4 ms·

What about circular rotates? a<<(x%32) & a>>(32-(x%32))

• thaumasiotes 12y ago
  having worked an example to see how this could possibly work, I feel compelled to note for other people who might have wondered that a circular rotate looks like "(a << (x % 32)) | (a >> (32 - (x % 32)))"; as written, you're replacing every bit with zero.
  • nostrebored 12y ago
    a = 1111 0000 0000 1111 1111 0000 0000 1111 a << 37 == a << (37 % 32) == a << 5 == 0000 0001 1111 1110 0000 0001 1110 0000 a >> (32 - (37 % 32)) == a >> 27 == 0000 0000 0000 0000 0000 0000 0001 1110 anding the two together you get == 0000 0001 1111 1110 0000 0001 1111 1110 which is a circular rotation of the bit string...
    • thaumasiotes 12y ago
      0 & 1 is not 1, it's 0. 0000 0001 1111 1110 0000 0001 1110 0000 & 0000 0000 0000 0000 0000 0000 0001 1110 ----------------------------------------- 0000 0000 0000 0000 0000 0000 0000 0000
      • nostrebored 11y ago
        You're right! My bad. Definitely meant to | it.