baseConverter.s

* m68k Binary Search, Michael Green
* ----------------------------------------------------------------------
* 
ORG     $0
DC.L    $3000           *  Stack pointer value after a reset
DC.L    start           *  Program counter value after a reset
ORG     $3000           *  Start at location 3000 Hex
* 
* ----------------------------------------------------------------------
* 
#minclude /home/ma/cs237/bsvc/iomacs.s
#minclude /home/ma/cs237/bsvc/evtmacs.s
* 
* ----------------------------------------------------------------------
* 
*  Register use
*  D1 = LENGTH OF BUFFER 1
*  D2 = NUMBER TO CONVERT IN ASCII, THEN QUOTIENT
*  D3 = LENGTH OF BUFF 2, THEN REMAINDER
*  D4 = TWOS COMPLIMENT NUMBER TO CONVERT
*  D5 = ASCII BASE
*  D6 = BASE
* 
* ----------------------------------------------------------------------
*

start:  	initIO
setEVT
			lineout title
			lineout skipln
			lineout p1

*  Read what number we need to convert to the base
*  we don't need to convert it yet
*  because we can test things with ascii

in1:		linein	buff1
			move.l	D0,D1       *  Put the length into D1 so we can save it for later
			tst.l	D1        
			BEQ		eb1			*  Jump to error if D1 is zero
			subq	#1,D1
			lea		buff1,A1

* Now we need to check if all the digits are in base
* 10 and if they aren't we need to return an error
* We can cycle through the addresses of each ascii
* character and compare it to the imediate values
* of ascii characters 0-9 if not return error

check1:		move.b (A1)+,D2    	*  start cycle check if above ASCII 30
			cmpi.b #$30,D2
			BLT    eb1    
check2:		cmpi.b #$39,D2      *  check if below ASCII 39
			BGT    eb1
			dbra   D1,check1
			addq   #1,D1

* At this point we know there are digits in the buffer
* and they are in base 10, now we need to check
* if the input is in the correct range max number
* we can convert is 65536 in base 10

check3:		cvta2  buff1,D0     *  lets put it in twos compliment
			move.l D0,D4        *  lets save the value for future use in D4    
			cmpi.l #0,D0        
			BLT    eb5          * if value is less than 0 then we need to go to error
			cmpi.l #65535,D0
			BGT    eb5          * if over 2^16 then we need to go to error

*  Finished checking if the first input is correct
*  We can now move onto the second input    

lineout		p2
in2:		linein	buff2
			move.w	D0,D3
			tst.w	D3
			BEQ		eb2
			subq.w  #1,D3
			lea    	buff2,A2

check1a:	move.b  (A2)+,D5    * Same checks as the first check
			cmpi.b  #$30,D5
			BLT    	eb2
check1b:	cmpi.b  #$39,D5
			BGT     eb2
			dbra    D3,check1a
			addq    #1,D3

* Now instead we need to check for 2-16 rather than
* 0-2^16

check1c:	cvta2   buff2,D0
			move.l  D0,D6
			cmpi.w  #2,D0       *  check if base is in 2-16
			BLT     eb2
			cmpi.w  #16,D0         
			BGT     eb2

			lea    	temp,A5     * We have to load this now because
			clr.l  	D1          * We can't have it being in the loop
			BRA    	go          * Gotta skip the carry the first time we run it


carry:		move.l  D2,D4
go:			divu.w  D6,D4       * Store quotient and remainder in D4
			clr.l   D2          * get rid of any junk in D2 we don't need it
			move.w  D4,D2       * put quotient in D2
			swap    D4
			cmpi.w  #10,D4        
			BGE     great       * branch if >= 10 else go to less


less:		addi.w  #$30,D4
			move.b  D4,(A5)+
			addi.w  #1,D1
			tst.l   D2
			BNE     carry
			BEQ     reverse


great:		addi.w  #$37,D4
			move.b  D4,(A5)+
			addi.w  #1,D1
			tst.l  	D2
			BNE     carry
			BEQ     reverse


reverse:	lea     ans,A6
reverse2:	subq    #1,D1
			move.b  -(A5),(A6)+
			cmpi.w  #0,D1
			BNE     reverse2
			move.b  #0,(A6)     * null terminating the string
			BRA     answer


eb1:  	  	lineout e1
			BRA    in1
eb2: 	   	lineout e3
			BRA    in2
eb3:	    lineout e2
			BRA    in2
eb4:    	lineout e3
			BRA    in2
eb5:	    lineout e2
			BRA    in1

answer:		lineoutanstxt


break
* 
* ----------------------------------------------------------------------
*        Storage declarations

title: 		dc.b    'Base Converter, Michael Green',0
skipln:		dc.b    0
buff1: 		ds.b    80
buff2:		ds.b    80
bufft: 		ds.b    80
p1:    		dc.b    'Enter a base 10 number:',0
p2:    		dc.b    'Enter the base to convert to:',0
anstxt:		dc.b    'The answer is: '
ans:   		ds.b    20
temp:  		ds.b    20
e1:    		dc.b    'Error: Enter a number that is in base ten.',0
e2:    		dc.b    'Error: Only input a number between 0 and 65535',0
e3:    		dc.b    'Error: Only input a base ten number between 2 and 16.',0


end