Hi All
does anybody have link to a simple Memory Map for a CPC 464 Please, Where the basic rom sits etc
Thanks Ray
Quote from: Audronic on 04:47, 28 May 15
Hi All
does anybody have link to a simple Memory Map for a CPC 464 Please, Where the basic rom sits etc
Thanks Ray
If OS is paged in, it sits at &0000-&3fff.
If an expansion ROM is paged in (this includes BASIC), it sits at &c000-&ffff.
You can disable the roms and have the full ram from &0000-&ffff.
If you are using firmware (includes using BASIC) the lower jumpblock is at &0000-&0040 and the upper jumpblock is around &b900-&bf00.
Stack is around &bf00-&c000 (starts &c000 and grows down).
If you are choosing to write your own programs that take control of the hardware, implement your own interrupt handler, setup the hardware as you need it and you can use all the memory as you wish, you can relocate the screen and stack.
Standard interrupt vector is at &0038.
Basic starts at &170 and grows up to HIMEM. HIMEM is set around &b0ff for 464, but lower for other systems with extra roms attached because roms can reserve some ram if they need it.
Screen is normally at &c000-&ffff.
Here is a copy from soft 968 explaining the memory map
@ arnoldemu
Thanks for the info.
@ Johnny Olsen
Thanks for the Table and the other info.
Ray
bump
I've got a question, where are the hardware registers located?
In the chips themselves. Z80 doesn't have a zero page.
yeah, but I mean registers responsible for i/o,sound and graphics, I've found a note that "The recommended I/O [Gate Array] port address is &7Fxx." does it mean it can be changed? I would really need to change the location of Gate Array
Right, these registers responsible for sound, graphics, etc., are still in the chips themselves though :D
The Z80 has an out command. So, rather than writing into memory, you would use out &hardware-register-address, value
There is a list on the wiki of the IO ports.
No idea what you mean with you want to change the location of the Gate Array - soldering iron?? :laugh:
ah, so does it mean the memory RAM isn't altered when the program calls the registers by OUT instruction?
Quote from: gorgh on 17:26, 20 August 18
ah, so does it mean the memory RAM isn't altered when the program calls the registers by OUT instruction?
Yes, that's right. It is not like on the 6502, which has memory mapped IO. Z80 has separate memory address and IO port address bus - no conflict.
Of course, there are some "registers" in memory, but those are for the software (BASIC stackpointer, OS, keyboard input buffers, etc... ) If you overwrite these, the system crashes unless you know what you are doing ;)
wow, that's great, I'm surprised how well Amstrad is designed, the more I study the architecture the more I like it
(thanks)
Quote from: gorgh on 17:40, 20 August 18
wow, that's great, I'm surprised how well Amstrad is designed, the more I study the architecture the more I like it
(thanks)
You are welcome. Even though this aspect is mainly caused by the Z80 CPU. All other Z80 system are similar in that regard.A great read is this one if you want to learn more about Z80 computers and their design:"Build Your Own Z80 Computer" by Steve Ciarcia. You should be able to find the PDF online.
I guess what you search is here:
http://www.cpcwiki.eu/index.php/I/O_Port_Summary
thanks!