有人知道递归使用生成器时send()函数的工作原理吗?我原以为值会传递到当前生成器,然后再传递到递归生成器...但似乎不是这样的?以下是一些示例代码:
def Walk(obj):
recurse = (yield obj)
if not recurse:
print 'stop recurse:', recurse
return
if isinstance(obj, list):
print 'is list:', obj
for item in obj:
print 'item loop:', item
walker = Walk(item)
for x in walker:
print 'item walk:', x
recurse = (yield x)
print 'item walk recurse:', recurse
walker.send(recurse)
root = ['a', ['b.0', ['b.0.0']]]
walker = Walk(root)
for i, x in enumerate(walker):
print i, x
print 'send true'
walker.send(True)
期望的输出应该是每个递归级别的每个值:
0 ['a', ['b.0', ['b.0.0']]]
1 'a'
2 ['b.0', ['b.0.0']]
3 'b.0'
4 ['b.0.0']
5 'b.0.0'
最终发生的情况是:
0 ['a', ['b.0', ['b.0.0']]]
send true
is list: ['a', ['b.0', ['b.0.0']]]
item loop: a
item walk: a
item walk recurse: None
stop recurse: None
似乎内部循环中的“recurse =(yield)”不等待发送值。或者说,不清楚内部循环“recurse”值如何变为“None”,因为调用者确实调用了“send()”。最终,目标基本上是递归地遍历树结构,但顶层调用者能够指定何时不递归进入子结构。例如:
walker = Walk(root)
for node in walker:
if CriteriaMet(node):
walker.send(True)
else:
walker.send(False)
(cdr(car(cdr(cdr(car(car lst)))))))
吗? - cwallenpoole