How can I write a leap-year program in one line using PHP?

Question

Please define what you mean by a leap-year program? This could mean many things. — a'r, Feb 26 '10 at 12:32
OK Define Leapyear: If Year % 400=0 Its Leap year If Year % 4 =0 Its also Leapyear and if Year % 100 its not a Leap year — streetparade, Feb 26 '10 at 12:36

Sarfraz · Accepted Answer · 2010-02-26T13:57:04.747

15

This is how to know it using one line of code :)

print (date("L") == 1) ? "Leap Year" : "Not Leap Year";

edited Feb 26 '10 at 13:57

answered Feb 26 '10 at 12:35

Sarfraz

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Even shorter: `$isLeapYear = (date("L") == 1);` – Gordon Feb 26 '10 at 12:41
2

@Gordon: That's true, but i just wanted to print the output :) – Sarfraz Feb 26 '10 at 12:43
2

@Gordon - date("L") returns bool already. `$isLeapYear = date("L");` is all you need. – thetaiko Feb 26 '10 at 13:21
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@thetaiko Not bool. String. `var_dump(date('L')); // string(1) "0"`. But since 1 and 0 would be typecasted as needed, yes, I agree it can be shortened even further. – Gordon Feb 26 '10 at 13:50

score 5 · Answer 2 · answered Feb 26 '10 at 12:34

5

if (($year % 400 === 0) || (($year % 100 !== 0) && ($year % 4 === 0))) echo "leap";

answered Feb 26 '10 at 12:34

Xr.

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From http://stackoverflow.com/questions/3220163/how-to-find-leap-year/#11595914 `if (year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0) { /* leap year */ }` – Kevin P. Rice Jul 21 '12 at 23:55
I would not want to maintain that code. Bit operations for something that is only arithmetic are cryptic. The original answer you quote considered performance, so it might be relevant in some cases. But for all other cases, I'd rather keep a readable expression. – Xr. Jul 23 '12 at 10:13
I usually place a URL in a comment when something is cryptic; or, place the code in a function like "IsLeapYear" where it is clear what it does. Since the Gregorian Calendar isn't changing any time soon, I don't think "maintainability" is much of a concern. This code can be more than twice as fast, and even ten times faster in 8-bit embedded system CPUs where there is no divide instruction. – Kevin P. Rice Jul 23 '12 at 17:36

score 3 · Answer 3 · answered Feb 26 '10 at 12:33

Since there is no limit in how long a line of code is, unless you are using a code convention like that of Zend Framework, you can use whatever works and write into one line. Of course, depending on the functionality of your leap-year program, this will likely be hard to maintain. I've seen legacy code running over 800 chars with PHP, HTML and CSS intermingled. Eye-bleeding, I can tell you.

Bastiaan Linders · Answer 4 · 2010-02-26T12:50:03.337

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$nextyear  = mktime(0, 0, 0, date("m"),   date("d"),   date("Y")+1);

edited Feb 26 '10 at 12:50

answered Feb 26 '10 at 12:32

Bastiaan Linders

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Ah! I didn't know the meaning of leap-year, sorry. But I'll keep this "answer" here in case people end up here with the question that I thought you asked! – Bastiaan Linders Feb 26 '10 at 12:39

score 2 · Answer 5 · answered Feb 26 '10 at 12:33

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echo date("L");

answered Feb 26 '10 at 12:33

Select0r

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score 1 · Answer 6 · answered Feb 28 '11 at 13:19

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date("L"); is even shorter.

if(date("L")){
   //leap year
} else {
   //not leap year
}

answered Feb 28 '11 at 13:19

OBL

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How can I write a leap-year program in one line using PHP?

6 Answers6