentation of the current implementation of refinements in ruby here:
http://ruby-doc.org//core-2.2.0/doc/syntax/refinements_rdoc.html,
but there are some strange corner cases.
First, include module is orthogonal to using module (one include the instance method of module while the other activates the refinement). But there is a trick to include a refinement module itself, see
Better way to turn a ruby class into a module than using refinements?.
def to_module(klass)
Module.new do
#note that we return the refinement module itself here
return refine(klass) {
yield if block_given?
}
end
end
class Base
def foo
"foo"
end
end
class Receiver
include to_module(Base) {
def foo
"refined " + super
end
}
end
Receiver.new.foo #=> "refined foo"
Strangely this refinement module can't be used with using!
m=to_module(Base) {}
m.class #=> Module
using m
#=>TypeError: wrong argument type Class (expected Module)
So using only work on the enclosing module of the refinement modules.
Secondly I wanted to use the above yield trick to be able to pass a Proc to refine (even through it only accepts a block), without resorting to converting the Proc back to source as in
https://www.new-bamboo.co.uk/blog/2014/02/05/refinements-under-the-knife/.
But using yield as in the include example does not work:
def ref_module1(klass)
Module.new do
refine(klass) {
yield
}
end
end
class Receiver1
using ref_module1(Base) {
def foo
"refined " + super
end
}
def bar
Base.new.foo
end
end
Receiver1.new.bar #=> NoMethodError: super: no superclass method `foo'
We see that Receiver1 still use Bar#foo and not the refined method.
Howewer we can use module_eval instead:
def ref_module2(klass,&b)
Module.new do
refine(klass) {
module_eval(&b)
}
end
end
class Receiver2
using ref_module2(Base) {
def foo
"refined " + super
end
}
def bar
Base.new.foo
end
end
Receiver2.new.bar #=> "refined foo"
I don't quite understand why module_eval works here and not the yield method. Inside the refinement block, the 'default_definee' is the refinement module, so module_eval which puts the 'default_definee' to self='the refinement module' should not affect it. And indeed in the 'include' example at the beginning, I get the same result when I use module_eval or a direct yield.
Can anyone explain this behavior?
Context (or binding) is the reason why module_eval works and yield doesn't in your last set of examples. It actually has nothing to do with refinements, as demonstrated below.
Starting with module_eval:
class Foo
def run(&block)
self.class.module_eval(&block)
end
end
foo = Foo.new
foo.run {
def hello
"hello"
end
}
puts foo.hello # => "hello"
puts hello => # '<main>': undefined method 'hello' for main:Object (NameError)
In Foo#run we call module_eval on Foo. This switches the context (self) to be Foo. The result is much like we had simple defined hello inside of class Foo originally.
Now let's take a look at yield:
class Foo
def run
yield
end
end
foo = Foo.new
foo.run {
def hello
"hello"
end
}
puts hello # => "hello"
puts foo.hello # => '<main>': private method 'hello' called for ...
yield simply invokes the block in its original context, which in this example would be <main>. When the block is invoked, the end result is exactly the same as if the method were defined at the top level normally:
class Foo
def run
yield
end
end
foo = Foo.new
def hello
"hello"
end
puts hello # => "hello"
puts foo.hello # => '<main>': private method 'hello' called for ...
You might notice that foo seems to have the hello method in the yield examples. This is a side effect of defining hello as a method at the top level. It turns out that <main> is just an instance of Object, and defining top level methods is really just defining private methods on Object which nearly everything else ends up inheriting. You can see this by opening up irb and running the following:
self # => main
self.class # => Object
def some_method
end
"string".method(:some_method) # => #<Method: String(Object)#some_method>
Now back to your examples.
Here's what happens in the yield example:
def ref_module1(klass)
Module.new do
refine(klass) {
yield
}
end
end
class Receiver1
# like my yield example, this block is going to
# end up being invoked in its original context
using ref_module1(Base) {
def foo
"I'm defined on Receiver1"
end
}
def bar
# calling foo here will simply call the original
# Base#foo method
Base.new.foo
end
end
# as expected, if we call Receiver1#bar
# we get the original Base#foo method
Receiver1.new.bar # => "foo"
# since the block is executed in its original context
# the method gets defined in Receiver1 -- its original context
Receiver1.new.foo # => "I'm defined on Receiver1"
As for module_eval, it works in your examples because it causes the block to be run in the context of the new module, rather than on the Receiver1 class.
I have the following code:
class A
def self.scope
yield
end
def self.method_added method
self.instance_eval %{
# do something involving the added method
}
end
end
class B < A
scope do
def foo
end
end
end
When the method_added hook is fired, will the code inside instance_eval run within the same scope as the method that was added? Or, will it run outside of it?
