| Title | How Did MS-DOS Decide that Two Seconds Was the Amount of Time to Keep the Floppy Disk Cache Valid? | |
| Date | Wednesday September 25 2019, @10:02AM | |
| Author | Fnord666 | |
| Topic | ||
| from the MS-what? dept. | ||
Submitted via IRC for Bytram
How did MS-DOS decide that two seconds was the amount of time to keep the floppy disk cache valid?
MS-DOS 2.0 contained a disk read cache, but not a disk write cache. Disk read caches are important because they avoid having to re-read data from the disk. And you can invalidate the read cache when the volume is unmounted.
But wait, you don't unmount floppy drives. You just take them out.
IBM PC floppy disk drives of this era did not have lockable doors. You could open the drive door and yank the floppy disk at any time. The specification had provisions for reporting whether the floppy drive door was open, but IBM didn't implement that part of the specification because it saved them a NAND gate. Hardware vendors will do anything to save a penny.
[...] Mark Zbikowski led the MS-DOS 2.0 project, and he sat down with a stopwatch while Aaron Reynolds and Chris Peters tried to swap floppy disks on an IBM PC as fast as they could.
They couldn't do it under two seconds.
So the MS-DOS cache validity was set to two seconds. If two disk accesses occurred within two seconds of each other, the second one would assume that the cached values were still good.
I don't know if the modern two-second cache flush policy is a direct descendant of this original office competition, but I like to think there's some connection.
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printed from SoylentNews, How Did MS-DOS Decide that Two Seconds Was the Amount of Time to Keep the Floppy Disk Cache Valid? on 2023-05-26 12:24:55