Does enumerate() produce a generator object?

While the Python documentation says that enumerate is functionally equivalent to:

def enumerate(sequence, start=0):
    n = start
    for elem in sequence:
        yield n, elem
        n += 1

The real enumerate function returns an iterator, but not an actual generator. You can see this if you call help(x) after doing creating an enumerate object:

>>> x = enumerate([1,2])
>>> help(x)
class enumerate(object)
 |  enumerate(iterable[, start]) -> iterator for index, value of iterable
 |  
 |  Return an enumerate object.  iterable must be another object that supports
 |  iteration.  The enumerate object yields pairs containing a count (from
 |  start, which defaults to zero) and a value yielded by the iterable argument.
 |  enumerate is useful for obtaining an indexed list:
 |      (0, seq[0]), (1, seq[1]), (2, seq[2]), ...
 |  
 |  Methods defined here:
 |  
 |  __getattribute__(...)
 |      x.__getattribute__('name') <==> x.name
 |  
 |  __iter__(...)
 |      x.__iter__() <==> iter(x)
 |  
 |  next(...)
 |      x.next() -> the next value, or raise StopIteration
 |  
 |  ----------------------------------------------------------------------
 |  Data and other attributes defined here:
 |  
 |  __new__ = <built-in method __new__ of type object>
 |      T.__new__(S, ...) -> a new object with type S, a subtype of T

In Python, generators are basically a specific type of iterator that’s implemented by using a yield to return data from a function. However, enumerate is actually implemented in C, not pure Python, so there’s no yield involved. You can find the source here: http://hg.python.org/cpython/file/2.7/Objects/enumobject.c