Lab2 - 6502 Assembly Language Lab(Modifying code)

-

 In the course Software Portability and Optimization we have given a code that is for 6502 Assembly language and we need to modify it and get the outcomes as follows: 

    1) Change the code to fill the display with light blue instead of yellow. 

        For this it is were simple to do it as we just change LDA #$07 to LDA #$0E which is a binary code for changing the yellow colour to light blue. As we now that lda means that something is going to loaded in accumulator and we are passing the command for colour to change to light blue.

    2) Change the code to fill the display with a different colour on each page.

        For this one in simple word we need to make four loop as they will make the part or page with different color. As shown in class with the simple code like loading a byte with some color in  y and then store it in byte after that it is incremented and make the loop till the 256 bytes are fully completed and it become 1 loop. We are repeating this thing for four time and then we are printing the different color every time. and this is simple way. Now from this code if we need to change it using the original bas code that we just need to add ASL command just below  1st bne loop; is completed this will increment or shift the value to left and change the color as in some default values. we are putting this as the first bne loop is done when they print 1st page with some color and after that we are passing command to change pages after that as we can printing in 4 pages or the whole bitmap. 

So, this is the solution for this two questions.

The base code is as follow : 

lda #$00	; set a pointer in memory location $40 to point to $0200
	sta $40		; ... low byte ($00) goes in address $40
	lda #$02	
	sta $41		; ... high byte ($02) goes into address $41

	lda #$07	; colour number

	ldy #$00	; set index to 0

loop:	sta ($40),y	; set pixel colour at the address (pointer)+Y

	iny		; increment index
	bne loop	; continue until done the page (256 pixels)

	inc $41		; increment the page
	ldx $41		; get the current page number
	cpx #$06	; compare with 6
	bne loop	; continue until done all pa
Code For 1)
        lda #$00	; set a pointer in memory location $40 to point to $0200
	sta $40		; ... low byte ($00) goes in address $40
	lda #$02	
	sta $41		; ... high byte ($02) goes into address $41

	lda #$0E	; colour number

	ldy #$00	; set index to 0

loop:	sta ($40),y	; set pixel colour at the address (pointer)+Y

	iny		; increment index
	bne loop	; continue until done the page (256 pixels)

	inc $41		; increment the page
	ldx $41		; get the current page number
	cpx #$06	; compare with 6
	bne loop	; continue until done all pages
Code For 2)
a)
	lda #$0E	
	ldy #$00	
loop:	sta $0200,y	
	iny		
	bne loop
	
	lda #$07	
	ldy #$00	
loop2:	sta $0300,y	
	iny		
	bne loop2
	
	lda #$05	
	ldy #$00	
loop3:	sta $0400,y	
	iny		
	bne loop3
	
	lda #$0A	
	ldy #$00	
loop4:	sta $0500,y	
	iny		
	bne loop4
Code For 2)
b)
        lda #$00	; set a pointer in memory location $40 to point to $0200
	sta $40		; ... low byte ($00) goes in address $40
	lda #$02	
	sta $41		; ... high byte ($02) goes into address $41

	lda #$07	; colour number

	ldy #$00	; set index to 0

loop:	sta ($40),y	; set pixel colour at the address (pointer)+Y

	iny		; increment index
	bne loop	; continue until done the page (256 pixels)
asl	
        inc $41		; increment the page
	ldx $41		; get the current page number
	cpx #$06	; compare with 6
	bne loop	; continue until done all pages


Comments

Post a Comment

Popular posts from this blog

ifunc function

-
IFUNC What is IFUNC? With the aid of a resolver function that was also written by the developer, numerous implementations of a particular function can be created using the GNU indirect function support (IFUNC) feature of the GNU toolchain. During early startup, the dynamic loader calls the resolver function to determine which implementation the application will use. Once a decision on implementation is made, it cannot be modified for the duration of the procedure. How to use it? As you can see, there are already a few limitations on how IFUNC can be used.                                                             It is t he implementations and the resolver must both be defined in the same translation unit. functions that cannot be weakly specified.    Since the resolution of an IFUNC might occur ver...
Read more

Lab5 Part 2 Algorithm 2

-
 According to the loudness factor, another algorithm simply shifted everything by an order of magnitude to the right. I haven't had time to talk about the outcomes of this algorithm with the group or with individuals, so I'm at a loss for words. I'd say it's definitely off if I had to guess which would be the best option. Multiplication takes longer than bit shifting, which is one of the reasons I say this. Why then is it so slow? This is so that the CPU doesn't merely shuffle bits around in memory but also compute values. The term implies that this is a bit shift. I would anticipate that bit shifting would also be favourable in terms of memory utilisation. Other techniques call for the lookup table to be stored and accessed somewhere in memory. Even though we're only working with 16-bit numbers, the largest possible values are 65, 536, etc., and the majority of contemporary PCs can hold 64k bits in registers, making them simple to store in registers. Without ac...
Read more

Lab2 - 6502 Assembly Language Lab(Calculation)

-
Image
 In this lab we need to find how much is time need to complete the process of code. So here we need to find the number of the cycles that are need to complete the process of one line of code to processed. So as shown in photo  The load accumulator need two cycle to complete the process and only one time this code is run. After that store accumulator will run and same as load accumulator it will run ones with two cycle. After all the function will be run as shown above but when the loop will be there t process will be change there are 256 bytes in one page and there are total 4 page to this will run 12 time and as shown above it will be executed.  And the time will be 5935us when there was 1Mhz clock speed is there. To change the speed we need to make change like we need to change the loop of pages to only one  which can change the part of time but I am not able to make that one. 
Read more