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  • Why are you asking about 6 bytes, and not 5? Why not 7? Commented Nov 15, 2024 at 10:53
  • @Flater 48 bits (6 bytes) is the halfway between 2 common types, namely 32 bit and 64 bit integers, providing half the differential benefits and detriments of each. Commented Jan 12 at 16:51
  • Careful with your definition of "half". The benefit here is the size of the number you can store. A data type that can store half of what a 64 bit integer can store is a 63 bit integer, not a 48 bit one. But more to the point, my question wasn't asking for you to explain that 6 is the midpoint, but rather to imply that 6 seems to be an arbitrary number with no real basis in binary (it's not a power of two) - making it no less arbitrary than 5 or 7. Commented Jan 12 at 23:21
  • @Flater - "half of the differential benefits and detriments", not half the number. For example, 8 bytes is the largest in the series we are discussing and has the most benefit, while 4 bytes is the smallest and has the least benefit, in terms of possible uses. 6 is the halfway. Commented Jan 14 at 2:09
  • The benefits do not increase linearly, they increase exponentially. "Half" is only in the exact middle for a linear increase. Expressing the benefits of the size of a number type as a linear function makes no sense - doubling the amount of bytes does not double the benefit. What you're doing is the same as arguing that inbetween one thousand (4 digits) and one billion (10 digits), the halfway point is one million (7 digits) because it has 7 digits. But that's not relevant when discussing the value that these numbers express. Commented Jan 14 at 2:10