How to add delta to python datetime.time?

How would this work? datetime.datetime.now().time() returns only hours, minutes, seconds and so on, there is no date information what .time() returns, only time.

Then, what should 18:00:00 + 8 hours return?
There's not answer to that question, and that's why you can't add a time and a timedelta.

In other words:

18:28:44, Sep. 16, 2012 + 8 hours #makes sense: it's 2:28:44, Sep. 17, 2012
18:28:44 + 8 hours # Doesn't make sense.

datetime.time objects do not support addition with datetime.timedeltas.

There is one natural definition though, clock arithmetic. You could compute it like this:

import datetime as dt
now = dt.datetime.now()
delta = dt.timedelta(hours = 12)
t = now.time()
print(t)
# 12:39:11.039864

print((dt.datetime.combine(dt.date(1,1,1),t) + delta).time())
# 00:39:11.039864

dt.datetime.combine(...) lifts the datetime.time t to a datetime.datetime object, the delta is then added, and the result is dropped back down to a datetime.time object.

All the solutions above are too complicated, OP had already shown that we can do calculation between datetime.datetime and datetime.timedelta, so why not just do:

(datetime.now() + timedelta(hours=12)).time()

Here is a function that adds a timedelta to a time:

def time_plus(time, timedelta):
    start = datetime.datetime(
        2000, 1, 1,
        hour=time.hour, minute=time.minute, second=time.second)
    end = start + timedelta
    return end.time()

This will provide the expected result so long as you don't add times in a way that crosses a midnight boundary.