blocks|key|1373730|text|不，没有这样的方法。迭代结束时会出现一个异常。请参阅documentation。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1373731|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#iterator-types^0|Q|D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

No, there is no such method. The end of iteration is indicated by an exception. See the <a href="http://docs.python.org/library/stdtypes.html#iterator-types" rel="noreferrer">documentation</a>.

blocks|key|1373932|text|通过使用next(iterator,+default_value)，有一种替代StopIteration的方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1373933|对于exapmle：|1373934|>>>+a+=+iter('hi')
>>>+print+next(a,+None)
h
>>>+print+next(a,+None)
i
>>>+print+next(a,+None)
None|code-block|syntax|javascript|1373935|因此，如果您不想使用异常方法，可以检测for+None或其他预先指定的迭代器末尾的值。|1373936|entityMap^0|4|T|13|D|0|0|0|N|4|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

There's an alternative to the <code>StopIteration</code> by using <code>next(iterator, default_value)</code>.

For exapmle:

<pre><code>&gt;&gt;&gt; a = iter('hi')
&gt;&gt;&gt; print next(a, None)
h
&gt;&gt;&gt; print next(a, None)
i
&gt;&gt;&gt; print next(a, None)
None
</code></pre>

So you can detect for <code>None</code> or other pre-specified value for end of the iterator if you don't want the exception way.

blocks|key|1158559|text|除了提到StopIteration之外，Python+"for“循环只做您想要做的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1158560|>>>+it+=+iter("hello")
>>>+for+i+in+it:
...+++++print+i
...
h
e
l
l
o|code-block|syntax|javascript|1158561|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In addition to all the mentions of StopIteration, the Python "for" loop simply does what you want:

<pre><code>&gt;&gt;&gt; it = iter("hello")
&gt;&gt;&gt; for i in it:
... print i
...
h
e
l
l
o
</code></pre>

blocks|key|1158645|text|尝试来自任何迭代器对象的__length_hint__()方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1158646|iter(...).__length_hint__()+>+0|code-block|syntax|javascript|1158647|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try the __length_hint__() method from any iterator object:

<pre><code>iter(...).__length_hint__() &gt; 0
</code></pre>

blocks|key|1158612|text|可以使用、itertools.tee检查迭代器，并在teed迭代器上检查StopIteration。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1158613|entityMap^0|5|D|10|D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

You can <code>tee</code> the iterator using, <code>itertools.tee</code>, and check for <code>StopIteration</code> on the teed iterator.

blocks|key|1373640|text|不是的。最相似的概念很可能是StopIteration+exception.|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1373641|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/exceptions.html#exceptions.StopIteration^0|E|O|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

No. The most similar concept is most likely a <a href="http://docs.python.org/library/exceptions.html#exceptions.StopIteration" rel="nofollow noreferrer">StopIteration exception.</a>

blocks|key|1158479|text|我相信python只是有了next()，根据文档，它抛出了一个异常，因为没有更多的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1158480|http://docs.python.org/library/stdtypes.html#iterator-types|offset|length|1158481|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1N|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

I believe python just has next() and according to the doc, it throws an exception is there are no more elements.

<a href="http://docs.python.org/library/stdtypes.html#iterator-types" rel="nofollow noreferrer">http://docs.python.org/library/stdtypes.html#iterator-types</a>

blocks|key|1158797|text|导致我搜索它的用例如下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1158798|def+setfrom(self,f):
++++"""Set+from+iterable+f"""
++++fi+=+iter(f)
++++for+i+in+range(self.n):
++++++++try:
++++++++++++x+=+next(fi)
++++++++except+StopIteration:
++++++++++++fi+=+iter(f)
++++++++++++x+=+next(fi)
++++++++self.a[i]+=+x+|code-block|syntax|javascript|1158799|在可以使用hasnext()的地方，可以这样做|1158800|def+setfrom(self,f):
++++"""Set+from+iterable+f"""
++++fi+=+iter(f)
++++for+i+in+range(self.n):
++++++++if+not+hasnext(fi):
++++++++++++fi+=+iter(f)+#+restart
++++++++self.a[i]+=+next(fi)|1158801|对我来说更干净。显然，您可以通过定义实用程序类来解决问题，但接下来会发生的情况是，您有20多种几乎相同的变通方法，每个变通方法都有各自的怪癖，如果您希望重用使用不同变通方法的代码，则必须在单个应用程序中使用多个近乎相同的变通方法，或者选择并重写代码以使用相同的方法。“做一次就做好”这句格言失败了。|1158802|此外，迭代器本身需要有一个内部的“hasnext”检查来运行，以查看它是否需要引发异常。然后隐藏此内部检查，以便需要通过尝试获取项、捕获异常并在抛出时运行处理程序来对其进行测试。这是不必要的隐藏IMO。|1158803|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The use case that lead me to search for this is the following

