- 论坛徽章:
- 2
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class anyiter():
def __init__(self,data,safe=False):
self.safe=safe
self.iter=iter(data)
def __iter__(self):
return self
def next(self,howmany=1):
retval=[]
for eachitem in range(howmany):
try:
retval.append(self.iter.next())
except StopIteration:
if self.safe:
break
else:
raise
return retval
>>> a=anyiter(range(10))
>>> i=iter(a)
>>> i.next(1)
对这个代码有几个疑问:
1 i.next(1)这个用法是什么意思,同样用iter方法:提示如下错误
>>> tuple=(1,2,3)
>>> i=iter(tuple)
>>> i.next(2)
Traceback (most recent call last):
File "<pyshell#54>", line 1, in <module>
i.next(2)
TypeError: expected 0 arguments, got 1
2 对这段代码不明白什么意思,如果howmay为1的话,for循环只能执行一次,那么retval.append如何生效呢
for eachitem in range(howmany):
try:
retval.append(self.iter.next())
except StopIteration:
if self.safe:
break
else:
raise
return retval
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