Pythons Project for beginners - Post 35: Leap Year Checker
- Anubhav Somani
- Sep 6
- 2 min read
Let's build a Leap Year Checker. This is a classic programming exercise that is perfect for beginners to master nested conditional logic (if/elif/else).
First, here is the explanation of how the project is built.
How the "Leap Year Checker" is Made
The Core Idea: The goal is to create a program that takes a year from the user and correctly determines whether that year is a leap year based on a specific set of rules.
The Rules for a Leap Year: A year is a leap year if it meets the following criteria:
It is evenly divisible by 4.
However, if the year is also evenly divisible by 100, it is not a leap year, unless...
...the year is also evenly divisible by 400. Then it is a leap year. For example, 2000 and 2400 are leap years, but 1800 and 1900 are not.
Getting User Input Safely: The program needs a year to check. We'll use input() to get the number from the user and int() to convert it to a whole number. This is placed inside a try...except ValueError block to ensure the program can handle cases where the user enters text instead of a valid year.
The Core Logic with Nested Conditionals: The rules for a leap year translate perfectly into a set of nested ifstatements. We check the conditions in order of precedence:
First if: Is the year divisible by 4? (year % 4 == 0). If not, we know right away it's not a leap year.
Nested if: If it is divisible by 4, we then check if it's also divisible by 100 (year % 100 == 0).
Deepest Nested if: If it is divisible by 100, we must make one final check: is it also divisible by 400? (year % 400 == 0). Only then is it a leap year. This structure directly models the logic of the rules.
Displaying the Result: Based on which condition is met, the program will print a clear message to the user, stating whether the year they entered is a leap year or not.
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