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Results 1 - 1 of 1 for 0b1100_0000 (0.12 sec)

  1. lib/fips140/v1.0.0.zip

    bits. // Setting the top two bits, rather than just the top bit, means that // when two of these values are multiplied together, the result isn't // ever one bit short. if excess := len(b)*8 - bits; excess < 7 { b[0] |= 0b1100_0000 >> excess } else { b[0] |= 0b0000_0001 b[1] |= 0b1000_0000 } // Make the value odd since an even number certainly isn't prime. b[len(b)-1] |= 1 // We don't need to check for p >= √2 × 2^(bits-1) (steps 4.4 and 5.4) // because we set the top two bits above, so // // p > 2^(bits-1)...
    Registered: Tue Sep 09 11:13:09 UTC 2025
    - Last Modified: Wed Jan 29 15:10:35 UTC 2025
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