There is an implied 6th bit that is zero. Timestamps have a two-second minimum resolution.
this post was submitted on 04 Oct 2024
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That makes sense. Presumably the missing bit is the least significant one, and DOS is rounding them and storing only even-numbered seconds.
Sounds like DOS doesn't keep time in increments any smaller than 2-second intervals. Double your 0-to-29 value whenever asked to provide time with seconds. Done.
Note: this is off the top of my head, with no in-depth knowlege of actual DOS time-keeping beyond that provided in the OP. I'm interested to see how many versions DOS went through with this time-keeping method, and what value any of this provides beyond querying the system's Real Time Clock.
So you only have seconds in even numbers?
Could do only-odd numbers if you wanted to be squirrely about it, but I think most people would get more inquisitive upon never seeing a zero.
This seems to be the likely answer. I'm assuming that it has something to do with the technological limitation of 16 bit. 1981 saw the first 32-bit non-x86 microprocessor (Intel iAPX 432), and MS-DOS was released for 16-bit in mind, like the 8086. Perhaps, the highest size of the integer was limited to 16 bit as well, and with that in mind, they had to make sure to create a non-padded struct for time, in which the hour were allotted 5 bits (= 32 ≈ 24hour), minutes were allotted 6 bits (= 64 ≈ 60mins). The last remaining 5 bits were assigned to seconds, and with the remaining bit-fields, the best precision they could come up with gave a 2-second interval. Is that a fair reasoning?
That's how I read it, yeah.
Actually versions of MS-DOS were released for the MSX platform, which had a 8-bit Zilog Z80 CPU.
The number of bits mentioned when referring to processors usually refer to the size of the internal registers. You'll find that it doesn't actually matter how big the internal registers are. This just matters for the number of bits possible to process at the same time. So in order to process more bits, it just takes more steps, but it isn't impossible.
Pretty sure no one here was worried about what was or wasn't possible, just the method that was used in this specific instance.
If you have ~~32~~ 5 bits to work with you keep one bit to determine which half of the 30 seconds you are on. If that bit is a 0, you are counting 1-30 and if that bit is 1, then you are counting 31-60.
But during the release of MS-DOS, there were only 16-bit microprocessors, right? 32-bit x86 processors came way later around 1985, I think?
I don't know the specifics, but why would 5 bits be a problem on a 16bit machine? Shit, my mistake, I should have said 'If you have 5 bits to work with'. I will correct it.
I don’t know the specifics, but why would 5 bits be a problem on a 16bit machine?
Because I'm assuming that there's only 16 bits, right, and hence these have to be divided between hours, minutes and seconds:
- 5 bits for the hours (2^5^ = 32, to represent numbers from 0-23)
- 6 bits for the minutes (2^6^ = 64, can represent numbers from 0-59)
- the 5 bits for the seconds (2^5^ = 32, can only represent numbers from 0-31, what about seconds 32-59?)
5 + 6 + 5 = 16 bits in total. This is why the last one makes no sense, unless the timeformat precision was restricted to 2 seconds?
Yeah, I have been trying to work out a table for just the seconds, and yeah, the 5 bits isn't enough. I wasn't thinking it through. Another bit would be needed to be able to capture the flip, and in that case the extra bit would just allow you to store the actual 0-59 value. That's what I get for speaking before I have worked it out in my head.
Probably need to go as deep as the ALU operations available and clock cycles to really see what is going on.
I know nearly nothing of value in this space, but have seen lots of little tricks like this in passing with hobbyist hardware or ROMs. Intuitively, it feels like one of those kinds of situations. You might even find there is a clever use of a carry flag or something of that nature where there is essentially a free bit on an operation. I'm not sure how x86 does the ALU or whatnot here or even if this advice is relevant with an overlay tree. Feel free to inform/correct.