Tracking leap.py years for gregorian calendar (Exercism org)

[DeaD_EyE]

(A leap year is divisible by 4) and (a leap year is not divisible by 100 or a leap year divisible by 400).
Natural language is unsuitable for programming tasks because ambiguities and inaccuracies are possible.
I think it's still possible to understand this wrong.

The other problem were many newcomers are struggling with, are expressions in if-conditions.
if wants to have a Boolean. If it's not a boolean, then it's converted to a boolean.
A precise description of this is here: https://docs.python.org/3/library/stdtyp...ue-testing

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some_value = 42 by4 = 4 by3  = 3   # bool(0) == False # bool(-1) == True # bool(1) == True   # if by4, will call implicit if bool(by4), because by4 is not a boolean. # if requires a True or False # if an object is of the type Boolean, # then bool is called with the object, to get a boolean back     #          False                    or  (3) if some_value % by4 == 0 or by3:     print("Always True because by3 is 3 and bool(3) is True")     print("The first expression is False")     print("And the second expression is the bug")     # You want: if some_value % by4 == 0 and some_value % by3 == 0:     print("You can divide 42 by 3 and 42 by 4")     # no, you can't!

[betinajessen]

If you look in Lib/datetime.py in your Python distribution, you’ll find this function, which I wrote decades ago :wink:.

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def _is_leap(year):     "year -> 1 if leap year, else 0."     return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

Note that 2023 calendar printable and binds more tightly than or: the parentheses are vital there, although I’d use them (for clarity) even if they weren’t necessary.