Question

I have the following Ruby code:

# func1 generates a sequence of items derived from x
# func2 does something with the items generated by func1
def test(x, func1, func2)
    func1.call(x) do | y |
        func2.call(y)
    end
end

func1 = lambda do | x |
    for i in 1 .. 5
        yield x * i
    end
end

func2 = lambda do | y |
    puts y
end


test(2, func1, func2) # Should print '2', '4', '6', '8', and '10'

This does not work, of course.

test.rb:11: no block given (LocalJumpError)
    from test.rb:10:in `each'
    from test.rb:10
    from test.rb:4:in `call'
    from test.rb:4:in `test'
    from test.rb:20

Answer 1

Lambdas don’t implicitly accept blocks like regular methods do, so your func1 can’t yield. Do this instead:

func1 = lambda do |x, &blk|
  for i in 1 .. 5
    blk.call(x * i)
  end
end

Specifically, I believe this is because yield would send control back to the caller‘s block, which would not include lambda invocations. So the following code works like you “expect”:

def foo
  (lambda { |n| yield(n) }).call(5)
end
foo { |f| puts f }  # prints 5