A leap year check in three instructions

With the following code, we can check whether a year 0 ≤ y ≤ 102499 is a leap year with only about 3 CPU instructions: bool is_leap_year_fast(uint32_t y) { return ((y * 1073750999) & 3221352463) <= 126976; } How does this work? The answer is surprisingly complex. This article explains it, mostly to have some fun with bit-twiddling; at the end, I'll briefly discuss the practical use. This is how a

[AKIMOTO]

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