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dgellow/snake6502.asm

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Created January 5, 2016 15:58
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snake6502.asm
; ___ _ __ ___ __ ___
; / __|_ _ __ _| |_____ / /| __|/ \_ )
; \__ \ ' \/ _` | / / -_) _ \__ \ () / /
; |___/_||_\__,_|_\_\___\___/___/\__/___|
; An annotated version of the snake example from Nick Morgan's 6502 assembly tutorial
; on http://skilldrick.github.io/easy6502/ that I created as an exercise for myself
; to learn a little bit about assembly. I **think** I understood everything, but I may
; also be completely wrong :-)
; Change direction with keys: W A S D
; $00-01 => screen location of apple, stored as two bytes, where the first
; byte is the least significant.
; $10-11 => screen location of snake head stored as two bytes
; $12-?? => snake body (in byte pairs)
; $02 => direction ; 1 => up (bin 0001)
; 2 => right (bin 0010)
; 4 => down (bin 0100)
; 8 => left (bin 1000)
; $03 => snake length, in number of bytes, not segments
;The screens is divided in 8 strips of 8x32 "pixels". Each strip
;is stored in a page, having their own most significant byte. Each
;page has 256 bytes, starting at $00 and ending at $ff.
; ------------------------------------------------------------
;1 | $0200 - $02ff |
;2 | |
;3 | |
;4 | |
;5 | |
;6 | |
;7 | |
;8 | |
; ------------------------------------------------------------
;9 | $03 - $03ff |
;10 | |
;11 | |
;12 | |
;13 | |
;14 | |
;15 | |
;16 | |
; ------------------------------------------------------------
;17 | $04 - $03ff |
;18 | |
;19 | |
;20 | |
;21 | |
;22 | |
;23 | |
;24 | |
; ------------------------------------------------------------
;25 | $05 - $03ff |
;26 | |
;27 | |
;28 | |
;29 | |
;30 | |
;31 | |
;32 | |
; ------------------------------------------------------------
jsr init ;jump to subroutine init
jsr loop ;jump to subroutine loop
init:
jsr initSnake ;jump to subroutine initSnake
jsr generateApplePosition ;jump to subroutine generateApplePosition
rts ;return
initSnake:
;start the snake in a horizontal position in the middle of the game field
;having a total length of one head and 4 bytes for the segments, meaning a
;total length of 3: the head and two segments.
;The head is looking right, and the snaking moving to the right.
;initial snake direction (2 => right)
lda #2 ;start direction, put the dec number 2 in register A
sta $02 ;store value of register A at address $02
;initial snake length of 4
lda #4 ;start length, put the dec number 4 (the snake is 4 bytes long)
;in register A
sta $03 ;store value of register A at address $03
;Initial snake head's location's least significant byte to determine
;where in a 8x32 strip the head will start. hex $11 is just right
;of the center of the first row of a strip
lda #$11 ;put the hex number $11 (dec 17) in register A
sta $10 ;store value of register A at address hex 10
;Initial snake body, two least significant bytes set to hex $10
;and hex $0f, one and two places left of the head respectively
lda #$10 ;put the hex number $10 (dec 16) in register A
sta $12 ;store value of register A at address hex $12
lda #$0f ;put the hex number $0f (dec 15) in register A
sta $14 ;store value of register A at address hex $14
;the most significant bytes of the head and body of the snake
;are all set to hex $04, which is the third 8x32 strip.
lda #$04 ;put the hex number $04 in register A
sta $11 ;store value of register A at address hex 11
sta $13 ;store value of register A at address hex 13
sta $15 ;store value of register A at address hex 15
rts ;return
generateApplePosition:
;Th least significant byte of the apple position will determine where
;in a 8x32 strip the apple is placed. This number can be any one byte value because
;the size of one 8x32 strip fits exactly in one out of 256 bytes
lda $fe ;load a random number between 0 and 255 from address $fe into register A
sta $00 ;store value of register A at address hex 00
;load a new random number from 2 to 5 into $01 for the most significant byte of
;the apple position. This will determine in which 8x32 strip the apple is placed
lda $fe ;load a random number from address $fe into register A
;AND: logical AND with accumulator. Apply logical AND with hex $03 to value in
;register A. Hex 03 is binary 00000011, so only the two least significant bits
;are kept, resulting in a value between 0 (bin 00000000) and 3 (bin 00000011).
