def get_year():
print("Please input year:")
year = int(input())
return year
def if_else(year):
if (year % 4 == 0):
print("It's a leap year")
elif (year % 400 == 0):
print("It's a leap year")
elif (year % 100 == 0):
print("It's not a leap year")
else:
print("There is an error")
def main():
years = get_year()
if_else(years)

when I input 2400(not a leap year) it says that its a leap year :( Please help

Please, wrap your code in code tags.

OK, this Larz60+ added code tags for you. Anyway, in the future always wrap your code in code tags.

2400 IS a leap year - it's exactly divisible by 400.

Quote:Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400. For example, the years 1700, 1800, and 1900 were not leap years, but the years 1600 and 2000 were.

Hello!

Put print() in each loop to see where it returns True.

For the year 2400, this will happen in the first loop because 2400 % 4 == 0 is True and will print that this year leap.

Change the conditions

def get_year():
print("Please input year:")
year = int(input())
return year
def if_else(year):
if year % 4 == 0 and year % 100 != 0:
print("It's a leap year")
elif year % 400 == 0 and year % 100 != 0:
print("It's a leap year")
elif (year % 100 == 0):
print("It's not a leap year")
else:
print("There is an error")
def main():
years = get_year()
if_else(years)

Or put

`if year % 100 ==0:`

on top

Also, if one enter something that cannot be converted to in like text for example the script will stop with ValueError

(Mar-09-2017, 07:07 AM)wavic Wrote: [ -> ]Or put if year % 100 ==0: on top

if he puts this on the top then it will not catch years divisible by 400 - this will be false negative!

The code works as expected, but as you mention he needs to catch incorrect user input, but also can have function that checks leap year or not and return True/False respectively. This will allow to reuse this function in the future, instead of

`calendar.isleap()`

, which is available.

Yes, I miss that. Put it on the top only is not going to work

Actually you have to do the test the other way: test for 400 then 100 then 4.

@MartinEvtimov: use raw_input() instead of input() if you use Python v2, or make sure you use Python v3 (your `print(...)`

with parentheses are normally the v3 syntax)

The truth table is:

Output:

Truth Table
----------------------------------------------------
| Div by 4? | Div by 100? | Div by 400? | Is Leap? |
----------------------------------------------------
| T | T | T | T |
----------------------------------------------------
| T | T | F | F |
----------------------------------------------------
| T | F | T | T |
----------------------------------------------------
| T | F | F | T |
----------------------------------------------------
| F | T | T | F |
----------------------------------------------------
| F | T | F | F |
----------------------------------------------------

A "truth" table that shows a number divisible by 400 and not by 4 or not by 100, hmmph. Welcome to "alternate mathematics" :)

Otherwise since programming languages do not support "unless"(*), `If ConditionA unless ConditionB`

and be rewritten `if ConditionB else if Condition A`

.

(*) of course some smart-aleck will come up with one..

:-)

from calendar import isleap
def is_leap(year):
return not (year%400 and not year%100 or year%4)
for year in (1900, 2017, 2000, 2016):
print year, is_leap(year), isleap(year) # year, is_leap(), calendar.isleap()

Output:

1900 False False
2017 False False
2000 True True
2016 True True
[Finished in 0.04s]

So you choose the easy way?