Go read the parent again. The author claims that sorting in this case takes time proportional to the size of largest element, and I'm saying, if it takes time proportional to the space consumption of in the largest element (presumably where the cost of the sort comes from), you can't define a computational model where comparison takes constant time -- it still has to read the digits, which we know takes linear time.
If your point is that you can define a model of computation with contradictions in it, then you should rethink whether what you are saying here is even relevant to the thread at all.
Short circuiting comparisons is still linear for integers of unbounded size. But besides that, it's incredibly common to assume that 64-bit primitives are done in constant time, because to the asymptotic analysis, what matters is that the integers are _bounded_ in size. As long as it's bounded, the asymptotic analysis here doesn't care -- you can pick any size you want, as long as it's bounded.
One thing to keep in mind here (which it seems you are maybe confused about) is that these bounds are _asymptotics_ of the input size generally, and have very little to say about any particular choice of input size. 64-bit, 32-bit, whatever, it doesn't matter, if the size of the operands is bounded, the they're the same thing to the asymptotic analysis. As long as they're bounded, you can say that the operations on those operands is proportional to a constant times the (bounded) size of the operands.
The consequence of this is that when you deal with operands of _unbounded_ size, all of this asymptotic analysis goes right out the window. You can't possibly have a meaningful comparison operator, for example, that operates in constant time on unbounded integers.
Short circuting cuts that to log(log(N)) on average and log(N) worst case.
It's actually generally faster to compare X bytes of Vary large numbers than X Bytes of long int's. Degenerate case being 2 numbers of x/2 bytes.
Anyway, the point is treating log(log(N)) as constant is generally reasonable especially when N is small.
We don't really deal with unbound integers worst case is something like 2^(2^1,000,000,000,000,000,000)) before we can't actually store them.
Even then log of log of N makes random numbers of that size practically constant time to compare. Put another way there is less than 1 in 2^64 you need to do more than 3 comparisons and less than one in 2^128 you need more than 4. One for length, one for the chunk which might be just one bit and one more for the next 64 bits. Sure that might happen, but reolistically random input is unlikely to need many comparisons.
Stings on the other hand are more of an issue.
Do you realize that the whole point of asymptotic analysis is that _we do not care_ what integers we "really" deal with? These bounds specifically deal with sorting arbitrary data of unbounded size, and _no other assumptions_.
Do you understand? No assumption that we can use parallelism. No games with integers you see in "real life". None of that is relevant. You can't play games with practicalities to get a better bound. This is the general bound, and if you fiddle with it to get something else, you are changing the scope of the question to be something else entirely.
If you want to have a discussion about those other models, fine, but that's not the discussion we're having.
However, by changing your assumptions you can still use O notation with more complex models.
So again, the issue is not that you can't compare numbers of arbitrary magnitude in constant time -- I never said that. The issue is that it is a contradiction to say that it takes linear time to inspect all the digits of a number AND that it only takes constant time to compare O(n) digits of two numbers.
You're essentially saying that you can axiomatize your mathematical universe with nonsense axioms, which yes, you could do, but that's not useful to point out.
Your statement was "If the numbers can be arbitrary in size, then you can't compare them in constant time". I was merely pointing out that this is not a true statement.
Again, for the third time: you cannot "assume" that reading the digits of a number takes O(n), but reading the digits of two numbers to compare them takes O(1). That is absolutely, obviously true. Your response, that you can "assume" the latter is true as part of the model of computation is just wrong, plain and simple. There's nothing else to say about it.
The original claim directly implies this, and if you can't address that point, then just save us both the time and don't respond.