;Add 2 to the result, giving a random value between 2 and 5
and #$03 ;mask out lowest 2 bits
clc ;clear carry flag
adc #2 ;add to register A, using carry bit for overflow.
sta $01 ;store value of y coordinate from register A into address $01
rts ;return
loop:
;the main game loop
jsr readKeys ;jump to subroutine readKeys
jsr checkCollision ;jump to subroutine checkCollision
jsr updateSnake ;jump to subroutine updateSnake
jsr drawApple ;jump to subroutine drawApple
jsr drawSnake ;jump to subroutine drawSnake
jsr spinWheels ;jump to subroutine spinWheels
jmp loop ;jump to loop (this is what makes it loop)
readKeys:
;for getting keypresses, the last address ($ff) in the zero page contains
;the hex code of the last pressed key
lda $ff ;load the value of the latest keypress from address $ff into register A
cmp #$77 ;compare value in register A to hex $77 (W)
beq upKey ;Branch On Equal, to upKey
cmp #$64 ;compare value in register A to hex $64 (D)
beq rightKey ;Branch On Equal, to rightKey
cmp #$73 ;compare value in register A to hex $73 (S)
beq downKey ;Branch On Equal, to downKey
cmp #$61 ;compare value in register A to hex $61 (A)
beq leftKey ;Branch On Equal, to leftKey
rts ;return
upKey:
lda #4 ;load value 4 into register A, correspoding to the value for DOWN
bit $02 ;AND with value at address $02 (the current direction),
;setting the zero flag if the result of ANDing the two values
;is 0. So comparing to 4 (bin 0100) only sets zero flag if
;current direction is 4 (DOWN). So for an illegal move (current
;direction is DOWN), the result of an AND would be a non zero value
;so the zero flag would not be set. For a legal move the bit in the
;new direction should not be the same as the one set for DOWN,
;so the zero flag needs to be set
bne illegalMove ;Branch If Not Equal: meaning the zero flag is not set.
lda #1 ;Ending up here means the move is legal, load the value 1 (UP) into
;register A
sta $02 ;Store the value of A (the new direction) into register A
rts ;return
rightKey:
lda #8 ;load value 8 into register A, corresponding to the value for LEFT
bit $02 ;AND with current direction at address $02 and check if result
;is zero
bne illegalMove ;Branch If Not Equal: meaning the zero flag is not set.
lda #2 ;Ending up here means the move is legal, load the value 2 (RIGHT) into
;register A
sta $02 ;Store the value of A (the new direction) into register A
rts ;return
downKey:
lda #1 ;load value 1 into register A, correspoding to the value for UP
bit $02 ;AND with current direction at address $02 and check if result
;is zero
bne illegalMove ;Branch If Not Equal: meaning the zero flag is not set.
lda #4 ;Ending up here means the move is legal, load the value 4 (DOWN) into
;register A
sta $02 ;Store the value of A (the new direction) into register A
rts ;return
leftKey:
lda #2 ;load value 1 into register A, correspoding to the value for RIGHT
bit $02 ;AND with current direction at address $02 and check if result
;is zero
bne illegalMove ;Branch If Not Equal: meaning the zero flag is not set.
lda #8 ;Ending up here means the move is legal, load the value 8 (LEFT) into
;register A
sta $02 ;Store the value of A (the new direction) into register A
rts ;return
illegalMove:
;for an illegal move, just return, so the keypress is ignored
rts ;return
checkCollision:
jsr checkAppleCollision ;jump to subroutine checkAppleCollision
jsr checkSnakeCollision ;jump to subroutine checkSnakeCollision
rts ;return
checkAppleCollision:
;check if the snake collided with the apple by comparing the least significant
;and most significant byte of the position of the snake's head and the apple.
