8086 Instructions hex code

MNEMONICS HEX CODE MNEMONICS HEX CODE
MOV [SI], CX 89 0C INC AX 40
MOV [SI], DX 8914 INC BX 43
MOV SI, 2000H BE 00 20 INC CX 41
MOV DI, 2000H BF 00 20 INC DX 42
MOV AX, [DI] 8B 05 INC SI 46
MOV BX, [DI] 8B 1D INC DI 47
MOV CX, [DI] 8B 0D DEC AL FE C8
MOV DX, [DI] 8B 15 DEC BL FE CB
MOV [DI], AX 89 05 DEC CL FE C9
MOV [DI], BX 89 1D DEC DL FE CA
MOV [DI], CX 89 0D DEC AH FE CC
MOV [DI], DX 89 15 DEC BH FE CF
MOV AX, 2001 B8 01 20 DEC CH FE CD
MOV BX, 2001 BB 01 20 DEC DH FECE
MOV [2001], AX A3 01 20 DEC AX 48
MOV [2001], BX 89 1E 0120 DEC BX 4B
MOV [2001], CX 89 0E 0120 DEC CX 49
MOV [2001], DX 89 16 0120 DEC DX 4A
MOV [2001], SI 89 36 0120 DEC SI 4E
MOV [2001], DI 89 3E 0120 DEC DI 4F
INC AL FE C0 MOV CX,16 bit B9 04 00
INC BL FE C3 MOV SI, 16 bit BE 00 30
INC CL FE C1 MOV DI, 16 bit BF 16 bit
INC DL FE C2 MOV AX [SI] 8B 04
INC AH FC C4 ADC [DI], AX 11 05
INC BH FC C7 MOV AH,00 B4 00
INC CH FC C5 MOV AH[SI] 8A 24
INC DH FC C6 MOV [SI],AH 88 24
MNEMONICS HEX CODE MNEMONICS HEX CODE
MOV BX [SI] 8B 1C JNZ (a) 75 EB
MOV CX [SI] 8B 0C LODSB AC
MOV AL, [SI] 8A 04 ADD 03 C3
MOV [SI], AL 88 04 STOSB AA
MOV CX, 0008 B9 08 00 DIV DL F6 F2
MOV [DI],00 C6 05 00 LOOP NZ (1) E0 F5
MOV [DI],AL 88 05 HLT F4
MOV BL,AL 8A D8 MUL DL F6 E2
MOV DL,34 B2 34 NEG AL F6 D8
MOV DL,36 B2 34 STD FD
MOV DI,0308 BF 08 03 REPNE MOV SW F2 A5
MOV CX,FFFF B9 FF FF XLAT D7
MOV AL,03 B0 03 MOV CL,04 B1 04
MOV BX,0300 BB 00 03 MOV AX[SI] 8B 04
INC CL FE C1 MOV BX[SI] 8B 1C
MOV [DI],CL 89 0D JNC L1 73 02
MUL BX F7 E3 DEC CL FE C9
MOV [DI],AL 88 05 MOV [DI],DX 89 15
CMP AL[SI] 3A 04 MOV AL,[SI] 8A 04
MOV [DI],AX 89 05 MOV BX,AX 8B D8
XCHG AL[SI] 86 04 MOV CX[SI] 8B 0C
LOOP E2 F1 MOV [SI],AL 88 04
MOV CL 00 B1 00 LOOP NZ E0 FA
MOV DX,FFFE BA FE FF    
CMP AL,80 3C 80    
MOV AL,83 B0 83    


Most Important Question for Exam

1. The mnemonic that is placed before the arithmetic operation is performed is




Answer is A)
Explanation: The AAD instruction converts two unpacked BCD digits in AH and AL to the equivalent binary number in AL.


2. The Carry flag is undefined after performing the operation




Answer is B)
Explanation: Since the operation, AAD is performed before division operation is performed, the carry flag, auxiliary flag and overflow flag are undefined.


3. The instruction that performs logical AND operation and the result of the operation is not available is




Answer is D)
Explanation: In the TEST instruction, the logical AND operation is performed and the result is not stored but flags are affected

4. In the RCL instruction, the contents of the destination operand undergo function as




Answer is D)
Explanation: In RRTC(Rotate right through carry), for each operation, the carry flag is pushed into LSB and the MSB of the operand is pushed into carry flag.

