How does Python manage a 'for' loop internally?

Question

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)
<         


        

Answer 1

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

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

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)

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;

Although I don't know how return NULL; eventually translates into a StopIteration.

Answer 2

We can easily see the sequence of events by using a little helper function foo:

def foo():
    for i in l:
        l.pop()

and dis.dis(foo) to see the Python byte-code generated. Snipping away the not-so-relevant opcodes, your loop does the following:

          2 LOAD_GLOBAL              0 (l)
          4 GET_ITER
    >>    6 FOR_ITER                12 (to 20)
          8 STORE_FAST               0 (i)

         10 LOAD_GLOBAL              0 (l)
         12 LOAD_ATTR                1 (pop)
         14 CALL_FUNCTION            0
         16 POP_TOP
         18 JUMP_ABSOLUTE            6

That is, it get's the iter for the given object (iter(l) a specialized iterator object for lists) and loops until FOR_ITER signals that it's time to stop. Adding the juicy parts, here's what FOR_ITER does:

PyObject *next = (*iter->ob_type->tp_iternext)(iter);

which essentially is:

list_iterator.__next__()

this (finally^*) goes through to listiter_next which performs the index check as @Alex using the original sequence l during the check.

if (it->it_index < PyList_GET_SIZE(seq))

when this fails, NULL is returned which signals that the iteration has finished. In the meantime a StopIteration exception is set which is silently suppressed in the FOR_ITER op-code code:

if (!PyErr_ExceptionMatches(PyExc_StopIteration))
    goto error;
else if (tstate->c_tracefunc != NULL)
    call_exc_trace(tstate->c_tracefunc, tstate->c_traceobj, tstate, f);
PyErr_Clear();  /* My comment: Suppress it! */

so whether you change the list or not, the check in listiter_next will ultimately fail and do the same thing.

^{*For anyone wondering, listiter_next is a descriptor so there's a little function wrapping it. In this specific case, that function is wrap_next which makes sure to set PyExc_StopIteration as an exception when listiter_next returns NULL.}

Answer 3

The reason why you shouldn't do that is precisely so you don't have to rely on how the iteration is implemented.

But back to the question. Lists in Python are array lists. They represent a continuous chunk of allocated memory, as opposed to linked lists in which each element in allocated independently. Thus, Python's lists, like arrays in C, are optimized for random access. In other words, the most efficient way to get from element n to element n+1 is by accessing to the element n+1 directly (by calling mylist.__getitem__(n+1) or mylist[n+1]).

So, the implementation of __next__ (the method called on each iteration) for lists is just like you would expect: the index of the current element is first set at 0 and then increased after each iteration.

In your code, if you also print b, you will see that happening:

a = [3,4,5,6,7]
for b in a:
    print a, b
    a.pop(0)

Result :

[3, 4, 5, 6, 7] 3
[4, 5, 6, 7] 5
[5, 6, 7] 7

Because :

• At iteration 0, a[0] == 3.
• At iteration 1, a[1] == 5.
• At iteration 2, a[2] == 7.
• At iteration 3, the loop is over (len(a) < 3)

Answer 4

AFAIK, the for loop uses the iterator protocol. You can manually create and use the iterator as follows:

In [16]: a = [3,4,5,6,7]
    ...: it = iter(a)
    ...: while(True):
    ...:     b = next(it)
    ...:     print(b)
    ...:     print(a)
    ...:     a.pop(0)
    ...:
3
[3, 4, 5, 6, 7]
5
[4, 5, 6, 7]
7
[5, 6, 7]
---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-16-116cdcc742c1> in <module>()
      2 it = iter(a)
      3 while(True):
----> 4     b = next(it)
      5     print(b)
      6     print(a)

The for loop stops if the iterator is exhausted (raises StopIteration).