<pre><code>def setfrom(self,f):
 """Set from iterable f"""
 fi = iter(f)
 for i in range(self.n):
 try:
 x = next(fi)
 except StopIteration:
 fi = iter(f)
 x = next(fi)
 self.a[i] = x 
</code></pre>

where hasnext() is available, one could do

<pre><code>def setfrom(self,f):
 """Set from iterable f"""
 fi = iter(f)
 for i in range(self.n):
 if not hasnext(fi):
 fi = iter(f) # restart
 self.a[i] = next(fi)
</code></pre>

which to me is cleaner. Obviously you can work around issues by defining utility classes, but what then happens is you have a proliferation of twenty-odd different almost-equivalent workarounds each with their quirks, and if you wish to reuse code that uses different workarounds, you have to either have multiple near-equivalent in your single application, or go around picking through and rewriting code to use the same approach. The 'do it once and do it well' maxim fails badly.

Furthermore, the iterator itself needs to have an internal 'hasnext' check to run to see if it needs to raise an exception. This internal check is then hidden so that it needs to be tested by trying to get an item, catching the exception and running the handler if thrown. This is unnecessary hiding IMO.

blocks|key|1374293|text|还可以实现一个辅助生成器，它包装任何迭代器，并回答问题，如果它有下一个值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1374294|Try+it+online!|offset|length|1374295|def+has_next(it):
++++first+=+True
++++for+e+in+it:
++++++++if+not+first:
++++++++++++yield+True,+prev
++++++++else:
++++++++++++first+=+False
++++++++prev+=+e
++++if+not+first:
++++++++yield+False,+prev

for+has_next_,+e+in+has_next(range(4)):
++++print(has_next_,+e)|code-block|syntax|javascript|1374296|以下哪项输出：|1374297|True+0
True+1
True+2
False+3|1374298|这种方法的主要缺点可能也是唯一的缺点是，它会预读更多的元素，对于大多数任务，它是完全可以的，但对于某些任务，它可能是不允许的，特别是如果has_next()的用户不知道这个预读逻辑，并且可能会错过它。|style|CODE|1374299|上面的代码也适用于无限迭代器。|1374300|实际上，对于我曾经编程过的所有情况，这样的has_next()是完全足够的，并且不会造成任何问题，事实上，它非常有用。你只需要知道它的预读逻辑。|1374301|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##bU/LCsIwELz3K/aYQG96Erz6Bd5LoBu7UJKQrGK/Pk26bVRwjvNgZsLCk3ennEe0MJk0OHyzItaXDgosxcRwhXt8ohA@AgI5IBZHBVlwnsX8YSsWwnnc0j2EiK8m4pzw13pU3UyRmlJDhRTif490bLG9pKsrjzNDL4PbuWjcA9VZ7xdDJMfq2627nFc^0|0|0|E|0|0|0|0|0|1W|A|0|0|L|A|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|F|3|G|5|H|7|19|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|1B|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|1C|8|@$D|1D|E|1E|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1G|8|@$D|1H|E|1I|Q|R]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

It is also possible to implement a helper generator that wraps any iterator and answers question if it has next value:
<a href="https://tio.run/##bU/LCsIwELz3K/aYQG96Erz6Bd5LoBu7UJKQrGK/Pk26bVRwjvNgZsLCk3ennEe0MJk0OHyzItaXDgosxcRwhXt8ohA@AgI5IBZHBVlwnsX8YSsWwnnc0j2EiK8m4pzw13pU3UyRmlJDhRTif490bLG9pKsrjzNDL4PbuWjcA9VZ7xdDJMfq2627nFc" rel="nofollow noreferrer" title="Python 3 – Try It Online">Try it online!</a>
<pre><code>def has_next(it):
 first = True
 for e in it:
 if not first:
 yield True, prev
 else:
 first = False
 prev = e
 if not first:
 yield False, prev

for has_next_, e in has_next(range(4)):
 print(has_next_, e)
</code></pre>
Which outputs:
<pre><code>True 0
True 1
True 2
False 3
</code></pre>
The main and probably only drawback of this method is that it reads ahead one more element, for most of tasks it is totally alright, but for some tasks it may be disallowed, especially if user of <code>has_next()</code> is not aware of this read-ahead logic and may missuse it.
Code above works for infinite iterators too.
Actually for all cases that I ever programmed such kind of <code>has_next()</code> was totally enough and didn't cause any problems and in fact was very helpful. You just have to be aware of its read-ahead logic.

