Engineering 4862 MICROPROCESSORS
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1. CURSOR PROC NEAR MOV AH 02 MOV BH 00 MOV DL 00 columns MOV DH 00 TOWS INT 10H RET CURSOR ENDP CDSEG ENDS END MAIN2. O00 it is R M field and represent AL ROL AL 1 E8 CALL Near proc Indicates that there is a call to a subroutine in the same segment The next two bytes IP INC Lo and IP INC Hi give a signed 16 bit displacement from the current value of IP Disp E2C8 negative 1110001011001000b gt 0001 1101001 11000b 1D38H 7480D CALL IP 7480 F9 STC 4 Usefull segment definition write a DOS compatible program that a clears the screen b set the cursor to screen position row 10 and column 5 c displays the prompt Please enter an 8 digit number d get the keyboard input and save the number to a buffer area in the memory you define e sort the number on its ascending order and save them to another buffer for display For example if the input number is 29034765 then after your sort the result should be 02345679 You can assume that the number for each digit is non repeat but actually the repeated case is the same f after your sort change to the start of next new line output The sorted number is and the number g exit use DOS function 4CH Write task a and b using subroutines Test your code on PC by yourself Tasks 1 Clear the screen use subroutine 2 Set the cursor to ROW 10 and COLUMN 5 on the screen use subroutine 3 Output a prompt string Please enter an 8 digit number 4 Accept keyboard input put to buffer INPUT_BUF 5 Sort the number on its ascending order 6 mov
3. 78h 2 Convert the following instructions to machine code give your answers in hexadecimal State what each of the bit fields mean i PUSH BX MOV SI 490 SP OUT DX AL POPF AND AX BX DI 2Dh ADD DS BP DX Note you will have to add a displacement XOR AL BX DI 36H MOV DI 476 ES 5 o B 8B FP wu a PUSH BX two possible answers Memory or Register Operand 11111111 mod 110 r m mod 11 Register Mode r m is treated as a reg field r m 011 Register BX Answer 1 FF F3 Register Operand 01010 reg reg 011 Register BX Answer 2 53 b MOV SI 490 SP Memory or Register Operand to from Register Operand 100010 d w mod reg r m disp lo disp hi d 0 From register w l Word operands SP is a 2 byte register mod 10 Memory Mode 16 bit displacement follows reg 100 Register SP r m 100 EA SI D16 disp 1EAh Displacement 490 1EAh Answer 89 A4 EA 01 c OUT DX AL Variable Port 1110111w w 0 Byte operand AL is a 1 byte register Answer EE d POPF 10011101 Answer 9D e AND AX BX DI 2Dh Memory or Register Operand with Register Operand 001000 d wl mod reg r m disp lo d To register w Word operands AX is 2 bytes large mod 01 Mem
4. ALL MemError Error call subroutine Exit INT 6 Notes I stated to at least one person that loading register CX with 0 before execution of the REP STOSB instruction would provide for automatically storing to 65536 64K locations However that is not the case if you try the above program in DEBUG with CX set to 0 then STOSB will not execute at all This corroborates the description of REP in the Intel Handbook which states that REP adds a do while structure to string operation On the other hand LOOP functions differently It first decrements CX and then checks to see if it is 0 If not it jumps Thus if CX is 0 then LOOP will jump 65536 times as required by the question In your implementation you do not need to call INT 6 after you call the subroutine 1 Read Intel 8086 88 User Manual page 6 53 to 6 55 understand how the instruction is encoded to and decoded from machine code Use the two tables that followed table 6 22 and 6 23 do the following two questions Convert the following hexadecimal machine codes to assembly language mnemonics State what each of the byte fields mean Table 6 23 from page 6 64 to page 6 69 a B8 00 20 b 8E D8 c 46 d 90 e 89 7C FE f 75 F7 g E2 EF h 26 80 07 78 a B8 00 20 B8 MOV AX IMMED16 00 Data lo 20 Data hi Answer MOV AX 2000h b 8E D8 8E MOV SEGREG REG16 MEM16 D8 2 byte MOD 0 SR R M 11 0 11 000 MOD 11 Register Mode
5. Memorial University of Newfoundland Engineering 4862 MICROPROCESSORS Assignment 4 Solution Unless otherwise noted please show all relevant calculations and explain your answers where appropriate 0 Write a sequence of instructions that is not a full program with TITLE ASSUME etc that checks whether the memory in a full 64KB segment is functioning correctly Assume that register AX contains the segment to be checked Use the string instruction STOSB to load the byte AAH into all 64K locations and then use LODSB to check that AAH was correctly stored at each of the locations If the wrong value is read then call a fictitious subroutine MemError Otherwise call Function 4CH of DOS INT 21H to exit Instruction sequence to check 64KB of memory using STOSB and LODSB Initialize MOV DS AX Set segments MOV ES AX CLD Set direction flag for increments Store 64K byte values to memory MOV DI 0 Set starting offset MOV AL OAAh Set test value MOV CX OFFFFh Set for 64K 1 locations REP STOSB Store test value STOSB Extra store for total of 64KB Check that values have been stored correctly MOV SI 0 Set starting offset MOV AH OAAh Set compare pattern MOV CX 0 Set maximum loops 64KB Again LODSB Load value from memory CMP AH AL Is value valid JNE HandleError No there is an error in memory LOOP Again Continue comparing JMP Exit All loaded values are valid HandleError C
