[问题] 请问generator的send方法

楼主: Tomcat (我想我疯了)   2018-01-19 19:23:42
小弟正在学习PYTHON,看到generator时,下面这支是看得懂...
gen.send.preparation.stop.py
stop = False
def counter(start=0):
n = start
while not stop:
yield n
n += 1
c = counter()
print(next(c)) # prints: 0
print(next(c)) # prints: 1
stop = True
print(next(c)) # raises StopIteration
不过到了下面这支后就卡关了...
我卡关的点在于... 为什么执行# C步骤时,只有印出0;
而执行# D时却是印出:
<class 'str'> Wow!
1
为什么不是先印出1呢? 明明result = yield n 就是在前面啊...
如果按照这逻辑的话,步骤# F应该也要先印出3才对...
可是却先执行# B 印出<class 'str'> Q 后产生StopIteration
麻烦各位大大赐教了,感谢!!
gen.send.py
def counter(start=0):
n = start
while True:
result = yield n # A
print(type(result), result) # B
if result == 'Q':
break
n += 1
c = counter()
print(next(c)) # C
print(c.send('Wow!')) # D
print(next(c)) # E
print(c.send('Q')) # F
执行结果:
$ python gen.send.py
0
<class 'str'> Wow!
1
<class 'NoneType'> None
2
<class 'str'> Q
Traceback (most recent call last):
File "gen.send.py", line 14, in <module>
print(c.send('Q')) # F
StopIteration
作者: seLain (建筑的永恒之道)   2018-01-19 20:23:00
C->A->C(0)->D->A->B(Wow)->A->D(1)->....another example: https://goo.gl/5VVhSZ
作者: zerof (猫橘毛发呆雕像)   2018-01-19 22:34:00
next(g) == g.send(None)yield 右边的值会先出来,左值会在下次 send 的时候 assign& 这个其实是 coroutine 了,见 PEP342

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