blocks|key|1374344|text|也许只有我这样认为，但虽然我喜欢https://stackoverflow.com/users/95810/alex-martelli的答案，但我发现这个更容易读懂：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1374345|from+collections.abc+import+Iterator++#+since+python+3.3+Iterator+is+here

class+MyIterator(Iterator):++#+need+to+subclass+Iterator+rather+than+object
++def+__init__(self,+it):
++++self._iter+=+iter(it)
++++self._sentinel+=+object()
++++self._next+=+next(self._iter,+self._sentinel)
++++
++def+__iter__(self):+
++++return+self
++
++def+__next__(self):++++++++#+__next__+vs+next+in+python+2
++++if+not+self.has_next():
++++++next(self._iter)++#+raises+StopIteration

++++val+=+self._next
++++self._next+=+next(self._iter,+self._sentinel)
++++return+val
++
++def+has_next(self):
++++return+self._next+is+not+self._sentinel|code-block|syntax|javascript|1374346|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/95810/alex-martelli^0|G|1F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Maybe it's just me, but while I like <a href="https://stackoverflow.com/users/95810/alex-martelli">https://stackoverflow.com/users/95810/alex-martelli</a> 's answer, I find this a bit easier to read:
<pre><code>from collections.abc import Iterator # since python 3.3 Iterator is here

class MyIterator(Iterator): # need to subclass Iterator rather than object
 def __init__(self, it):
 self._iter = iter(it)
 self._sentinel = object()
 self._next = next(self._iter, self._sentinel)
 
 def __iter__(self): 
 return self
 
 def __next__(self): # __next__ vs next in python 2
 if not self.has_next():
 next(self._iter) # raises StopIteration

 val = self._next
 self._next = next(self._iter, self._sentinel)
 return val
 
 def has_next(self):
 return self._next is not self._sentinel
</code></pre>

blocks|key|1374384|text|这种基于处理"StopIteration“执行的方法已经解决了这个问题，为了读取所有迭代，这个方法非常简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1374385|++++end_cursor+=+False
++++while+not+end_cursor:
++++++++try:
++++++++++++print(cursor.next())
++++++++except+StopIteration:
++++++++++++print('end+loop')
++++++++++++end_cursor+=+True
++++++++except:
++++++++++++print('other+exceptions+to+manage')
++++++++++++end_cursor+=+True|code-block|syntax|javascript|1374386|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The way has solved it based on handling the &quot;StopIteration&quot; execption is pretty straightforward in order to read all iterations :
<pre><code> end_cursor = False
 while not end_cursor:
 try:
 print(cursor.next())
 except StopIteration:
 print('end loop')
 end_cursor = True
 except:
 print('other exceptions to manage')
 end_cursor = True
</code></pre>

blocks|key|1374433|text|我认为当你想要某种has_next功能时，有一些有效的用例，在这种情况下，你应该用定义的has_next来装饰迭代器。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1374434|结合这个问题的答案，这里是我的实现，对我来说，这是一个很好的简明解决方案(python+3.9)：|1374435|_EMPTY_BUF+=+object()


class+BufferedIterator(Iterator[_T]):
++++def+__init__(self,+real_it:+Iterator[_T]):
++++++++self._real_it+=+real_it
++++++++self._buf+=+next(self._real_it,+_EMPTY_BUF)

++++def+has_next(self):
++++++++return+self._buf+is+not+_EMPTY_BUF

++++def+__next__(self)+->+_T_co:
++++++++v+=+self._buf
++++++++self._buf+=+next(self._real_it,+_EMPTY_BUF)
++++++++if+v+is+_EMPTY_BUF:
++++++++++++raise+StopIteration()
++++++++return+v|code-block|syntax|javascript|1374436|主要区别在于has_next只是一个布尔表达式，并且还处理具有None值的迭代器。|1374437|将其添加到带有测试和示例用法的gist+here中。|1374438|entityMap|0|LINK|mutability|MUTABLE|url|https://gist.github.com/tkutcher/d00ac4ec90b21bdc1fe02b8c4b910e99^0|9|8|18|8|0|0|0|6|8|V|4|0|K|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

