I tested the code in an 8080 emulator. Are you trying this on real hardware or in emulation?
Me bad. I was testing this on an 8080 emulator, not actually under CP/M. I forgot that CP/M automatically echoes the input.
Scelbi did not make up the mnemonics.
Most schools that I know of use x86, 68000, MIPS, ARM or possibly RISCV to teach assembly language.This almost sounds like some homework assignment, except what school uses 8080 anymore
I played a similar game called RATRUN on a SuperPET at the university computer center.
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I played a similar game called RATRUN on a SuperPET at the university computer center.
Puzzles like
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Daily Medium Ballsort Puzzle for Thursday 26th September 2024
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would be really dazzling on a Dazzler...
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;
; Writes a "zero" into all the RAM from 0C00 to 0FFF.
;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
BDOS equ 5 ; CP/M function entry point
org 100h ; It runs in the TPA
Start:
lxi H,1C00h ; Point HL to 0C00h
; Note that although the instruction
; says "H", it means HL. This is
; common in 8080 assembly.
lxi D,0FFFh-0C00h+1 ; Load number of bytes
; This is calculating the number of
; bytes using the constant expression
; within the assembler. The CPU
; does not perform this calculation
; at runtime.
; Also note, that the D here means
; DE register pair. The result here
; is a 16 bit number. In this case,
; 0x400, or 1024 bytes.
Loop:
mvi M,0 ; Store a zero at [HL]
; It says M here, but means "where
; HL is pointing". So, since HL
; was initially set to 0x1C00 above,
; the first 0 will be stored in the
; address 0x1C00
inx H ; Point HL to next location
; Again, H mean HL. Since HL is a 16
; bit register, INX will bump it by
; 1 and handle the 16 bath for us.
; The first time through, HL will
; be incremented to 0x1C01.
dcx D ; Decrement count
; Similarly, this is decrementing the
; DE register pair.
mov A,D ; Has count reached 0?
ora E
; This is a bit of idiomatic code
; here, and it's not intutive at its
; face from the simple reading of the
; mnemonics.
; What is happening here is that we
; have the DE register pair, which
; contains the count. The MOV copies
; the 8-bit D register into the
; accumulator. Unlike above where D
; or H mean DE and HL, respectively,
; here D means D.
; Next, it does an OR of the A
; with the other 8-bit register E
; ORing two registers is quick way
; to tell if they're BOTH zero.
; 1 or 0 is 1, 0 or 1 is 1. The only
; way you get zero is 0 or 0.
; You'll note that since it's ORing
; E into A, the result is in A.
; Anytime the A register is set to
; zero, the Z flag is set, otherwise
; it is reset.
; So, if D and E are both 0, which
; the combination DE register pair is
; 0, then that means we have consumed
; all of the count that was stored in
; DE and now we can exit the loop.
jnz Loop ; No
; DE is not 0, so Z flag is reset, so
; we jump to the Loop label.
; Otherwise, DE is zero, thus A is
; zero, thus the Z flag is set, and
; we continue past to end the routine.
; And we're done, back to CP/M using
; BDOS
mvi C,0 ; Exit to CP/M
call BDOS
end Start
; fill memory from 0C00-0FFF with zero
ORG 0100H ; start of Transient Program Area where CP/M programs run
start:
LXI H,0C00H ; start address of memory block
MVI C,0 ; value to fill memory with (if it needs to be changeable)
loop:
MOV M,C ; store value at HL (or just use MVI M,0 or whatever)
INX H ; increment HL pointer
MOV A,H ; get H
CPI 10H ; = 1000H / 256, compare the high byte of HL against the address of the first byte after the memory block
JNZ loop ; could also use JNC
; don't check L because the end is aligned 256
; but can also use MOV A,L, CPI, JNZ/JNC if the low byte of the address needs to be checked too
MVI C,0 ; exit to Command Control Processor
CALL 5 ; BDOS
END start