How do you find Leapyear in VBA?

Question

What is a good implementation of a IsLeapYear function in VBA?

Edit: I ran the if-then and the DateSerial implementation with iterations wrapped in a timer, and the DateSerial was quicker on the average by 1-2 ms (5 runs of 300 iterations, with 1 average cell worksheet formula also working).

Solution 1

Public Function isLeapYear(Yr As Integer) As Boolean  

    ' returns FALSE if not Leap Year, TRUE if Leap Year  

    isLeapYear = (Month(DateSerial(Yr, 2, 29)) = 2)  

End Function  

I originally got this function from Chip Pearson's great Excel site.

Pearson's site

Solution 2

public function isLeapYear (yr as integer) as boolean
    isLeapYear   = false
    if (mod(yr,400)) = 0 then isLeapYear  = true
    elseif (mod(yr,100)) = 0 then isLeapYear  = false
    elseif (mod(yr,4)) = 0 then isLeapYear  = true
end function

Wikipedia for more... http://en.wikipedia.org/wiki/Leap_year

Solution 3

If efficiency is a consideration and the expected year is random, then it might be slightly better to do the most frequent case first:

public function isLeapYear (yr as integer) as boolean
    if (mod(yr,4)) <> 0 then isLeapYear  = false
    elseif (mod(yr,400)) = 0 then isLeapYear  = true
    elseif (mod(yr,100)) = 0 then isLeapYear  = false
    else isLeapYear = true
end function

Solution 4

Late answer to address the performance question.

TL/DR: the Math versions are about 5x faster

---

I see two groups of answers here

1. Mathematical interpretation of the Leap Year definition
2. Utilize the Excel Date/Time functions to detect Feb 29 (these fall into two camps: those that build a date as a string, and those that don't)

I ran time tests on all posted answers, an discovered the Math methods are about 5x faster than the Date/Time methods.

---

I then did some optimization of the methods and came up with (believe it or not Integer is marginally faster than Long in this case, don't know why.)

Function IsLeapYear1(Y As Integer) As Boolean
    If Y Mod 4 Then Exit Function
    If Y Mod 100 Then
    ElseIf Y Mod 400 Then Exit Function
    End If
    IsLeapYear1 = True
End Function

For comparison, I came up (very little difference to the posted version)

Public Function IsLeapYear2(yr As Integer) As Boolean
    IsLeapYear2 = Month(DateSerial(yr, 2, 29)) = 2
End Function

The Date/Time versions that build a date as a string were discounted as they are much slower again.

The test was to get IsLeapYear for years 100..9999, repeated 1000 times

Results

• Math version: 640ms
• Date/Time version: 3360ms

---

The test code was

Sub Test()
    Dim n As Long, i As Integer, j As Long
    Dim d As Long
    Dim t1 As Single, t2 As Single
    Dim b As Boolean

    n = 1000

    Debug.Print "============================="
    t1 = Timer()
    For j = 1 To n
    For i = 100 To 9999
        b = IsYLeapYear1(i)
    Next i, j
    t2 = Timer()
    Debug.Print 1, (t2 - t1) * 1000

    t1 = Timer()
    For j = 1 To n
    For i = 100 To 9999
        b = IsLeapYear2(i)
    Next i, j
    t2 = Timer()
    Debug.Print 2, (t2 - t1) * 1000
End Sub

Solution 5

I found this funny one on CodeToad :

Public Function IsLeapYear(Year As Varient) As Boolean
  IsLeapYear = IsDate("29-Feb-" & Year)
End Function 

Although I'm pretty sure that the use of IsDate in a function is probably slower than a couple of if, elseifs.

View more solutions

Comments

[nathaniel]

this one might even be more efficient. i like that it specifically takes the definition of leap year and works it into the answer.

[Lance Roberts]

the variable isLeap isn't being used

[rp.]

Being evenly divisible by 4 doesn't a leap year make! 2100 isn't a leap year. The division by 400 part of the test should come before the division by 4.

[Erik van Brakel]

creative solution! I wonder how it performs against the others posted.

[StingyJack]

That does not take into account all of the leap year rules.

[Lance Roberts]

Actually, if you study what they're doing, it always works. They check to see if the february month has 29 days, and that makes it a leapyear. It basically pawns the leapyear rules onto Microsoft. Chip has a lot of good solutions.

[Oorang]

If efficiency is the goal you can get rid of isLeapYear = false, as boolean values default to false:)

[Lance Roberts]

do you mean to use the DAY function instead of the MONTH function?