Using 'return' in a Ruby block

user_280258 0
Viewed 70269

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.

proc-object return lambda ruby
8 Answers

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
user_2192170

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
# ...
user_980380

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
}
user_1235270

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 }
user_132853880

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
user_1508720
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