sx28 memory question

Vertex78

Is this how the memory is laid out in the sx28

for each bank:

$0 - $f = global registers, indf,rtcc....
$10 - $20 = sram

Vertex78

let me clarify what I think I understand

for each bank

$0 = indf
$1 = rtcc
$2 = pc
$3 = status
$4 = fsr
$5 = ra
$6 = rb
$7 = rc
$8 = ?
$9 = ?
$a = ?
$b = ?
$c = ?
$d = ?
$e = ?
$f = ?
$10 = sram
$11= sram
$12= sram
$13= sram
$14= sram
$15= sram
$16= sram
$17= sram
$18= sram
$19= sram
$1a= sram
$1b= sram
$1c= sram
$1d= sram
$1e= sram
$1f= sram

Bean

The memory from $00 to $0F is global. From $08 to $0F is SRAM just like all the other memory. It's just unique in that it can always be accessed no matter what "bank" is selected.

All other memory is "bank switched" using FSR.
$10 to $1F is a bank - This is the bank used by SX/B for normal variables, all other banks are for arrays
$30 to $3F is a bank
$50 to $5F is a bank

So if you want to use memory from $30 to $3F you would set FSR to #$20.

It's really confusing, and one of the major reasons why SX/B was created.

Bean.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
“The United States is a nation of laws -· poorly written and randomly enforced.” - Frank Zappa

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
www.hittconsulting.com
·

Vertex78

ok i get it so you have

$0 - $f = globals
$10 - $1f = bank 0
$20 - $2f = globals
$30 - $3f = bank 1
$40 - $4f = globals
$50 - $5f = bank 2
$60 - $6f = globals
$70 - $7f = bank 3
$80 - $8f = globals
$90 - $9f = bank 4
$a0 - $af = globals
$b0 - $bf = bank 5
$c0 - $cf = globals
$d0 - $df = bank 6
$e0 - $ef = globals
$f0 - $ff = bank7

right?