What is the current value of a Python itertools counter

Another hack to get next value without advancing iterator is to abuse copy protocol:

>>> c = itertools.count()
>>> c.__reduce__()[1][0]
0
>>> next(c)
0
>>> c.__reduce__()[1][0]
1

Or just take it from object copy:

>>> from copy import copy
>>> next(copy(c))
1

Use the source, Luke!

According to module implementation, it's not possible.

typedef struct {
    PyObject_HEAD
    Py_ssize_t cnt;
    PyObject *long_cnt;
    PyObject *long_step;
} countobject;

Current state is stored in cnt and long_cnt members, and neither of them is exposed in object API. Only place where it may be retrieved is object __repr__, as you suggested.

Note that while parsing string you have to consider a non-singular increment case. repr(itertools.count(123, 4)) is equal to 'count(123, 4)' - logic suggested by you in question would fail in that case.

According to the documentation there is no way to access the current value of the function. itertools.count() is a generator method from the itertools module. As such, it is common practice to just simply assign the value of a generator's current value to a variable.

Simply store the the result of the next call:

current_value = x.next()

or ( Built-in python method for Python version ≥ 2.6 )

current_value = next(x)

You could make a wrapper function, or a utility decorator class if you would like some added syntactic sugar, but assignment is standard.