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Comment by OMGWTF | Hacker News Reader

How about 15 bytes? (x86, month in ecx, result in eax)

  // 28 + ((0x3bbeecc>>m>>m)&3);
  b8 cc ee bb 03    mov  eax,0x3bbeecc
  d3 f8             sar  eax,cl
  d3 f8             sar  eax,cl
  83 e0 03          and  eax,0x3
  83 c0 1c          add  eax,0x1c

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I think you can save a byte on the add with

    04 1c             add al, 0x1c

(but i've never written any x86 assembly)

You are right. I was lazy, I compiled it with gcc -O2 and extracted the assembler with objdump -Mintel -d.

In fact you can save 2 bytes if you only care about the result in AL, bringing it down to 13 bytes:

    b8 cc ee bb 03    mov  eax,0x3bbeecc
    d3 f8             sar  eax,cl
    d3 f8             sar  eax,cl
    24 03             and  al, 3
    04 1c             add  al, 28

I feel like you should be able to get away with a 2-byte constant of 0x3bec, but I haven't been able to simplify the bitshift amount enough, from (n >> 2 << 2) + 2 * (n % 2).

ETA: I think the simplest I can get it is:

  days(n) = 28 + (0x1df6 << (1^n&3) >> n) & 3

My intuition is that the 1^n&3 isn't going to be cheaper.

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