Fill the byte array with 0s until it fits the size required. byte[] bytes = new byte[255]{ 0x1F, 0x1A, 0x1B, 0x2C, 0x3C, 0x6D, 0x1E }; //7 bytes while(bytes.length < 8){ bytes.Concat(new byte[] { 0x00 }); } long res = BitConverter.ToUInt64(bytes, 0); Reference: BitConverter ...

c#,class,random,constructor,ulong

Random.Next() only generates 31 bits of true randomness: A 32-bit signed integer greater than or equal to zero and less than MaxValue. Since the result is always positive, the most significant bit is always 0. You could generate two ints and combine them to get a number with 62 bits...

c#,bitwise-operators,binary-operators,.net-reflector,ulong

How about casting num7 to a ulong? if (!flag && ((ulong)num7 < (((ulong) endLocation) - (((ulong) 4L) + num6)))) This has nothing to do with bitwise operators, by the way....

However, it seems if I multiply two ulongs together and it results in an overflow, I end up with zero instead of the "modulus(ulong.max)" of whatever I would have otherwise. You wouldn't get the result modulus(ulong.max) - you would get the result modulus(ulong.max + 1). For example: ulong x...

c#,bit-manipulation,bitwise-operators,ulong

1U is not ulong. It's uint. Declare test and t as ulong and you'll get what you need: var test = 1UL; var t1 = ~test; var t = 1UL; for (int i = 0; i < 64; i++) { Console.Write((t & t1) != 0 ? 1 : 0); t...