Using 'return' in a Ruby block

MetaFu picture MetaFu · Feb 24, 2010 · Viewed 58.9k times · Source

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.

Answer

MBO picture MBO · Feb 24, 2010

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