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

Lambda不会像常规方法那样隐式接受块，因此func1无法屈服。这样做：

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的块，这不包括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

Answer 2

仅限Ruby 1.9：

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

Answer 3

def test(x, func1, func2)
    func1.call(x) do | y |
        func2.call(y)
    end
end

#change func1 to a method
def func1 x
    for i in 1 .. 5
        yield x * i
    end
end

#func2 may be either a method or a lambda
#I changed it for consistency, but you don't have to
def func2 y
    puts y
end


test(2, method(:func1), method(:func2))

Answer 4

基于Nikita Misharin在这里的答案：[https://stackoverflow.com/a/45571976/2165560]，我喜欢这个：

def iterator(x)
  for i in 1 .. 5
    yield x * i
  end
end


iteratorWrapper = -> (m,&block) { iterator(m) {|n| block.call n}  }
iteratorWrapper.call(2) { |y| puts y }

它在这里回答了我的问题[In Ruby, can you use the lambda or or Proc call method to invoke an iterator?。

通过包装迭代器，它可以任意传递给其他方法并迭代它们的块。

在块/ lambda中产生麻烦

4 个答案: