Memorial University of Newfoundland
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1. TITLE Subtracting ASCII Numbers PAGE 60 132 STSEG SEGMENT DB 64 DUP STSEG ENDS DTSEG SEGMENT DATA1 DB 3546882164 DATA2 DB 2345611245 RESULT DB 10 DUP FJES DATA BCD DB 5 DUP DATA2 BCD DB 5 DUP RESULT BCD DB 5 DUP DTSEG ENDS CDSEG SEGMENT MAIN PROC FAR ASSUME CS CDSEG DS DTSEG SS STSEG MOV AX DTSEG MOV DS AX MOV BX OFFSET DATAI MOV DI OFFSET DATA BCD MOV CX 10 CALL CONVERT BCD MOV BX OFFSET DATA2 MOV DI OFFSET DATA2 BCD MOV CX 10 CALL CONVERT BCD CALL SUBTRACTION MOV SI OFFSET RESULT BCD MOV DI OFFSET RESULT MOV CX 5 CALL CONVERT ASC MOV AH 4CH INT 21H MAIN ENDP SUBROUTINE CONVERT ASCII NUMBERS TO BCD NUMBERS CONVERT BCD PROC NEAR REPO MOV AX BX HG AH AL D AX OFOFH USH CX V CL 4 L AH CL AL AH V DI AL DD BX 2 NC DI P CX OOP REPO RET CONVERT BCD ENDP T Zoa O VT O U H P O U x UP x O T SUBROUTINE PERFORM PACKED BCD NUMBER SUBTRACTION SUBTRACTION PROC NEAR MOV BX OFFSET DATA1 BCD MOV DI OFFSET DATA2 BCD MOV SI OFFSET RESULT BCD MOV CX 5 CLC REPL MOV AL BX 4 SBB AL DI 4 DAS MOV SI 4 AL DEC BX DEC DI DEC SI OOP REP1 RET SUBTRACTION ENDP SUBROUTINE CONVERT BCD NUMBERS TO ASCII NUMBERS2. CONVERT ASC PROC NEAR REP3 MOV AL SI AL D AX OFOOFH CX CL 4 AH CL AX 3030H HG AH AL V DI AX CSI D DI 2 P CX OOP REP3 RE CONVERT ASC ENDP O lt D ja al ocz n T V R N lt UPHEKXONK UD 02Z0Q O CDSEG ENDS ND MAIN m Write a program that converts an ASCII string saved by Old String to its uppercase in ASCII and save back to the New String Leave the space and period unchanged The Data Segment is defined as following DTSEG SEGMENT Old String DB This is THE String to be converted New String DB 35 DUP DTSEG ENDS 7 CI TITLE COV PAGE 60 132 STSEG SEGMENT DB 64 DUP STSEG ENDS RT LOWER CASE TO UPPER CASE DTSEG EGMENT d String DB This is THE String to be converted eString DB 35 DUP YS Z Os n MAIN PROC FAR ASSUME CS CDSEG DS DTS MOV AX DTSEG MOV DS AX Q N SS STSEG MOV SI OFFSET Old String MOV BX OFFSET New String MOV CX 35 REPO MOV AL SI CMP AL 61H IF LESS THAN a THEN EXIT JB OVER CMP AL JAH IF GREATER THAN z THEN EXIT A OVER ND AL 11011111B MASK d5 TO CONVERT TO UPPER CASE OV BX AL NC SI NC BX LOOP REPO Q D T OVER H H x
3. E MOV AH 4CH INT 21H MAIN ENDP CDSEG ENDS END MAIN
4. mov ds ax mov es ax in al DIPS mov ah al and al OFh call Convert xchg ah al mov cl 4 shr al cl call Convert xchg ah al int 6 cmp al 9 ja Letter add al 30h jmp Done add al 37h ret ENDS END Main ds myseg es myseg read switch values copy values mask out upper 4 bits subroutine to convert AL to ASCII Swap AH and AL for Convert Shift AL to right by 4 Convert AL to ASCII Swap AH and AL back Finished KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK KKKKKKKAKA K Convert converts AL to ASCII KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKAK and stores resulting byte in AL Is AL above 9 Yes so AL is a letter ASCII 30h to 39h are numeric ASCII 37h Ah 41h and ASCII 41h to 46h are A to F 5 Write a program that subtracts two multi digit ASCII numbers Datal Data2 The result should be saved back to Result in ASCII The Data Segment is defined as following DTSEG Datal DB Data2 DB Result DB DTSEG SEGMENT 3546882164 2345611245 10 DUP ENDS The approach I used is first convert the ASCII numbers to packed BCD numbers also stored in memory then perform multi byte packed BCD number subtraction result also stored in memory finally convert result to ASCII and save them to the location RESULT as required All these functions are placed in subroutines
5. 8 instruction set Write a sequence of instructions that function equivalently to PUSH DX You may use any other valid instruction but restore any registers you change that PUSH DX does not First read what the push instruction does page 3 132 it decrements SP by two and then moves the source to the memory location given my SS SP The problem is that there is no addressing mode that allows you to use SP directly The effective addresses allow different combinations of BP BX SI and DI We ll use BP as the microprocessor automatically uses the SS as the segment However we must be careful to not lose the existing value in BP Attempt 1 not guite right SUB SP 2 decrement stack pointer XCHG BP SP save BP and use value in SP MOV BP DX move data to memory at SS BP XCHG BP SP restore BP and SP This is not bad but unfortunately the SUB instruction modifies a number of flags and PUSH does not modify any flags A correct method is to use an instruction we did not look at in class LEA load effective address Read page 3 114 of the Intel User s Manual for details Essentially you give the source as a valid memory reference but the offset not the value is placed into the destination It affects no flags and uses no push instructions of any kind Attempt 2 XCHG BP SP save BP get SP LEA BP BP 2 set BP to new memory location MOV BP DX move data to memory XCHG BP SP restore BP and SP Other solut
