Small correction, it’s every fourth year except every hundreth year except every 400th year. I’m assuming the misconception comes from the last time it was a 100th also being a 1000th (2000) but the next time a leap year will end in 00 is 2400
Ahh right! Thanks for correcting me. Now that you mention it I remember too. It also makes sense, a year is roughly 365.2425 days long. Add 0.25 (one out of four), subtract 0.01 (one out of hundred), add another 0.0025 (2.5 out of thousand which is 1 out of 400)
look I’m not trying to be a dick or anything, but do you not know about leap years and which years they are?
edit: just realized it was the python and not ruby example, I was very tired and distracted when I was reading this thread.
Yes, and I have no idea…
https://youtu.be/-5wpm-gesOY
Leap years are each fourth year, except each hundredth year, except each
thousandthfourhundredth year.1896 leap year
1900 not leap year
1904 leap year
…
1996 leap year
2000 leap year
2004 leap year
…
2096 leap year
2100 not leap year
2104 leap year
Then you just arrange the 10 year window in different positions to overlap 1 to 3 leap years to reveal the three outcomes of the bug.
- / - - - / - - - /
- - / - - - / - - -
- - 0 - - - / - - -
- is a normal year, / is a leap year, 0 is an exceptional non-leap year.
Small correction, it’s every fourth year except every hundreth year except every 400th year. I’m assuming the misconception comes from the last time it was a 100th also being a 1000th (2000) but the next time a leap year will end in 00 is 2400
Ahh right! Thanks for correcting me. Now that you mention it I remember too. It also makes sense, a year is roughly 365.2425 days long. Add 0.25 (one out of four), subtract 0.01 (one out of hundred), add another 0.0025 (2.5 out of thousand which is 1 out of 400)