Does itertools.product evaluate its arguments lazily?

Question

The following never prints anything in Python 3.6

from itertools import product, count

for f in product(count(), [1,2]): 
    print(f)

Ins

Answer 1

The issue seems to be that product never returns an iterator

No, product is already "lazy".

The issue is thatcount() counts to infinity. From count's docs:

Equivalent to:

def count(start=0, step=1):
    # count(10) --> 10 11 12 13 14 ...
    # count(2.5, 0.5) -> 2.5 3.0 3.5 ...
    n = start
    while True:
        yield n
        n += step

You code is basically the same as doing:

def count():
    i = 0
    while True:
        yield i
        i += 1

count()