Python如何在内部管理“ for”循环? [英] How does Python manage a 'for' loop internally?
问题描述
我正在尝试学习Python,我开始玩一些代码:
I'm trying to learn Python, and I started to play with some code:
a = [3,4,5,6,7]
for b in a:
print(a)
a.pop(0)
输出为:
[3, 4, 5, 6, 7]
[4, 5, 6, 7]
[5, 6, 7]
我知道这不是 的好习惯,当我在它上循环时更改数据结构,但是我想了解Python如何在此管理迭代器
I know that's not a good practice change data structures while I'm looping on it, but I want to understand how Python manage the iterators in this case.
主要问题是:如果我更改 a的状态,它如何知道必须完成循环
?
The principal question is: How does it know that it has to finish the loop if I'm changing the state of a
?
推荐答案
kjaquier和Felix讨论了迭代器协议,我们可以在您的实践中看到它情况:
kjaquier and Felix have talked about the iterator protocol, and we can see it in action in your case:
>>> L = [1, 2, 3]
>>> iterator = iter(L)
>>> iterator
<list_iterator object at 0x101231f28>
>>> next(iterator)
1
>>> L.pop()
3
>>> L
[1, 2]
>>> next(iterator)
2
>>> next(iterator)
Traceback (most recent call last):
File "<input>", line 1, in <module>
StopIteration
由此我们可以推断出 list_iterator .__ next __
的代码行为类似于:
From this we can infer that list_iterator.__next__
has code that behaves something like:
if self.i < len(self.list):
return self.list[i]
raise StopIteration
它不能天真地得到物品。这将引发一个 IndexError
,它会冒泡到顶部:
It does not naively get the item. That would raise an IndexError
which would bubble to the top:
class FakeList(object):
def __iter__(self):
return self
def __next__(self):
raise IndexError
for i in FakeList(): # Raises `IndexError` immediately with a traceback and all
print(i)
实际上,查看
Indeed, looking at listiter_next
in the CPython source (thanks Brian Rodriguez):
if (it->it_index < PyList_GET_SIZE(seq)) {
item = PyList_GET_ITEM(seq, it->it_index);
++it->it_index;
Py_INCREF(item);
return item;
}
Py_DECREF(seq);
it->it_seq = NULL;
return NULL;
尽管我不知道如何返回NULL;
最终转换为 StopIteration
。
Although I don't know how return NULL;
eventually translates into a StopIteration
.
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