; This routine will hopefully scroll the screen,
; let's put it at 32768 in RAM
        org $8000
        ld a,7          ; Colour seven is white ink on black paper
        ld (23693),a    ; Sets the colour as above
        call 3503       ; Calls the routine to clear screen
        ld a,0          ; Zero is black
        call 8859       ; Sets the border colour
INFINITE
        ld a,22         ; 22 is the Sinclair ROM value for PRINT AT
        rst $10         ; Call the ROM PRINT routine as above
        ld a,0          ; So, we need an x and y co-ord for PRINT AT
        rst $10         ; like in BASIC
        rst $10         ; so we call it twice (PRINT AT 0,0;)
        ld bc,STRING    ; Points the register bc to STRING
LOOP
        ld a,(bc)       ; Puts the current byte at bc into a
        cp 0            ; Compares a to zero
        jr z,MOVEBYTE   ; If equal to zero, jump to MOVEBYTE
        rst $10         ; Outputs a to the screen
        inc bc          ; Increases bc by one
        jr LOOP         ; Goes back to LOOP
MOVEBYTE
        ld bc,STRING    ; Let's point bc at the start of STRING again
        ld de,OFFSET    ; And de at OFFSET, which is one byte before
        ld a,(bc)       ; Get the first byte at bc and put into a
        ld (de),a       ; the byte at OFFSET
SCROLL
        inc bc          ; Increase bc by one
        inc de          ; Increase de by one
        ld a,(bc)       ; Get the current location of bc at store in a
        cp 0            ; Compare a to zero
        jr z,AGAIN      ; We'll jump to AGAIN if it equals zero
        ld (de),a       ; Store a at de (one byte before bc)
        jr SCROLL   ; Jumps to SCROLL and does it all again
AGAIN
        ld a,(OFFSET)   ; Puts the byte at OFFSET into a
        ld (de),a       ; Store a at location de (end of STRING before zero)
        ld b,%00000101  ; Now we're using the register b
DELAY
        halt            ; Stop everything and wait for an interupt
        djnz DELAY      ; Loop back until interupt is found
        jr INFINITE     ; Jump back to INFINITE
OFFSET
        defb %11111111  ; byte used in MOVEBYTE and AGAIN
STRING         ; Main text to scroll
        defb 32,32,32,32,32,32,"Na Na Na Na Naah!",32,32,32,32,32,32,32,32,32
        defb 0      ; End

        ld b,%00000101  ; Now we're using the register b
DELAY
        halt            ; Stop everything and wait for an interupt
        djnz DELAY      ; Loop back until interupt is found

SCR EQU 16384           ; Screen Ram
   ORG $8000
   LD BC,ROLL   ; Hopefully, this is going to work...
   PUSH BC      ; Got the first location of ROLL and put it into BC
   LD A,(BC)   ; Now load A with the first byte of ROLL
   LD (23693),A   ; set screen colours
   CALL 3503   ; clear the screen
   LD A,1      ; One is blue
   CALL 8859   ; Set border to blue
BEGIN
   LD BC,STRING   ; BC=STRING location in memory
   LD DE,SCR    ; DE=SCR or 16384
LOOP
   LD A,(BC)   ; A=Value of BC
   CP 0      ; Compare A to zero (end of STRING)
   JR Z,NEXTCOL   ; If equal then branch to NEXTCOL label
   RST $10      ; Output A to screen
   INC BC      ; Increace BC by 1 to get the next location of the STRING data
   INC DE      ; Increade DE by 1 to get the next cursor position
   JR LOOP    ; Branch back to LOOP label
NEXTCOL         ; This is where we loop back to do the colour roll
   POP BC      ; Okay, so we need to call BC back from the stack
   INC BC      ; Now this will increase it by one
   PUSH BC      ; Right, so we'll push it back to the stack so that we read the next byte next time around
   LD A,16      ; Loading A with 16 and then calling RST $10 will tell the Speccy that the next byte will
   RST $10      ; be the INK colour, which we'll call from the current location stored at BC
   LD A,(BC)   ; so now A is the next byte in the ROLL bit of memory
   CP 0      ; Right, let's compare it to zero and...
   JR Z,RESTART   ; ... if it is zero then we're at the end of the colour roll data
   RST $10      ; And if not, then we output the text colour to screen so that when the text is redrawn,
         ; it's a new colour, innit?
   LD A,22      ; Okay, so if A=22 and there's an RST $10 after it then that means we're going to tell the
         ; Speccy that we're relocating the curson
   RST $10      ; and therefore we want to relocate it at position
   LD A,0      ; zero twice, or 0,0 in the PRINT X,Y thingy
   RST $10      ; so A is now equal to zero, and as the X and Y locations need to be that to redraw over
   RST $10      ; the text, we just call RST $10 twice so the text is redrawn in the new colour
   JR BEGIN   ; Now, let's go back to the label BEGIN to redraw the text in the new colour
RESTART
   LD BC,ROLL   ; Oh no! Zero has been detected, so what are we going to do about it? Well, obviously we
   PUSH BC      ; don't want to keep reading bytes of memory after the ROLL data or we will hit an error eventually
   LD A,(BC)   ; as the INK colour will not be valid, so this makes BC equal to the start of the data segment in
   RST $10      ; memory, pushes it to the stack, makes A equal to the start again and throws out the colour
   LD A,22      ; at the start of the ROLL bit.
   RST $10      ; We need to remember to set the PRINT AT 0,0 thingy again, or it'll redraw after the first string
   LD A,0      ; which will eventually fill the screen and ask you if you want to Scroll? No, we don't want that.
   RST $10      ; PRINT AT 0,0 is effectively set in this code now
   RST $10      ;
   JR BEGIN   ; So, we'll go back to the beginning and start again. Infinite loops are good :-)
ROLL
   DEFB 7,7,6,6,5,5,4,4,3,3,2,2,1,1,2,2,3,3,4,4,5,5,6,6,0
STRING
   DEFB "Na Na Na Na Naah!"   ; Here is the data
   DEFB 0            ; 0 is the end of data