Hi All,
After reading info on using Peek/Poke commands to access extra memory locations (or assigning more variables beyond the B0 to B55 in 28X2/40X2 picaxes).
My understanding is:
The 28X2 & 40X2 devices have: 256 bytes of user RAM (+1024 more in scratchpad)
That means 256+1024 = 1280 bytes we can access via peek/poke commands.
(Obviously the first 56 bytes are equally accessable via b0 to b55 (or w0 to w27) variables too).
So, if a SLOT in a 40X2 is 4096 bytes, Is it possible to still access the bytes above the 1280 byte range?
My question is, if a slot has 4096 bytes available for 'code', then why not access almost all 4096 bytes for a huge storage of variables or whatever?
(Subract a handful of bytes to configure the slot to switch back to other slots for the main program (run 0, run 1 etc)
As long as we confirm the code in each slot is xx bytes long (let's say upto 96 bytes of housekeeping to go back to Slot 0, that leaves 4000 bytes available, per slot.
So, in affect, Slot 0 could be the main program, and slot 1,2 & 3 could potentially be 3x 4k (12k total) of byte storage?
Am i totally off the rails here? or am i correct with the maths?
Thanks in advance.
After reading info on using Peek/Poke commands to access extra memory locations (or assigning more variables beyond the B0 to B55 in 28X2/40X2 picaxes).
My understanding is:
The 28X2 & 40X2 devices have: 256 bytes of user RAM (+1024 more in scratchpad)
That means 256+1024 = 1280 bytes we can access via peek/poke commands.
(Obviously the first 56 bytes are equally accessable via b0 to b55 (or w0 to w27) variables too).
So, if a SLOT in a 40X2 is 4096 bytes, Is it possible to still access the bytes above the 1280 byte range?
My question is, if a slot has 4096 bytes available for 'code', then why not access almost all 4096 bytes for a huge storage of variables or whatever?
(Subract a handful of bytes to configure the slot to switch back to other slots for the main program (run 0, run 1 etc)
As long as we confirm the code in each slot is xx bytes long (let's say upto 96 bytes of housekeeping to go back to Slot 0, that leaves 4000 bytes available, per slot.
So, in affect, Slot 0 could be the main program, and slot 1,2 & 3 could potentially be 3x 4k (12k total) of byte storage?
Am i totally off the rails here? or am i correct with the maths?
Thanks in advance.