5. The instruction that is used as prefix to an instruction to execute it repeatedly until the CX register becomes zero is




Answer is C)
Explanation: The instruction to which the REP is prefix, is executed repeatedly until CX register becomes zero. When CX becomes zero, the execution proceeds to the next instruction in sequence.

6. The instructions that are used to call a subroutine from the main program and return to the main program after execution of called function are




Answer is B)
Explanation: At each CALL instruction, the IP and CS of the next instruction are pushed onto the stack, before the control is transferred to the procedure. At the end of the procedure, the RET instruction must be executed to retrieve the stored contents of IP & CS registers from a stack.

7. instruction that unconditionally transfers the control of execution to the specified address is




Answer is C)
Explanation: In this the control transfers to the address specified in the instruction and flags are not affected by this instruction.

8. The assembler directives which are the hints using some predefined alphabetical strings are given to




Answer is B)
Explanation: These directives help the assembler to correctly understand the assembly language programs to prepare the codes.

9. The logic required for implementing a program can be expressed in terms of




Answer is B)
Explanation: The logic required for implementing a program must be visualized clearly which is possible by flowchart and algorithm.

10. The operands, source and destination in an instruction cannot be




Answer is C)
Explanation: Only one memory operand can be specified in one instruction

11. The instruction that is not possible among the following is




Answer is A)
Explanation: Both the operands cannot be memory operands

12. Both the operands source and destination of an instruction cannot be




Answer is B)
Explanation: Since destination operand should not be immediate data.

13. The registers that cannot be used as operands for arithmetic and logical instructions are




Answer is C)
Explanation: Segment registers are not allowed as operands for arithmetic and logical instructions.

14. The operands of an instruction cannot be




Answer is C)
Explanation: Both the operands should not be immediate operands and memory operands

15. If the processor is executing the main program that calls a subroutine, then after executing the main program up to the CALL instruction, the control will be transferred to




Answer is C)
Explanation: Since subroutine is called, to start the execution of the subroutine, the control is transferred to the subroutine address

16. The stack is useful for




Answer is A)
Explanation: Stack is used for temporary storage of contents of registers and memory locations, status of registers.

17. The Stack is accessed using




Answer is B)
Explanation: The stack is accessed using a pointer that is implemented using SP and SS registers.

18. As the storing of data words onto the stack is increased, the stack pointer is




Answer is A)
Explanation: The data is stored from top address of the stack and is decremented by 2.

19. While retrieving data from the stack, the stack pointer is




Answer is C)
Explanation: The data in the stack, may again be transferred back from a stack to register. At that time, the stack pointer is incremented by 2.

20. The stack pointer register contains




Answer is B)
Explanation: The stack pointer register contains the offset of the address of the stack segment.

21. PUSH operation




Answer is D)
Explanation: Each PUSH operation decrements the SP (Stack Pointer) register.

22. The register or memory location that is pushed into the stack at the end must be




Answer is B)
Explanation: The data can be retrieved by POP operation and as in stack; the data that is pushed at the end must be popped off first.

23. . In the instruction, ASSUME CS : CODE, DS : DATA, SS : STACK the ASSUME directive directs to the assembler the




Answer is A)
Explanation: The directive ASSUMES facilitates to name the segments with the desired name that is not a mnemonic or keyword.

24. 8086 does not support




Answer is C)
Explanation: The 8086 microprocessor does not support direct BCD packed operations.

25. For 8086 microprocessor, the stack segment may have a memory block of a maximum of




Answer is C)
Explanation: In 8086 microprocessor, the memory segments each have a memory of 64K bytes.

26. If an interrupt is generated from outside the processor then it is an




Answer is C)
Explanation: If an external device or a signal interrupts the processor from outside then it is an external interrupt.

27. Example of an external interrupt is




Answer is C)
Explanation: Since the keyboard is external to the processor, it is an external interrupt.

28. The type of the interrupt may be passed to the interrupt structure of CPU from




Answer is A)
Explanation: After an interrupt is acknowledged, the CPU computes the vector address from the type of the interrupt that may be passed to the internal structure of the CPU from an interrupt controller in case of external interrupts.

29. During the execution of an interrupt, the data pushed into the stack is the content of




Answer is A)
Explanation: The contents of IP, CS and PSW are pushed into the stack during the execution.

30. All the functions of the ports of 8255 are achieved by programming the bits of an internal register called




Answer is B)
Explanation: By programming the bits of control word register, the operations of the ports are specified.

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