在块/ lambda中产生麻烦

Question

我有以下Ruby代码:

# 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'

当然,这不起作用.

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不会像常规方法那样隐式接受块,所以你func1不能屈服.改为:

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

具体来说,我认为这是因为yield会将控制权发送回caller块,而不包括lambda调用.所以下面的代码就像你"期待"一样:

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