Cleanest and most Pythonic way to get tomorrow's date?

What is the cleanest and most Pythonic way to get tomorrow's date? There must be a better way than to add one to the day, handle days at the end of the month, etc.

datetime.date.today() + datetime.timedelta(days=1) should do the trick

timedelta can handle adding days, seconds, microseconds, milliseconds, minutes, hours, or weeks.

>>> import datetime
>>> today = datetime.date.today()
>>> today
datetime.date(2009, 10, 1)
>>> today + datetime.timedelta(days=1)
datetime.date(2009, 10, 2)
>>> datetime.date(2009,10,31) + datetime.timedelta(hours=24)
datetime.date(2009, 11, 1)

As asked in a comment, leap days pose no problem:

>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=1)
datetime.date(2004, 2, 29)
>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=2)
datetime.date(2004, 3, 1)
>>> datetime.date(2005, 2, 28) + datetime.timedelta(days=1)
datetime.date(2005, 3, 1)

No handling of leap seconds tho:

>>> from datetime import datetime, timedelta
>>> dt = datetime(2008,12,31,23,59,59)
>>> str(dt)
'2008-12-31 23:59:59'
>>> # leap second was added at the end of 2008, 
>>> # adding one second should create a datetime
>>> # of '2008-12-31 23:59:60'
>>> str(dt+timedelta(0,1))
'2009-01-01 00:00:00'
>>> str(dt+timedelta(0,2))
'2009-01-01 00:00:01'

darn.

EDIT - @Mark: The docs say "yes", but the code says "not so much":

>>> time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")
(2008, 12, 31, 23, 59, 60, 2, 366, -1)
>>> time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))
1230789600.0
>>> time.gmtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")))
(2009, 1, 1, 6, 0, 0, 3, 1, 0)
>>> time.localtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")))
(2009, 1, 1, 0, 0, 0, 3, 1, 0)

I would think that gmtime or localtime would take the value returned by mktime and given me back the original tuple, with 60 as the number of seconds. And this test shows that these leap seconds can just fade away...

>>> a = time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))
>>> b = time.mktime(time.strptime("2009-01-01 00:00:00","%Y-%m-%d %H:%M:%S"))
>>> a,b
(1230789600.0, 1230789600.0)
>>> b-a
0.0

Even the basic time module can handle this:

import time
time.localtime(time.time() + 24*3600)

For people who are dealing with servers Time Stamp

To get yesterday Time Stamp:

yesterdaytimestamp = datetime.datetime.today() + datetime.timedelta(days=-1)

To get Today Time Stamp:

currenttimestamp = datetime.datetime.now().timestamp()

To get Tomorrow Time Stamp:

tomorrowtimestamp = datetime.datetime.today() + datetime.timedelta(days=1)

To print:

print('\n Yesterday TimeStamp is : ', yesterdaytimestamp.timestamp(), 
    '\n Today TimeStamp is :', currenttimestamp, 
    '\n Tomorrow TimeStamp is: ', tomorrowtimestamp.timestamp())

The output:

Yesterday TimeStamp is :  1632842904.110993 
Today TimeStamp is       :  1632929304.111022 
Tomorrow TimeStamp is    :  1633015704.11103

Cleanest and most Pythonic way to get tomorrow's date?

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