6. Memorial University of Newfoundland Engineering 4862 MICROPROCESSORS Assignment 3 Solution 0 For each of the CF PF ZF SF and OF flags briefly describe the meaning when it is set Give conditional jump instructions that can be used to test each one Flag Description when set Conditional Jumps CF High order bit carry or borrow JC JNC JAE JNB JB JNAE PF L ow order 8 bits or result contain even number JP JPE JNP JPO of 1 bits ZF Result is zero JE JZ JNE JNZ SF Result is positive if a signed number JS JNS OF Signed result cannot be expressed within the JO JNO number of bits of destination operand Notes a The description for CF states a high order bit carry because the bit examined depends on whether the operands are 8 bit or 16 bit On the addition of two 8 bit numbers CF is set if there is a carry out of bit 7 where the bits are numbered from 0 to 7 For two 16 bit numbers the examined bit is number 15 b The microprocessor does not know if you are adding or subtracting two numbers that are signed or unsigned Thus OF is set or cleared whether or not signed numbers are used Since the range for an 8 bit signed number is from 128 to 127 if two 8 bit numbers are added to get a value greater than 127 or less than 128 you will have to write code to convert the result to a 16 bit signed number 1 a Determine the contents of register BX and the six conditional status fl
7. ags after each of the following instructions executes If a flag or register contents are unknown indicate with a CLC MOV BL 4DH SUB BL 3EH XOR BH BH MOV SI BX BX CF PF AF ZF SF OF CLC 2 0 MOV BL 4DH 2 4D 0 SUB BL 3EH 0F 0 1 1 0 0 0 XOR BH BH 000F 0 1 1 0 0 MOV SI BX 000F 0 1 1 0 0 Notes XOR automatically sets the CF and OF to 0 It does not matter what values is in BH as XORing will set the result to 0 4DH 0100 1101 3EH 0011 1110 0000 1111 AF is set to 1 because there is a borrow from high order 4 bits to the low order 4 bits between bits 3 and 4 CF is set to 0 because there is no borrow into the high order bit bit 7 b Read how to use debug in appendix A of the textbook For each instruction in a use DEBUG or some other program to determine the equivalent machine code The machine code is given in bold in the following capture from DEBUG Note that entering numbers in DEBUG automatically defaults to hexadecimal This can be seen in the first MOV instruction as the 4D is automatically translated into 4D as machine code 12A7 0100 F8 CLC 12A7 0101 B34D MOV BL 4D 12A7 0103 80EB3E SUB BL 3E 12A7 0106 30FF XOR BH BH 12A7 0108 891C MOV SI BX The 8086 88 user manual also tells you how many bytes that the instruction will be turned into and what machine code for each instruction Assume that the PUSH instruction does not exist in the 8086 808
8. gorithm is a simple repetitive addition Do_again Skip hi Done PUSH BX save registers and flags PUSH CX PUSHF MOV DX O clear DX amp AX used to store product MOV BX AX use BX to count the number of additions MOV AX O JCXZ Done If multiplier is zero so is the product ADD AX CX JNC Skip hi INC DX Carry out from AX so increment DX DEC BX JNZ Do again continue BX times POP CX Get flags Reset all flags except TF IF and DF AND CX 0000011100000000b CMP DX O is DX zero JE Set flags yes so leave OF amp CF as 0 DX is not zero so OF CF 1 OR CX 0000100000000001b set flags PUSH CX store flags POPF restore correct flag values POP CX restore registers POP BX RET Done 4 Write a MUN 88 compatible program that reads the contents of the DIP switches and then converts the 8 bit decimal value into two 8 bit ASCII values representing each hex digit Store the lower digit in AL and the upper digit in AH This should be a full program so include a title segment definitions etc as well as comments Example result Suppose that after reading the input port for the DIP switches AL is 9FH Your program should place 39H ASCII for 9 into AH and 46H ASCII for F into AL DIPS TITLE myseg Main Convert Letter Done myseg equ 30h DIP Converter SEGMENT ASSUME cs myseg mov ax cs
9. ions are trickier to implement If you use SUB then you must figure out how to restore the flags to their original values Note that PUSHF and POPF are available to your use XCHG BP SP SUB SPy 2 POPF restore SP and BP change SP to original SP 4 restore flags SP orig SP 2 Attempt 3 PUSHF save flags at original SP 2 PUSHF save flags again at orig SP 4 ADD SP 2 change SP to original SP 2 XCHG BP SP Swap SP and BP MOV BP DX save DX This final attempt is the simplest of all so far Use PUSHF to update SP but then overwrite the flags with DX because none of the flags will actually be changed Attempt 4 PUSHF save flags on stack XCHG BP SP swap SP and BP SP has been updated MOV BP DX save DX to SS BP XCHG BP SP restore SP and BP Write a subroutine to replace the multiplication instruction MUL CX You may use any valid 8086 8088 instructions other than MUL but take care to properly handle the flags and restore any registers that you use to store temporary values Start your subroutine with the label mul cx and end with the RET instruction MUL CX Notes on implementation Source CFand AF PF is a word CX so result will be placed in DX high word and AX low word OF are either set if DX is non zero or cleared DX is zero SF and ZF are undefined but we ll set them to zero We ll take care not to affect the other 3 flags TF IF and DF The following al
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