Department of Control and Systems Engineering Lecture 4 – Page
1
of 4 Third Year - Microprocessors
By Mr.WaleedFawwaz
Lecture 4- 8086 programming-Integer instructions and computations
Objective: 1. Data transfer instructions
2. Arithmetic instructions 3. Logic instructions
4. Shift instructions 5. Rotate instructions
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1. Data transfer instructions
a MOV instruction
b XCHG Instruct
Fig 1 a XCHG data transfer instruction b Allowed operands Example 1:For the figure below. What is the result of executing the following instruction?
XCHG AX , [0002]
Solution:
01007 01006
01005 01004
01003 01002
01001 01000
DS 0100
34 12
DF DD
30 00
A2 55
AX 9068
01007 01006
01005 01004
01003 01002
01001 01000
DS 0100
34 12
DF DD
90 68
A2 55
AX 3000
Before After
Department of Control and Systems Engineering Lecture 4 – Page
2
of 4 Third Year - Microprocessors
By Mr.WaleedFawwaz
c XLAT
Mnemonic Meaning
Format Operation
Flags affected XLAT
Translate XLAT AL + BX + DS 10
AL none
Fig 2 a XLAT data transfer instruction Example 2: For the figure below, what is the result of executing the following instruction?
XLAT
Solution:
d LEA, LDS, and LES instructions
Fig 3 a LEA, LDS and LES data transfer instruction Example 3: For the figure below, what is the result of executing the following instruction?
LEA SI , [ DI + BX +2H]
Solution:
SI= DI + BX + 2H = 0062H
DS 0100
SI F002
Before After
DI 0020
AX 0003
BX 0040
DS 0100
SI
0062
DI 0020
AX 0003
BX 0040
01047 01046
01045 01044
01043 01042
01041 01040
DS 0100
34 12
DF DD
90 68
A2 55
AX xx90
01047 01046
01045 01044
01043 01042
01041 01040
DS 0100
34 12
DF DD
90 68
A2 55
AX xx03
Before After
BX 0040
BX 0040
Department of Control and Systems Engineering Lecture 4 – Page
3
of 4 Third Year - Microprocessors
By Mr.WaleedFawwaz
For these threeinstructions LEA, LDS and LES the effective address could be formed of
all or any various combinations of the three elements in Fig 4
�� = ��� ���
+ ���
��� +
� 8
− ��� ������������ 16
− ��� �������������
Fig 4The three element used to compute an effective address
Example 4: For the figure below, what is the result of executing the following instruction?
LEASI , [ DI + BX +2H]
Solution:
SI= DI + BX +2H = 0062H
Example 5 : Instruction Sample
Result
LEA SI , [ BX + SI + 55 ] Valid
SI= BX + SI + 55 LEA SI , [ BX + SI ]
Valid SI= BX + SI
LEA BP , [ 890C ] valid
BP= 890C LEA AX , [ BX + SI + 20 ]
Valid AX = BX + SI + 20
LEA DI , [ BP + DI + 55 ] Valid
DI = BP + DI + 55 LEA DI , [ DI + DI + 55 ]
Not valid because EA doesn’t involve DI twice LEA CS , [ BP + DI + 55 ]
Not valid because destination cant be segment register LEA IP , [ BP +550C ]
Not valid because destination cant be instruction pointer LEA AX , [ CX + DI + 1D ]
Not valid because EA doesn’t involve CX LEA AL , [ DI + 103D ]
Not valid because destination must be 16 bit
Example 6:What is the result after executing each one of the next instructions? LEA BP, [F004]
MOV BP, F004 MOV BP, [F004]
DS 0100
SI F002
Before After
DI 0020
AX 0003
BX 0040
DS 0100
SI 0062
DI 0020
AX 0003
BX 0040
01047 01046
01045 01044
01043 01042
01041 01040
34 12
DF DD
90 68
A2 55
01047 01046
01045 01044
01043 01042
01041 01040
34 12
DF DD
90 68
A2 55
Department of Control and Systems Engineering Lecture 4 – Page
4
of 4 Third Year - Microprocessors
By Mr.WaleedFawwaz
Solution: Instruction
Result
LEA BP, [F004] The value F004 will be assigned to the Base Pointer
MOV BP, F004 The value F004 will be assigned to the Base Pointer
MOV BP, [F004] The wordat memory locations F004 and F005 in the current
Data Segment will be assigned to Base Pointer The instruction LES is similar to the instruction LDS except that it load the Extra Segment
Register instead of Data Segment Register
2. Arithmetic instruction