I think there are valid use cases for when you may want some sort of <code>has_next</code> functionality, in which case you should decorate an iterator with a <code>has_next</code> defined.
Combining concepts from the answers to this question here is my implementation of that which feels like a nice concise solution to me (python 3.9):
<pre class="lang-py prettyprint-override"><code>_EMPTY_BUF = object()


class BufferedIterator(Iterator[_T]):
 def __init__(self, real_it: Iterator[_T]):
 self._real_it = real_it
 self._buf = next(self._real_it, _EMPTY_BUF)

 def has_next(self):
 return self._buf is not _EMPTY_BUF

 def __next__(self) -&gt; _T_co:
 v = self._buf
 self._buf = next(self._real_it, _EMPTY_BUF)
 if v is _EMPTY_BUF:
 raise StopIteration()
 return v

</code></pre>
The main difference is that <code>has_next</code> is just a boolean expression, and also handles iterators with <code>None</code> values.
Added this to a gist <a href="https://gist.github.com/tkutcher/d00ac4ec90b21bdc1fe02b8c4b910e99" rel="nofollow noreferrer">here</a> with tests and example usage.

blocks|key|1374187|text|非常有趣的问题，但是这个"hasnext“设计已经被放入leetcode：https://leetcode.com/problems/iterator-for-combination/|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1374188|下面是我的实现：|1374189|class+CombinationIterator:

def+__init__(self,+characters:+str,+combinationLength:+int):
++++from+itertools+import+combinations
++++from+collections+import+deque
++++self.iter+=+combinations(characters,+combinationLength)
++++self.res+=+deque()


def+next(self)+->+str:
++++if+len(self.res)+==+0:
++++++++return+''.join(next(self.iter))
++++else:
++++++++return+''.join(self.res.pop())


def+hasNext(self)+->+bool:
++++try:
++++++++self.res.insert(0,+next(self.iter))
++++++++return+True
++++except:
++++++++return+len(self.res)+>+0|code-block|syntax|javascript|1374190|entityMap|0|LINK|mutability|MUTABLE|url|https://leetcode.com/problems/iterator-for-combination/^0|11|1J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

very interesting question, but this &quot;hasnext&quot; design had been put into leetcode:
<a href="https://leetcode.com/problems/iterator-for-combination/" rel="nofollow noreferrer">https://leetcode.com/problems/iterator-for-combination/</a>
here is my implementation:
<pre><code>class CombinationIterator:

def __init__(self, characters: str, combinationLength: int):
 from itertools import combinations
 from collections import deque
 self.iter = combinations(characters, combinationLength)
 self.res = deque()


def next(self) -&gt; str:
 if len(self.res) == 0:
 return ''.join(next(self.iter))
 else:
 return ''.join(self.res.pop())


def hasNext(self) -&gt; bool:
 try:
 self.res.insert(0, next(self.iter))
 return True
 except:
 return len(self.res) &gt; 0
</code></pre>

blocks|key|1158856|text|到目前为止，我解决问题的方法是保持迭代的对象数量的计数。我想使用对一个实例方法的调用来迭代一个集合。因为我知道集合的长度，以及到目前为止计算的项目数，所以我实际上有了一个hasNext方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1158857|我的代码的一个简单版本：|1158858|class+Iterator:
++++#+s+is+a+string,+say
++++def+__init__(self,+s):
++++++++self.s+=+set(list(s))
++++++++self.done+=+False
++++++++self.iter+=+iter(s)
++++++++self.charCount+=+0

++++def+next(self):
++++++++if+self.done:
++++++++++++return+None
++++++++self.char+=+next(self.iter)
++++++++self.charCount+%2B=+1
++++++++self.done+=+(self.charCount+<+len(self.s))
++++++++return+self.char

++++def+hasMore(self):
++++++++return+not+self.done|code-block|syntax|javascript|1158859|当然，该示例是一个玩具示例，但您可以理解其中的含义。这在无法获得迭代器长度的情况下不起作用，比如生成器等。|1158860|entityMap^0|2D|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

The way I solved my problem is to keep the count of the number of objects iterated over, so far. I wanted to iterate over a set using calls to an instance method. Since I knew the length of the set, and the number of items counted so far, I effectively had an <code>hasNext</code> method.

A simple version of my code:

<pre><code>class Iterator:
 # s is a string, say
 def __init__(self, s):
 self.s = set(list(s))
 self.done = False
 self.iter = iter(s)
 self.charCount = 0

 def next(self):
 if self.done:
 return None
 self.char = next(self.iter)
 self.charCount += 1
 self.done = (self.charCount &lt; len(self.s))
 return self.char

 def hasMore(self):
 return not self.done
</code></pre>

Of course, the example is a toy one, but you get the idea. This won't work in cases where there is no way to get the length of the iterable, like a generator etc.

问Python迭代器中的hasNext？
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