6. Register Operand with Register Operand 001100 d w mod reg r m disp lol d 1 To register w 0 Byte operands AL is 1 byte large mod 01 Memory Mode 8 bit displacement follows reg 000 Register AL r m 001 BX DI D8 disp Displacement is 36h which can fit in an 8 bit 2 s complement form 5 36h 0011 0110b gt 2 s comp gt 11001010b CAh disp Answer 32 41 CA h MOV DI 476 ES Segment Register to Memory or Register Operand 10001100 mod 0 reg r m disp lo disp hi mod 10 Memory Mode 16 bit displacement follows Note that 476 is too large for an 8 bit signed number reg 00 Register ES note that this is a segment register and thus is 2 bits large r m 101 DI D16 disp 01DC Displacement is 47619 1DCi6 Answer 8C 85 DC 01 3 The following bytes are found in order somewhere in memory Assuming they are machine codes decode the values into meaningful assembly language mnemonics B9 00 12 DO CO E8 C8 E2 F9 B9 MOV CX ImmeD 16 The next two bytes are the immediate 16 bit value loaded into CX 00 12 gt 1200H MOV CX 1200H DO One of eight possibilities ROL ROR RCL etc so use the next byte CO 11000000 The first two digit MSB 11 is MOD and 11 means that r m reg field The next three digits O00 indicates that ROL Reg8 1 The last three digits LSB is
7. SR 11 Segment register DS see page 3 57 operand 1 R M 000 Register AX operand 2 Answer MOV DS AX c 46 46 INC SI Answer INC SI d 90 90 NOP Answer NOP e 89 7C FE 89 MOV REG16 MEM16 REG16 note error in table 6 23 7C 2 byte MOD REG R M 01 111 100 MOD 01 Memory Mode 8 bit displacement follows REG 111 DI operand 2 R M 100 SD D8 operand 1 FE 8 bit signed displacement 11111110 2 s complement 00000010 2 Thus the displacement is 2 Answer MOV SI 2 DI f 75 F7 75 JNE JNZ short label F7 IP INC8 8 bit signed offset to add to IP 11110111 2 s complement 00001001 9 Thus the offset is 9 Answer JNE 9 that is jump back 9 machine code bytes if not equal or JNZ 9 g E2 EF E2 LOOP short label EF IP INC8 8 bit signed offset to add to IP 11101111 2 s complement 00010001 17 Thus the offset is 17 Answer LOOP 17 loop back 17 machine bytes h 26 80 07 78 26 Segment override ES 80 One of several choices look at bits 3 4 amp 5 of next byte 07 00000111 Bits 3 4 amp 5 are 000 so instruction is ADD REG8 MEM8 IMMED8 MOD 00 Memory mode no displacement follows R M 111 BX 78 8 bit immediate value Note instruction must have BYTE PTR to indicate an 8 bit operation Answer ADD BYTE PTR ES BX
8. e the sorted number to display buffer OUTPUT_BUF 7 Change to a new line and Output string The sorted number is and the number and then exit 6 TITLE Example PAGE 120 60 LF EQU ODH CR EQU OAH STSEG SEGMENT DB 64 DUP STSEG ENDS DTSEG SEGMENT PROMPT1 DB Please enter an 8 digit number PROMPT2 DB LF CR The sorted number is LF CR INPUT_BUF LABEL BYTE SIZE_IBUF DB 09H INUMBER DB 00H INPUT_DATA DB 9 DUP OFFH DTSEG ENDS CDSEG SEGMENT MAIN PROC FAR ASSUME CS CDSEG DS DTSEG SS STSEG ES DTSEG MOV AX DTSEG MOV DS AX MOV ES AX CALL CLEAR CALL CURSOR MOV AH 09H Prompt digit inputs MOV DX OFFSET PROMPT1 INT 21H MOV AH OAH get inputs MOV DX OFFSET INPUT_BUF INT 21H MOV BX 7 sort the number using natural sort method MOV SI OFFSET INPUT_DATA REPO MOV DI SI ADD DI 1 MOV CX BX REPI MOV AL SI MOV AH DI CMP AH AL JA CONTO MOV SI AH MOV DI AL CONTO INC DI LOOP REP1 INC SI DEC BX JNZ REPO MAIN 2 MOV AH 09H send the output prompt MOV DX OFFSET PROMPT2 INT 21H MOV DX OFFSET INPUT_DATA sprepare for the sorted sequence output MOV BX DX MOV BYTE PTR BX 8 MOV AH 09H output the sorted sequence INT 21H MOV AH 4CH INT 21H ENDP apace tates CLEAR THE SCREEN CLEAR PROC NEAR MOV AH 06 MOV AL 00 MOV BH 07 MOV Cx 0000 MOV DX 184FH INT 10H RET CLEAR ENDP s Sonte aes MOVE CURSOR TO THE POSITION
9. ory Mode 8 bit displacement follows 2Dh reg 000 Register AX r m 001 BX DI D8 disp 2D Displacement is 2Dh which can fit in a 1 byte signed number Answer 23 41 2D f ADD DS BP DX Segment override It is a bit tricky finding the prefix byte for segment overrides If you look in Table 6 22 of the Intel User s Manual on page 6 61 you will see that the last entry is SEGMENT Override prefix 001 reg 110 The reg field is actual a segment register field and you can use the Segment column of the reg Field Bit Assignments chart on page 3 57 to determine how to set it Segment DS Override 001 11 110 3E Note you will have to add a displacement The addition of a displacement to the memory reference that is BP is needed because there is no encoding for BP Logically the mod field should be 00 and the r m field should be 110 But that is a special case for when a direct address is used something like 1000h To encode you will have to rewrite the instruction into the functionally equivalent form ADD DS BP 0 DX Memory or Register Operand with Register Operand 000000 d wl mod reg r m disp lo d 0 From register w l Word operands DX is 2 bytes large mod 01 Memory Mode 8 bit signed displacement follows reg 010 Register DX r m 110 BP D8 disp 00 Displacement is 0 Answer 3E 01 56 00 g XOR AL BX DI 36H Memory or
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