Count your RAM banks

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Lorin Millsap
Posts: 193
Joined: Wed Apr 29, 2020 6:46 pm

Count your RAM banks

Post by Lorin Millsap »

is it even physically possible to have a non-power-of-two number of banks?  [emoji848]   Having an odd number of mem chips on the board somehow seems "off" to me (but i'm a noob regarding that so yeah)

Absolutely. There is nothing about the design that limits that. In fact the base 512k is just one.


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Wavicle
Posts: 282
Joined: Sun Feb 21, 2021 2:40 am

Count your RAM banks

Post by Wavicle »



3 hours ago, Elektron72 said:




Yes; it appears that the current version of MEMTOP assumes that the number of banks must be a power of two.



The behavior I've predicted is actually worse: I think it will report 2048K. After testing for the existence of bank 128 (which will succeed), it will left-shift the value in A, which will cause it to contain zero upon return (side note: a zero in A from MEMTOP indicates 256 banks).



At this point, I think it's clear that the entire MEMTOP routine needs to be rewritten. Someone should open a GitHub issue about this (I might do that later today).



If I put on my computer science hat, I would point out that a binary search has the same time complexity, O(log N), as the current implementation.

If I put on my software engineer hat, I would point out that a binary search is rather a lot of easy-to-get-wrong code just for counting available memory banks.

If I put on my computer engineer hat, I would point out that there is probably a minimum increment that could be checked, probably around 64K, which would significantly reduce the size of a linear search while still getting the right number.

Having close to zero 6502 experience, I'm not the right person to fix it in any case ?

Wavicle
Posts: 282
Joined: Sun Feb 21, 2021 2:40 am

Count your RAM banks

Post by Wavicle »



33 minutes ago, desertfish said:




is it even physically possible to have a non-power-of-two number of banks?  ?   Having an odd number of mem chips on the board somehow seems "off" to me (but i'm a noob regarding that so yeah)



In general parallel bus RAM chips usually have a power of 2 memory size. I'm not 100% certain of this, but I suspect that it is probably so that there are no invalid combinations of their address wires. Off the top of my head, I cannot recall seeing such a RAM chip that was not sized as a power of 2.

Seeing a non-power (or even multiple) of 2 RAM chips on a design is reasonably common. Usually all of the memory components are the same size, but this is mostly done to save on production costs.

rje
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Location: Dallas Area

Count your RAM banks

Post by rje »


Banked RAM consists of up to four RAM chips, each with 512K of static RAM, if I understand correctly.

Chip 1: Banks 0-63.  Always present (required, since Bank 0 is used by the OS).

Chip 2: Banks 64-127.

Chip 3: Banks 128-191.

Chip 4: Banks 192-255.

Thus, a test would check one bank as a proxy for each 512K segment (say, banks 65, 129, and 193), and see if it can write/read that bank.

 

I wonder if a 256K static RAM would fit in those spots... horrors...

 

rje
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Count your RAM banks

Post by rje »

Ed Minchau
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Joined: Sat Jul 11, 2020 3:30 pm

Count your RAM banks

Post by Ed Minchau »



4 hours ago, rje said:




Banked RAM consists of up to four RAM chips, each with 512K of static RAM, if I understand correctly.



Chip 1: Banks 0-63.  Always present (required, since Bank 0 is used by the OS).

Chip 2: Banks 64-127.

Chip 3: Banks 128-191.

Chip 4: Banks 192-255.



Thus, a test would check one bank as a proxy for each 512K segment (say, banks 65, 129, and 193), and see if it can write/read that bank.



 



I wonder if a 256K static RAM would fit in those spots... horrors...



 



Is 3 RAM chips even going to be an option? My understanding was they'd only be selling 512k, 1M, and 2M variants.

Elektron72
Posts: 137
Joined: Tue Jun 30, 2020 3:47 pm

Count your RAM banks

Post by Elektron72 »



3 minutes ago, Ed Minchau said:




Is 3 RAM chips even going to be an option? My understanding was they'd only be selling 512k, 1M, and 2M variants.



The way I interpreted the FAQ is that all Commander X16s will be sold with 512K; any additional RAM must be added by the user.

rje
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Count your RAM banks

Post by rje »


That's right.  The X16 only comes with 512K of the banked RAM.  It's socketed so it's easy to add more, and (if I recall right) they're less than $10 apiece.

Screech
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Joined: Fri Jan 01, 2021 5:35 pm

Re: Count your RAM banks

Post by Screech »

On the off chance someone is looking to do this in BASIC:
10 BC=0:I=0
15 BANK I
20 IF PEEK($A000)=160 THEN GOTO 40
30 BC=BC+1:I=I+1:GOTO 15
40 PRINT"TOTAL BANKS:";BC
50 PRINT"TOTAL HI-RAM:";(BC*8);"K"
It's not the most elegant way, and if you're planning on using that much ram in basic, you may want to consider assembly. But this worked for me.
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ahenry3068
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Re: Count your RAM banks

Post by ahenry3068 »

Screech wrote: Fri May 10, 2024 7:01 pm On the off chance someone is looking to do this in BASIC:
10 BC=0:I=0
15 BANK I
20 IF PEEK($A000)=160 THEN GOTO 40
30 BC=BC+1:I=I+1:GOTO 15
40 PRINT"TOTAL BANKS:";BC
50 PRINT"TOTAL HI-RAM:";(BC*8);"K"
It's not the most elegant way, and if you're planning on using that much ram in basic, you may want to consider assembly. But this worked for me.
This is a bit more elegant.

10 MT = $FF99 20 AR = $30C 30 XR = $30D 40 YR = $30E 50 PR = $30F 60 POKE PR, 1 : REM Set the Carry Flag 70 SYS MT : REM Call the Memory Address $FF99 (ROM MEMTOP) 80 BC = PEEK(AR) 90 IF BC=0 THEN BC=256 110 PRINT "TOTAL BANKS";BC 115 PRINT "HIGHEST BANK #";BC-1 120 PRINT BC*8192;" BYTES OF HIRAM"
Last edited by ahenry3068 on Fri May 10, 2024 11:28 pm, edited 2 times in total.
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