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Power of TWO

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Why???? that means 16 bits = 64k... addressing 60k STILL needs 16 bits... That would be a waste of 4k.

Yes, remaining 4k combination addresses for no location. Your decreasing the no. of registers used, and thus reducing the cost.
Thanks Ian, KISS, Nigel, 3v0; I hope this thread helped in understanding the concepts well, both for me and some of who didn't know this well. And this would also help others in future.
 
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One machine I knew of essentially could have 60K of ram. The upper 4K were just locations used to control peripherals.

As IAN said, it's to maximize the the use of the address buss. Any machine that isn't fully populated doesn't have all of it's marbles, so to speak. With Windows a certain amount of memory is reserved for the display. You don't have access to it.
 
One machine I knew of essentially could have 60K of ram. The upper 4K were just locations used to control peripherals.
Like the PDP11 series of mini-computers.
They* had a 16 bit address bus, and when fully populated with magnetic core memory there would be 64k bytes of memory fitted.
However, the top 4k of addresses was reserved for peripheral control registers, so there was 4k of memory which could never be used.

(* at least the older ones did, some of the later ones had more address lines)

JimB
 
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Hey, some posts disappeared!

e.g.

Dear KeepItSimpleStupid,

Electroenthusiast has just replied to a thread you have subscribed to entitled - Power of TWO - in the Members Lounge forum of Electronic Circuits Projects Diagrams Free.

This thread is located at:
https://www.electro-tech-online.com/threads/power-of-two.124759/

Here is the message that has just been posted:
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---Quote (Originally by KeepItSimpleStupid)---
EE: Capacities are different for disk drives: 1K =1000 for disc drives. See: https://en.wikipedia.org/wiki/Hard_disk_drive
---End Quote---
I'm not saying about a disk drive. It was for Memory Stick/Flash Drives.
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Exactly, but the USB emulates a hard disk, therefore it has to look like one from the interface.

Start here and start following some of the links. https://en.wikipedia.org/wiki/Camera_File_System

So, it's memory on one level and a disk drive on the other. So, the memory inside the USB stick will have to conform to the hard drive specifications. My 4 GB USB drive shows a capacity of 3.73 GB and the number of bytes is considerably higher. Remember that there could be bad block tables, file allocation maps likely included and places for S.M.A.R.T data. Remember the days of hard and soft sectoring?
 
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