How to resolve the algorithm Pointers and references step by step in the Z80 Assembly programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Pointers and references step by step in the Z80 Assembly programming language
Table of Contents
Problem Statement
In this task, the goal is to demonstrate common operations on pointers and references. These examples show pointer operations on the stack, which can be dangerous and is rarely done. Pointers and references are commonly used along with Memory allocation on the heap.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Pointers and references step by step in the Z80 Assembly programming language
Source code in the z80 programming language
ld a,(&C000) ;load the accumulator with the value from the memory address &C000.
ld hl,&D000
ld e,(hl) ;load the E register with the byte at memory address &D000.
ld bc,(&E000) ;load the register pair BC from memory address &E000. The Game Boy cannot use this addressing mode.
; This is the same as:
; ld a,(&E000)
; ld c,a
; ld a,(&E001)
; ld b,a
ld IX,&F000
ld a,(IX+3) ;load A with the byte stored at &F003, The Game Boy cannot use this addressing mode.
ld h,(bc) ;if (bc) or (de) is the operand of a LD command, the other operand must be A.
ld e,(&A000) ;if a constant pointer is the operand of a LD command, the other operand must be A.
inc (HL) ;increment the byte pointed to by HL
rrc (IX+5) ;take the byte pointed to by IX+5 and rotate it right. The value of IX is unchanged.
sub (IY-23) ;subtract the value of the byte that is 23 bytes before IY, from A.
;The value of that byte at that address is unchanged, as is IY.
ld hl,PointerTable
ld e,(hl)
inc hl
ld d,(hl)
; depending on what we need to do next, we might need this new pointer back in HL. There are a few ways to do this:
; ex de,hl
; push de
; pop hl
; ld L,E
; ld H,D
ld hl,PointerTable
ld a,(hl)
ld (SMC+1),a ;SMC+1 points to low byte of the operand
inc hl ;increment HL so we can fetch the high byte
ld a,(hl)
ld (SMC+2),a ;SMC+2 points to high byte of the operand
SMC:
ld a,&BEEF ;gets overwritten with the pointer stored in entry 0 of PointerTable.
ld hl,PointerTable
ld a,(index)
add a ;this is a table of 16-bit values so double our index
add L
ld L,a
ld a,H
adc 0
ld H,a ;offset HL by our index
ld e,(hl)
inc hl
ld d,(hl) ;get the Ath entry of PointerTable and store it into DE.
ld (referenceStorage),DE ;store DE into ram for later use as a reference.
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