To run code in EPROM will RAM need to contain a pattern with RAM[adress] = 7..0(adress).
The EPROM is stored with typical 0FFh in bank 0, or RST 38h. It means that the processor begins to execute a RST 38h. The typical value at 0038h is RST 38h too, and a 0039 pattern is now stored into memory due to the recursive process.
ORG 0h ; Rom bank 0, unused typical 0FFh
NOP
JR 0FF88h ; Jump to entering code.
ORG 38h
RST 38h
ORG 39h
EX (SP),HL
ORG 400h ; Rom bank 1, other adr. free.
LD (HL),L
ORG 439h
DEC HL
ORG 839h ; Rom bank 2, other adr. free
RST 0h
ORG 0FF88h ; Rom bank 63.
; Code here to initialize RAM, and begin execution of
; EPROM code.
The process continues until RAM is filled with 0039 downto
address 0038h. At this address is stored a EX (SP),HL in
EPROM, and this instruction loads HL with 0039, and store a
random instruction at 0038h. This is look'ed up and replaced
by a RST 38h too. Again is RST 38h executing until next time
0038h is reach, and EX (SP),HL executed again.
This time is 0039h stored into 0038h, and instructions at 0h and 39h in EPROM is executing. These are NOP and the EX (SP),HL. These two instructions continues until next EPROM bank is reached.
The next rom bank, located at 4xxh, contain a DEC HL, and LD (HL),L instructions at 400h and 439h. Now is memory from 0039 and down initialized with RAM[adr]:=7..0(adr). Addresses at high memory is initialized too.
This continues until next rom bank (1024 instructions). Next bank is entered in 839h and this address contains a RST 0h. After RST 0h will processor execute instructions at 0h. It begins with a NOP and then a JR to 0FF88h where code is to initialize memory.
The RAM is now initialized, and execution of EPROM is ready to begin.