lda $00 ;load value at address $00 (the least significant
;byte of the apple's position) into register A
cmp $10 ;compare to the value stored at address $10
;(the least significant byte of the position of the snake's head)
bne doneCheckingAppleCollision ;if different, branch to doneCheckingAppleCollision
lda $01 ;load value of address $01 (the most significant byte
;of the apple's position) into register A
cmp $11 ;compare the value stored at address $11 (the most
;significant byte of the position of the snake's head)
bne doneCheckingAppleCollision ;if different, branch to doneCheckingAppleCollision
;Ending up here means the coordinates of the snake head are equal to that of
;the apple: eat apple
inc $03 ;increment the value held in memory $03 (snake length)
inc $03 ;twice because we're adding two bytes for one segment
;create a new apple
jsr generateApplePosition ;jump to subroutine generateApplePosition
doneCheckingAppleCollision:
;the snake head was not on the apple. Don't do anything with the apple
rts ;return
checkSnakeCollision:
ldx #2 ;Load the value 2 into the X register, so we start with the first segment
snakeCollisionLoop:
lda $10,x ;load the value stored at address $10 (the least significant byte of
;the location of the snake's head) plus the value of the x register
;(2 in the first iteration) to get the least significant byte of the
;position of the next snake segment
cmp $10 ;compare to the value at address $10 (the least significant
;byte of the position of the snake's head
bne continueCollisionLoop ;if not equals, we haven't found a collision yet,
;branch to continueCollisionLoop to continue the loop
maybeCollided:
;ending up here means we found a segment of the snake's body that
;has a least significant byte that's equal to that of the snake's head.
lda $11,x ;load the value stored at address $11 (most significant byte of
;the location of the snake's head) plus the value of the x register
;(2 in the first iteration) to get the most significant byte
;of the position of the next snake segment
cmp $11 ;compare to the value at address $11 (the most significant
;byte of the position of the snake head)
beq didCollide ;both position bytes of the compared segment of the snake body
;are equal to those of the head, so we have a collision of the
;snake's head with its own body.
continueCollisionLoop:
;increment the value in the x register twice because we use two bytes to store
;the coordinates for snake head and body segments
inx ;increment the value of the x register
inx ;increment the value of the x register
cpx $03 ;compare the value in the x register to the value stored at
;address $03 (snake length).
beq didntCollide ;if equals, we got to last section with no collision: branch
;to didntCollide
;ending up here means we haven't checked all snake body segments yet
jmp snakeCollisionLoop;jump to snakeCollisionLoop to continue the loop
didCollide:
;there was a collision
jmp gameOver ;jump to gameOver
didntCollide:
;there was no collision, continue the game
rts ;return
updateSnake:
;collision checks have been done, update the snake. Load the length of the snake
;minus one into the A register
ldx $03 ;load the value stored at address $03 (snake length) into register X
dex ;decrement the value in the X register
txa ;transfer the value stored in the X register into the A register. WHY?
updateloop:
;Example: the length of the snake is 4 bytes (two segments). In the lines above
;the X register has been set to 3. The snake coordinates are now stored as follows:
;$10,$11 : the snake head
;$12,$13,$14,$15: the snake body segments (two bytes for each of the 2 segments)
;
;The loop shifts all coordinates of the snake two places further in memory,
;calculating the offset of the origin from $10 and place it in memory offset to
;$12, effectively shifting each of the snake's segments one place further:
;
;from: x===
;to: ===
lda $10,x ;load the value stored at address $10 + x into register A
sta $12,x ;store the value of register A into address $12
;plus the value of register X
dex ;decrement X, and set negative flag if value becomes negative
bpl updateloop ;branch to updateLoop if positive (negative flag not set)
;now determine where to move the head, based on the direction of the snake
;lsr: Logical Shift Right. Shift all bits in register A one bit to the right
;the bit that "falls off" is stored in the carry flag
lda $02 ;load the value from address $02 (direction) into register A
lsr ;shift to right
bcs up ;if a 1 "fell off", we started with bin 0001, so the snakes needs to go up
lsr ;shift to right
bcs right ;if a 1 "fell off", we started with bin 0010, so the snakes needs to go right
lsr ;shift to right
bcs down ;if a 1 "fell off", we started with bin 0100, so the snakes needs to go down
lsr ;shift to right
bcs left ;if a 1 "fell off", we started with bin 1000, so the snakes needs to go left
up:
lda $10 ;put value stored at address $10 (the least significant byte, meaning the
;position in a 8x32 strip) in register A
sec ;set carry flag
sbc #$20 ;Subtract with Carry: subtract hex $20 (dec 32) together with the NOT of the
;carry bit from value in register A. If overflow occurs the carry bit is clear.
