newfile 2ok.65s

在网上搜集的一些关于6502的资料,在win2000下调试通过了,有vb也有vc的程序,还有一些6502基础命令学习,对想学6502汇编的有一定的帮助,如何使用,详细看里面有说明文件:)

;==============================================================================
;                 Coding Examples for the Students of AER201S
;==============================================================================
;
;<===  Note the semicolon is a comment indicator
;
	.ORG $E000
	   SEI             ; INITIALIZING THE STACK POINTER
	   LDX #$FF
	   TXS
;
	   LDX #$00        ; Initial delay to allow spurious Resets to
	   LDY #$00        ; to subside.  At this point accept that this code
Delay      DEX             ; must appear at the beginning of every EPROM
	   BNE Delay       ; that will ever be used for the 6502 project board.
	   DEY
	   BNE Delay
;
;  We are now ready to try some simple examples
;
;
;  1) Add the numbers $65 and $27 and place the answer in the memory
;     location 'ANSWER'.
;
; First we must assign the location to the label  'ANSWER'
;
ANSWER = $0000
;
; OK now we do the addition:
	   LDA #$65       ; Load the number (#) $65 into the accumulator.
	   CLC            ; Clear the carry flag to prevent carries from a
			  ; previous and unrelated computation.
	   ADC #$27       ; Add the number (#) $27, the result goes in the
			  ; accumulator
	   STA ANSWER     ; Store the result into the memory location ANSWER
;
; Remarks:  Note that the # number sign preceded the numbers $65 and $27 to
;     indicate that these were in fact CONSTANTS, NUMBERS.  If the # sign were
;     omitted, TASM would interpret LDA $65 to mean 'load the accumulator from
;     the memory location $0065'.
;           Looking back at the assignment of ANSWER, we could have said:
;     ANSWER = $00.  This, when used in the statement STA ANSWER would still
;     be interpreted as referring to the memory location $0000 but it would
;     allow the 6502 to use something called 'Zero Page' addressing, a faster
;     method of accessing the memory locations between $0000-$00FF.
;
;
;
; 2) Subtract the number $27 from the number $65 and store the result in the
;    memory location 'ANS2'.
;
ANS2 = $01   ; We will use zero page addressing to refer to ANS2
;
	   LDA #$65       ; Load the number (#) $65 into the accumulator
	   SEC            ; Clear the C flag (used as the borrow flag in
			  ; subtraction) to prevent borrows from a previous
			  ; unrelated computation.
	   SBC #$27       ; Subtract the number (#) $27 from the accumulator
			  ; the result will be in the accumulator.
	   STA ANS2       ; Store the result in the memory location 'ANS2'
;
; Remarks:  Note that the carry or borrow flag has a inverse logic in the case
;     of subtraction: You have to SEC set the carry flag to prevent a borrow.
;
;
;
; 3)  Program a subroutine to turn on specific bits in on VIA 1, Port A which
; can be used to light LEDs  to  monitor  program  performance.   We  will
; assume that there are 8 LEDs connected to Port A.
;
; We must first initialize port A as an output port
;
	  ; LDA #%11111111     ; All ones means all bits are outputs
	   STA $A003          ; $A003 is the address of VIA1 Port A, data
			      ; direction register.  This register determines
			      ; whether a bit is input or output.
	 ;  LDA #%00000000     ;
	   STA $A001          ; $A001 is the address of VIA1 Port A I/O
			      ; register.  It is initialized to all zeros
			      ; to mean all bits output 0 volts.
 
