I've been scratching my head over how to add behaviour to a specific instance
of a class in Ruby. If I use instance_exec like below however the method that calls the proc does not execute all of the way through.
class HasSecret
def initialize(x)
#secret = x
end
def doTheThing(&block)
puts "before block"
instance_exec(#secret, &block)
puts "after block"
end
end
foo = HasSecret.new(5)
foo.doTheThing do |x|
puts "x"
return x
end
This example simply gives:
before block
5
Is there any perscribed approach to doing this?
Try this:
foo.doTheThing do |x|
puts x
end
Generally, you are not supposed to return in blocks. Calling return in a block will cause the method yielding (enclosing) to return, which is usually not the behaviour desired. It can also case a LocalJumpError.
In fact, because this scenario is so nuanced and uncommon, a lot of people say you can't return from a block. Check out the conversation in the comments of the answer on this question, for example: Unexpected Return (LocalJumpError)
In your case, you aren't seeing the "after block" puts because the return in the passed block is causing doTheThing to return before getting there.
Related
I defined this method:
# #return [Array<String>] items
def produce_items
["foo", "bar", "baz"]
end
A correct usage would be, obviously,
produce_items.each do |i|
puts i
end
But I wrote this, which silently does nothing:
produce_items do |i|
puts i
end
Is there a way to declare produce_items in such a way that my incorrect usage produces an error/exception? If MRI cannot do it, can an alternative interpreter do it? Can a static analysis tool like RuboCop or ruby-lint do it?
I suspect it may be hard because there are common idioms where methods take an optional block:
def block_optional_implicitly
if block_given?
puts "Got a block"
yield
else
puts "Did not get a block"
end
end
def block_optional_explicitly(&block)
unless block.nil?
puts "Got a block"
block.call
else
puts "Did not get a block"
end
end
(This is a reverse of the question How to require a block in Ruby?)
Yep, no way of doing this in ruby, except the block_given? checks all over the code.
Passing a block to a method that doesn't expect it does no harm. Besides something not happening (because you forgot the .each). This should be covered by unit tests. You kill two birds with one stone: make your code more regression-proof and don't get to insert fail "unexpected block" if block_given? into every method of your app.
I came across this code:
class RandomSequence
def initialize(limit,num)
#limit,#num = limit,num
end
def each
#num.times { yield (rand * #limit).floor }
end
end
i = -1
RandomSequence.new(10,4).each do |num|
i = num if i < num
end
Is it the case that the each method is called only once and will compute four different values, and then for each of those values we execute code block between do and end? Is my understanding of control flow correct?
Your understanding is close. The random number will be generated, then a block yielded, then another generated, etc 4 times. You can verify this easily by adding puts statements into you blocks to see when they are executed.
class RandomSequence
def initialize(limit,num)
#limit,#num = limit,num
end
def each
puts "in each"
#num.times { yield (rand.tap {|x| puts "Generated #{x}" } * #limit).floor }
end
end
i = -1
RandomSequence.new(10,4).each do |num|
puts "in block"
i = num if i < num
end
Outputs
in each
Generated 0.6724385316643955
in block
Generated 0.8906983274750662
in block
Generated 0.49038868732214036
in block
Generated 0.38100454011243456
in block
The each method on the RandomSequence class is called once. In it #num.times which creates an Enumerator. The Enumerator is iterated over and the block with the yield statement is called (with the argument to it ignored).
The yield statement calls the block that's passed to the each method passing the value of (rand * #limit).floor. In your code the block is not bound to a variable, i.e you could get a reference to the block by doing:
def each(&block)
#... do stuff with block, e.g. block.call("some args")
end
which can be useful at times.
A bit off topic, but one thing I found scary with Ruby starting out is that a return statement returns the flow of execution from where it was defined.
def create_proc
puts "Creating proc"
Proc.new do
puts "In proc!"
return "some value" # notice the explicit return
end
end
def do_stuff
my_proc = create_proc
my_proc.call # This will cause a runtime error
end
If the explicit return is removed everything works there is no error... Lesson being that in ruby you should probably avoid using explicit returns.
I'd like to write a method that yields values in one place and pass it as a parameter to another method that will invoke it with a block. I'm convinced it can be done but somehow I'm not able to find the right syntax.
Here's some sample (non-working) code to illustrate what I'm trying to achieve:
def yielder
yield 1
yield 2
yield 3
end
def user(block)
block.call { |x| puts x }
end
# later...
user(&yielder)
$ ruby x.rb
x.rb:2:in `yielder': no block given (yield) (LocalJumpError)
from x.rb:12:in `<main>'
FWIW, in my real code, yielder and user are in different classes.