What are the caveats and gotchas involved within this?
Your scope method is basically a no-op. When you pass a block to a method that yields, the block is evaluated in the current scope. Observe:
class A
def self.scope
yield
end
end
A.scope { p self }
# main
Since nothing is yielded to the block, and nothing is done with the return value of yield, any code run in the block will have the same effect run outside the scope block.
This isn't the case with instance_eval, however. When instance_eval runs a block, self in the block is set to the receiver (rather than whatever self is in the block's scope). Like this:
class A
end
A.instance_eval { p self }
# A
But note that this means that self.instance_eval { ... } is also a fancy no-op, because you're changing the block's self to the same self outside the block.
So your code is equivalent to this:
class A
def self.method_added method
# do something involving the added method
end
end
class B < A
def foo
end
end
Let's find out!
class A
def self.scope
yield
end
def self.method_added method
puts "In method_added, method = #{method}, self = #{self}"
instance_eval 'puts "In instance_eval, method = #{method}, self = #{self}"'
end
end
class B < A
scope do
puts "In scope's block, self = #{self}"
def foo
end
end
end
# In scope's block, self = B
# In method_added, method = foo, self = B
# In instance_eval, method = foo, self = B
Notice that you don't need self. in self.instance_eval.
class Test
class << self
attr_accessor :some
def set_some
puts self.inspect
some = 'some_data'
end
def get_some
puts self.inspect
some
end
end
end
Test.set_some => Test
puts Test.get_some.inspect => Test nil
Here above I could find self as Test itself but not returning the some_data as output.
But while I modified in following way it returns expected output
class Test
class << self
attr_accessor :some
def set_some
puts self.inspect
self.some = 'some_data'
end
def get_some
puts self.inspect
self.some
end
end
end
Test.set_some => Test
puts Test.get_some.inspect => Test some_data
What is the differences?
EDIT
Now in the first example if I set as get some method as
Test.some = 'new_data'
puts Test.some.inspect #=> new_data
Test.set_some
puts Test.get_some.inspect => new_data
Now it made me much more confused.
some = :foo makes ruby think it should create a new local variable with name some. If you want to call some=(), you have to use an explicit reciever - as in self.some = :foo. I once lost a bet on that... :-/
It's (local) variable in the first example
In the first example some is a local variable.
In the second one, some is a method of self. Why? Because attr_accessor :some is the same as:
def some= (val)
#some = val
end
def some
return #some
end
So, you have created the getter and setter methods for the instance variable #some (it's an instance variable of the object Test, as every class is also an object of class Class).
in the first method
def set_some
puts self.inspect
some = 'some_data'
end
some is a local variable.. its not the same as #some that is a instance variable (in this case a class instance variable) so the value disappears when the method ends.
if you want to call the setter method some or set #some to something then do this
#some = 'some_data'
or
self.some = 'some_data'
in the second method
def get_some
puts self.inspect
self.some
end
your calling the method some. which returns the instace variable #some.. and since at this point #some has no value.. returns nil..
Example 1 with no method override and no local variable
class Foo
def initialize
#foo = 'foo'
end
def print_foo
print #foo
print self.foo
print foo
end
end
#foo, self.foo, and foo will access instance variable #foo within the instance method:
Foo.new.print_foo #=> foofoofoo
Example 2 with method override
class Foo
def initialize
#foo = 'foo'
end
def foo
return 'bar'
end
def print_foo
print #foo
print self.foo
print foo
end
end
#foo will access the instance variable, but self.foo and foo will call the foo override method:
Foo.new.print_foo #=> foobarbar
Example 3 with method override and local variable
class Foo
def initialize
#foo = 'foo'
end
def foo
return 'bar'
end
def print_foo
foo = 'baz'
print #foo
print self.foo
print foo
end
end
#foo accesses instance variable, self.foo accesses override method, and foo accesses local variable:
Foo.new.print_foo #=> foobarbaz
Is it possible to execute a proc within the context of another object?
I know that normally you'd do proc.call(foo), and then the block should define a parameter. I was wondering though whether I could get "self" to bind to foo so that it's not necessary to have a block parameter.
proc = Proc.new { self.hello }
class Foo
def hello
puts "Hello!"
end
end
foo = Foo.new
# How can proc be executed within the context of foo
# such that it outputs the string "Hello"?
proc.call
foo.instance_eval &proc
instance_eval can take a block instead of a string, and the & operator turns the proc into a block for use with the method call.
This is for ruby 1.9:
class MyCat
def initialize(start, &block)
#elsewhere = start
define_singleton_method(:run_elsewhere, block) if block_given?
end
end
MyCat.new('Hello'){ #elsewehere << ' world' }.run_elsewhere