;This moves the snake up one row in its strip and checks for overflow
sta $10 ;store value of register A at address $10 (the least significant byte
;of the head's position)
bcc upup ;If the carry flag is clear, we had an overflow because of the subtraction,
;so we need to move to the strip above the current one
rts ;return
upup:
;An overflow occurred when subtracting 20 from the least significant byte
dec $11 ;decrement the most significant byte of the snake's head's position to
;move the snake's head to the next up 8x32 strip
lda #$1 ;load hex value $1 (dec 1) into register A
cmp $11 ;compare the value at address $11 (snake head's most significant
;byte, determining which strip it's in). If it's 1, we're one strip too
;(the first one has a most significant byte of $02), which means the snake
;hit the top of the screen
beq collision ;branch if equal to collision
rts ;return
right:
inc $10 ;increment the value at address $10 (snake head's least
;significant byte, determining where in the 8x32 strip the head is
;located) to move the head to the right
lda #$1f ;load value hex $1f (dec 31) into register A
bit $10 ;the value stored at address $10 (the snake head coordinate) is ANDed
;with hex $1f (bin 11111), meaning all multiples of hex $20 (dec 32)
;will be zero (because they all end with bit patterns ending in 5 zeros)
;if it's zero, it means we hit the right of the screen
beq collision ;branch to collision if zero flag is set
rts ;return
down:
lda $10 ;put value from address $10 (the least significant byte, meaning the
;position in a 8x32 strip) in register A
clc ;clear carry flag
adc #$20 ;add hex $20 (dec 32) to the value in register A and set the carry flag
;if overflow occurs
sta $10 ;store the result at address $10
bcs downdown ;if the carry flag is set, an overflow occurred when adding hex $20 to the
;least significant byte of the location of the snake's head, so we need to move
;the next 8x3 strip
rts ;return
downdown:
inc $11 ;increment the value in location hex $11, holding the most significatnt byte
;of the location of the snake's head.
lda #$6 ;load the value hex $6 into the A register
cmp $11 ;if the most significant byte of the head's location is equals to 6, we're
;one strip to far down (the last one was hex $05)
beq collision ;if equals to 6, the snake collided with the bottom of the screen
rts ;return
left:
;A collision with the left side of the screen happens if the head wraps around to
;the previous row, on the right most side of the screen, where, because the screen
;is 32 wide, the right most positions always have a least significant byte that ends
;in 11111 in binary form (hex $1f). ANDing with hex $1f in this column will always
;return hex $1f, so comparing the result of the AND with hex $1f will determine if
;the snake collided with the left side of the screen.
dec $10 ;subtract one from the value held in memory position $10 (least significant
;byte of the snake head position) to make it move left.
lda $10 ;load value held in memory position $10 (least significant byte of the
;snake head position) into register A
and #$1f ;AND the value hex $1f (bin 11111) with the value in register A
cmp #$1f ;compare the ANDed value above with bin 11111.
beq collision ;branch to collision if equals
rts ;return
collision:
jmp gameOver ;jump to gameOver
drawApple:
ldy #0 ;load the value 0 into the Y register
lda $fe ;load the value stored at address $fe (the random number generator)
;into register A
sta ($00),y ;dereference to the address stored at address $00 and $01
;(the address of the apple on the screen) and set the value to
;the value of register A and add the value of Y (0) to it. This results
;in the apple getting a random color
rts ;return
drawSnake:
ldx #0 ;set the value of the X register to 0
lda #1 ;set the value of the A register to 1
sta ($10,x) ;dereference to the memory address that's stored at address
;$10 (the two bytes for the location of the head of the snake) and
;set its value to the one stored in register A
ldx $03 ;set the value of the x register to the value stored in memory at
;location $03 (the length of the snake)
lda #0 ;set the value of the a register to 0
sta ($10,x) ;dereference to the memory address that's stored at address
;$10, add the length of the snake to it, and store the value of
;register A (0) in the resulting address. This draws a black pixel on the
;tail. Because the snake is moving, the head "draws" on the screen in
;white as it moves, and the tail works as an eraser, erasing the white trail
;using black pixels
rts ;return
spinWheels:
;slow the game down by wasting cycles
ldx #0 ;load zero in the X register
spinloop:
nop ;no operation, just skip a cycle
nop ;no operation, just skip a cycle
dex ;subtract one from the value stored in register x
bne spinloop ;if the zero flag is clear, loop. The first dex above wrapped the
;value of x to hex $ff, so the next zero value is 255 (hex $ff)
;loops later.
rts ;return
gameOver: ;game over is literally the end of the program
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