tokenize.py采用这种方法：调用者必须传一个 tokeneater 函数给 tokenize() ，当 tokenize() 找到下一个 token 时再调用。这使得 tokenize 能以自然的方式编码，但程序调用 tokenize 会变得极其复杂，因为它需要记住每次回调前最后出现的是哪个 token(s)。tabnanny.py中的 tokeneater 函数是处理得比较好的例子，它在全局变量中维护了一个状态机，用于记录已出现的 token 和预期会出现的 token 。这很难正确地工作，而且也挺难让人理解。不幸的是，它已经是最标准的解决方法了。def fib():
a, b = 0, 1
while 1:
yield b
a, b = b, a+byield_stmt：“yield”expression_list
future 声明【注释8】来进行引入：在早期版本中，若想使用生成器的模块，必须在接近头部处包含以下行（详见 PEP 236）：from __future__ import generators>>> def g():
... i = me.next()
... yield i
>>> me = g()
>>> me.next()
Traceback (most recent call last):
...
File "<string>", line 2, in g
ValueError: generator already executingreturn>>> def f1():
... try:
... return
... except:
... yield 1
>>> print list(f1())
[]>>> def f2():
... try:
... raise StopIteration
... except:
... yield 42
>>> print list(f2())
[42]>>> def f():
... return 1/0
>>> def g():
... yield f() # the zero division exception propagates
... yield 42 # and we'll never get here
>>> k = g()
>>> k.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in g
File "<stdin>", line 2, in f
ZeroDivisionError: integer division or modulo by zero
>>> k.next() # and the generator cannot be resumed
Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
>>>>>> def f():
... try:
... yield 1
... try:
... yield 2
... 1/0
... yield 3 # never get here
... except ZeroDivisionError:
... yield 4
... yield 5
... raise
... except:
... yield 6
... yield 7 # the "raise" above stops this
... except:
... yield 8
... yield 9
... try:
... x = 12
... finally:
... yield 10
... yield 11
>>> print list(f())
[1, 2, 4, 5, 8, 9, 10, 11]
>>># 二叉树类
class Tree:
def __init__(self, label, left=None, right=None):
self.label = label
self.left = left
self.right = right
def __repr__(self, level=0, indent=" "):
s = level*indent + `self.label`
if self.left:
s = s + "\n" + self.left.__repr__(level+1, indent)
if self.right:
s = s + "\n" + self.right.__repr__(level+1, indent)
return s
def __iter__(self):
return inorder(self)
# 从列表中创建 Tree
def tree(list):
n = len(list)
if n == 0:
return []
i = n / 2
return Tree(list[i], tree(list[:i]), tree(list[i+1:]))
# 递归生成器，按顺序生成树标签
def inorder(t):
if t:
for x in inorder(t.left):
yield x
yield t.label
for x in inorder(t.right):
yield x
# 展示：创建一棵树
t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
# 按顺序打印树的节点
for x in t:
print x,
print
# 非递归生成器
def inorder(node):
stack = []
while node:
while node.left:
stack.append(node)
node = node.left
yield node.label
while not node.right:
try:
node = stack.pop()
except IndexError:
return
yield node.label
node = node.right
# 练习非递归生成器
for x in t:
print x,
print
Both output blocks display:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Zreturn 3 and continue
return and continue 3
return generating 3
continue return 3
return >> , 3
from generator return 3
return >> 3
return << 3
>> 3
<< 3
* 3return 并不总是等同于 try-except 结构中的 raise StopIteration（参见“设计规格：Return”部分）。return expr 意味着“我已经完成”和“但我还有最后一个有用的值可以返回，这就是它”。 在初始阶段，不强制使用return expr的情况下，使用 yield 仅仅传递值，这很简单明了。