What Correction is Needed to Make a Python Function Determine Leap Years Correctly?

Determining Leap Years

You're attempting to write a program to ascertain if a specific year qualifies as a leap year. However, your current code returns "None" when executed in Python IDLE.

Recall that a leap year meets the following criteria:

• Divisible by 4
• Not divisible by 100 (exception: divisible by 400)

Examining your code:

<code class="python">def leapyr(n):
    if n%4==0 and n%100!=0:
        if n%400==0:
            print(n, "is a leap year.")
    elif n%4!=0:
        print(n, "is not a leap year.")

print(leapyr(1900))</code>

The issue lies in the last line: print(leapyr(1900)). When you call a function returning no value (such as leapyr in this case), the return value is always None. To resolve this, you can directly print the result within the function itself:

<code class="python">def leapyr(n):
    if n%4==0 and n%100!=0:
        if n%400==0:
            return n, "is a leap year."
    elif n%4!=0:
        return n, "is not a leap year."

result = leapyr(1900)
print(result)</code>

An Alternative Approach Using calendar.isleap

Python provides a built-in function named calendar.isleap that explicitly determines whether a given year is a leap year:

<code class="python">import calendar

print(calendar.isleap(1900))</code>

This function simplifies the task and returns a straightforward Boolean value.

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