Update
Thanks for your answers. As Andrew Grimm mentioned, I want the iterator method to take parameters. My original example left this detail out. This snippet provides an iterator that counts up to a given number. To make it work, I made the inner block explicit. It does what I want, but it's a bit ugly. If anyone can improve on this I'd be very interested in seeing how.
def make_iter(upto)
def iter(upto, block)
(1 .. upto).each do |v|
block.call(v)
end
end
lambda { |block| iter(upto, block) }
end
def user(obj)
obj.call Proc.new { |x| puts x }
end
# later...
user(make_iter(3))
This doesn't use a lambda or unbound method, but it is the simplest way to go...
def f
yield 1
yield 2
end
def g x
send x do |n|
p n
end
end
g :f
When you write &yielder, you're calling yielder and then trying to apply the & (convert-to-Proc) operator on the result. Of course, calling yielder without a block is a no-go. What you want is to get a reference to the method itself. Just change that line to user(method :yielder) and it will work.
I think this might be along the lines of what you want to do:
def yielder
yield 1
yield 2
yield 3
end
def user(meth)
meth.call { |x| puts x }
end
# later...
user( Object.method(:yielder) )
Some related info here: http://blog.sidu.in/2007/11/ruby-blocks-gotchas.html
As it has been pointed out the baseline problem is that when you try to pass a function as a parameter Ruby executes it – as a side effect of parenthesis being optional.
I liked the simplicity of the symbol method that was mentioned before, but I would be afraid of my future self forgetting that one needs to pass the iterator as a symbol to make that work. Being readability a desired feature, you may then wrap your iterator into an object, which you can pass around without fear of having code unexpectedly executed.
Anonymous object as iterator
That is: using an anonymous object with just one fuction as iterator. Pretty immediate to read and understand. But due to the restrictions in the way Ruby handles scope the iterator cannot easily receive parameters: any parameters received in the function iterator are not automatically available within each.
def iterator
def each
yield("Value 1")
yield("Value 2")
yield("Value 3")
end
end
def iterate(my_iterator)
my_iterator.each do |value|
puts value
end
end
iterate iterator
Proc object as iterator
Using a Proc object as iterator lets you easily use any variables passed to the iterator constructor. The dark side: this starts looking weird. Reading the Proc.new block is not immediate for the untrained eye. Also: not being able to use yield makes it a bit uglier IMHO.
def iterator(prefix:)
Proc.new { |&block|
block.call("#{prefix} Value 1")
block.call("#{prefix} Value 2")
block.call("#{prefix} Value 3")
}
end
def iterate(my_iterator)
my_iterator.call do |value|
puts value
end
end
iterate iterator(prefix: 'The')
Lambda as iterator
Ideal if you want to obfuscate your code so hard that no one else besides you can read it.
def iterator(prefix:)
-> (&block) {
block.call("#{prefix} Value 1")
block.call("#{prefix} Value 2")
block.call("#{prefix} Value 3")
}
end
def iterate(my_iterator)
my_iterator.call do |value|
puts value
end
end
iterate iterator(prefix: 'The')
Class as iterator
And finally the good ol' OOP approach. A bit verbose to initialize for my taste, but with little or none surprise effect.
class Iterator
def initialize(prefix:)
#prefix = prefix
end
def each
yield("#{#prefix} Value 1")
yield("#{#prefix} Value 2")
yield("#{#prefix} Value 3")
end
end
def iterate(my_iterator)
my_iterator.each do |value|
puts value
end
end
iterate Iterator.new(prefix: 'The')
I'm trying to use Ruby 1.9.1 for an embedded scripting language, so that "end-user" code gets written in a Ruby block. One issue with this is that I'd like the users to be able to use the 'return' keyword in the blocks, so they don't need to worry about implicit return values. With this in mind, this is the kind of thing I'd like to be able to do:
def thing(*args, &block)
value = block.call
puts "value=#{value}"
end
thing {
return 6 * 7
}
If I use 'return' in the above example, I get a LocalJumpError. I'm aware that this is because the block in question is a Proc and not a lambda. The code works if I remove 'return', but I'd really prefer to be able to use 'return' in this scenario. Is this possible? I've tried converting the block to a lambda, but the result is the same.
Simply use next in this context:
$ irb
irb(main):001:0> def thing(*args, &block)
irb(main):002:1> value = block.call
irb(main):003:1> puts "value=#{value}"
irb(main):004:1> end
=> nil
irb(main):005:0>
irb(main):006:0* thing {
irb(main):007:1* return 6 * 7
irb(main):008:1> }
LocalJumpError: unexpected return
from (irb):7:in `block in irb_binding'
from (irb):2:in `call'
from (irb):2:in `thing'
from (irb):6
from /home/mirko/.rvm/rubies/ruby-1.9.1-p378/bin/irb:15:in `<main>'
irb(main):009:0> thing { break 6 * 7 }
=> 42
irb(main):011:0> thing { next 6 * 7 }
value=42
=> nil
return always returns from method, but if you test this snippet in irb you don't have method, that's why you have LocalJumpError
break returns value from block and ends its call. If your block was called by yield or .call, then break breaks from this iterator too
next returns value from block and ends its call. If your block was called by yield or .call, then next returns value to line where yield was called
You cannot do that in Ruby.