; Before we proceed with writing the subroutine we need to send the program
; to a point somewhere after the subroutine we are about to write.  We will
; use a JuMP statement:
;
	   JMP Down_Past  ; Down_Past is a program position label we will
			  ; define later.
;
;
;
;
; ============================================================================
;  It is a good idea to put these bars in to make subroutines obvious
; ============================================================================
;
; Ok know we need to have some Specifications for this subroutine:
;
;  Subroutine Name:  Show
;  Input Registers:  .A  contains a '1' in the bit position that you want to
;                        have turned on
;  Output Registers: None
;
Show       ORA $A001      ; 'Show' against the margin indicates a program
			  ; position label.  ORA $A001 takes the bit pattern
			  ; in .A and logically OR's it with the pattern of
			  ; bits in memory location $A001, the VIA 1, Port A
			  ; I/O register.
;
; Say $A001 contains the bits:  %abcdefgh  (a through h are unknown 0 or 1)
; and we or it with .A :     OR %00001000
;                           -------------
; We will get:                  %abcd1fgh
;
; Because:   0 or 0 = 0,  1 or 0 = 1,  0 or 1 = 1,  1 or 1 = 1
;   the following is true:  u or 0 = u   and   u or 1 = 1
;   where u is an unknown level.
;
	   STA $A001      ; Store the new bit levels back into the I/O Port
	   RTS            ; Return from Subroutine
;
; Remarks:  This subroutine can be called by setting specific bits in the .A
;   register and calling the subroutine:
;          LDA #%00000001  ; Set bit 0
;          JSR Show        ; Show this on the VIA port
;
;           Subroutines can be nested IE one subroutine can call others.  Care
;   should be exercised when one subroutine calls itself because there are no
;   'local' variables in assembly language (unless you simulate them) since
;   the memory in the 6502 is global to all programs or subroutines.
;   Subroutines can significantly compact a program and can lend to readability
;   of the code if properly documented.
;
;
;
Down_Past          ; This is the program position label that marks where the
		   ; JMP Down_Past goes to.
;
;
; 4) Now that we have done an output subroutine, lets do some comparisons and
;    light leds to show the results.  Check if ANSWER - ANS2 = 2*#$27.
;
RHS = $03         ; Stores the value of the right hand side of the equation
;
	   LDA ANSWER    ; Calculate the RHS
	   SEC
	   SBC ANS2
	   STA RHS
;
	   LDA #$27      ; To calculate 2*#$27 you use #$27+#$27
	   CLC
	   ADC #$27      ; Got the LHS in the accumulator
;
	   CMP RHS
	   BNE Not_Equal ; Don't Show any LED's
;
; The equation is true, light LED connected to bit 0
	;   LDA #%00000001
	   JSR Show
;
Not_Equal
;
;
;
; 5)  See if ANSWER and ANS2 have any bits in common;  light LED 1 if they do.
;
	   LDA ANSWER
	   AND ANS2      ; AND Tests to see if a bit in both arguments is
			 ; set.  If it is the resulting bit is set, otherwise
			 ; the resulting bit is cleared.
			 ; If two arguments have no bits in common the result
			 ; will have no bits set or will be zero.  In this case
			 ; the AND instruction will set the zero flag.
	   BEQ No_Bits   ; The BEQ branches on result equal for comparisons
			 ; or result zero for arithmetic and logical
			 ; OpCodes.
	;   LDA #%00000010
	   JSR Show      ; This turns on bit 1 if there were any bits in common
;
; There were no bits in common, the AND resulted in a Z=1, zero
No_Bits
;
;
;
; 6) See if ANSWER and ANS2 have bit #1 in common; light LED 2 if they do.
;
	   LDA ANSWER
	;   AND #%00000010  ; This picks off the value of bit 1.  You can think
			   ; of a 1 in a constant of an AND statement like a
			   ; delta function that zeros picks off one value
			   ; and zeros the rest of the domain.  As a result
			   ; ANDing with #%11111111 does not change a number.
	   AND ANS2        ; We know that the .A has all zeros except possibly
			   ; bit 1.  ANDing with ANS2 will result in zero
			   ; unless bit 1 was set in both ANSWER and ANS2
	   BEQ Bit1_Not_Set  ; If the result is Z=1, don't show LEDs
	 ;  LDA #%00000100  ; otherwise Show LED 2
	   JSR Show
;
Bit1_Not_Set
;
;
;
; 7) Do an endless loop to alternate the pattern of indicators now present
; on the LEDs with the binary representation of ANSWER.
;
Endless    LDX $A001       ; Grab the present bit pattern
	   LDA ANSWER      ; Grab the new bit pattern
	   STA $A001       ; Put the new pattern on the Port
	   STX ANSWER      ; Put the old pattern into ANSWER
;
	   LDX #$00
	   LDY #$00
D1         DEX             ; A double indexed wait loop to waste time
	   BNE D1
	   DEY
	   BNE D1
	   JMP Endless     ; Next time the old bit pattern becomes the new
			   ; by  the  way it was stored, rotating through
			   ; the ANSWER==>.A==>Port==>.X==>ANSWER loop.
;

 
; The last entry in every program must  set  the  reset  vector  to  point
; to  the  beginning  of  the  program.  If there is an interrupt routine
; then the interrupt vector must be set as well.  In this  case  there  is
; only a main program.
;
	.ORG $FFFC                 ; Set pointer to Reset Vector Location
	.WORD $E000                ; Store $E000 as the reset vector
	.END