The return keyword always returns from the method or lambda in the current context. In blocks, it will return from the method in which the closure was defined. It cannot be made to return from the calling method or lambda.
The Rubyspec demonstrates that this is indeed the correct behaviour for Ruby (admittedly not a real implementation, but aims full compatibility with C Ruby):
describe "The return keyword" do
# ...
describe "within a block" do
# ...
it "causes the method that lexically encloses the block to return" do
# ...
it "returns from the lexically enclosing method even in case of chained calls" do
# ...
I admire the answer of s12chung. Here is my little improvement of his answer. It lets avoid cluttering the context with method __thing.
def thing(*args, &block)
o = Object.new
o.define_singleton_method(:__thing, block)
puts "value=#{o.__thing}"
end
thing { return 6 * 7 }
You are looking it from the wrong point of view.
This is an issue of thing, not the lambda.
def thing(*args, &block)
block.call.tap do |value|
puts "value=#{value}"
end
end
thing {
6 * 7
}
I had the same issue writing a DSL for a web framework in ruby... (the web framework Anorexic will rock!)...
anyway, I dug into the ruby internals and found a simple solution using the LocalJumpError returned when a Proc calls return... it runs well in the tests so far, but I'm not sure it's full-proof:
def thing(*args, &block)
if block
block_response = nil
begin
block_response = block.call
rescue Exception => e
if e.message == "unexpected return"
block_response = e.exit_value
else
raise e
end
end
puts "value=#{block_response}"
else
puts "no block given"
end
end
the if statement in the rescue segment could probably look something like this:
if e.is_a? LocalJumpError
but it's uncharted territory for me, so I'll stick to what I tested so far.
I found a way, but it involves defining a method as an intermediate step:
def thing(*args, &block)
define_method(:__thing, &block)
puts "value=#{__thing}"
end
thing { return 6 * 7 }
Where is thing invoked? Are you inside a class?
You may consider using something like this:
class MyThing
def ret b
#retval = b
end
def thing(*args, &block)
implicit = block.call
value = #retval || implicit
puts "value=#{value}"
end
def example1
thing do
ret 5 * 6
4
end
end
def example2
thing do
5 * 6
end
end
end
I believe this is the correct answer, despite the drawbacks:
def return_wrap(&block)
Thread.new { return yield }.join
rescue LocalJumpError => ex
ex.exit_value
end
def thing(*args, &block)
value = return_wrap(&block)
puts "value=#{value}"
end
thing {
return 6 * 7
}
This hack allows users to use return in their procs without consequences, self is preserved, etc.
The advantage of using Thread here is that in some cases you won't get the LocalJumpError - and the return will happen in the most unexpected place (onside a top-level method, unexpectedly skipping the rest of it's body).
The main disadvantage is the potential overhead (you can replace the Thread+join with just the yield if that's enough in your scenario).
class MyClass
def test
...
end
end
tmp = MyClass.new
tmp.test do |t|
"here"
end
Why am I getting the error
multiple values for a block parameter (0 for 1)
Here is a slightly longer example, based on your code:
class MyClass
def test
yield self
end
def my_own_puts s
puts s
end
end
tmp = MyClass.new
tmp.test do |t|
t.my_own_puts "here"
end
Running this code will output "here".
What is happening is there is a method test that can take a block of code, so you can call it with the do .. end syntax. Because it is passing an arg to yield, that arg is available to the block, so you give this to the block using the do |some_arg_name| ... end syntax.
The yield is where the block gets executed in the test method, and in this case I to yield I pass in self. Since the block now has access to self (an instance of MyClass), the block can call the my_own_puts method on it, and print out "here".
if test is defined with a yield statement, when that statement is reached and if there is a parameter on the yield statement, that parameter will be put into the block variable t. Thus if you have:
def test
.....
yield x
......
end
then x will be the value of t when yield is executed.
With your help, I was able to get the code working like this
class MyClass
def test
a = yield self
puts a
end
end
tmp = MyClass.new
tmp.test do |t|
"here"
end
Thanks, I had to tweak your code a bit but it works the way I wanted to now.
Passing a block to a function (as Bob shows in his answer) is overkill in this case. If you are reading in a string and printing it out, all you should need is something like:
class MyClass
def test(a)
puts a
end
end
tmp = MyClass.new
tmp.test("here")
Using a block might function correctly, but you are calling a lot of unnecessary code and making the true nature of your code unclear.
Proper block usage aside, let me address the particular error message you are seeing. When you say tmp.test do |t|, Ruby is expecting tmp.test to yield a single value which it will temporarily call t and pass to the block (think like the block is a function and you are passing it the argument your yield statement as a parameter). In your case, the method test method must not be yield-ing anything, thus the message "(0 for 1)" implies that it is seeing zero objects yielded when it is expecting to see one. I don't know what your code for test does, but check and make sure that test yields exactly one value.