OR
Explain the following addressing modes of 8086.
i) Direct addressing mode.
ii) Based
Index addressing mode. iii) Index addressing mode. (6 M)
Ans
: When the instructions are executed in 8086 ,it performs on the
specified data i.e. Operand . the operand may me resides in the one of the
internal register of the microprocessor, may be stored at an address in memory,
to access this different types of
operand, the 8086 is provided with various Addressing
Modes. Addressing modes indicates a way of locating data or operands. The
addressing modes are classified into basic three types :
1) Register
operand addressing
2) Immediate
operand addressing
3) Memory
operand addressing
1) Register
Operand Addressing :
The
operand to be accessed is specified as residing in an internal registers of
8086.( The instruction performed on the operands are stored in the registers.)
Operand may be byte or word.
e.g. MOV AX , BX
In the above instruction the operand is stored in the
register BX (source) which has to be move to the register AX (destination).
2) Immediate Operand Addressing :
If an operand is a part of the
instruction instead of the contents of the register or memory location, that
operand is said to immediate operand (operand is directly given in
instruction.) And the addressing mode use for these operands are immediate
addressing mode. The operand may be byte or word.
e.g. MOV AL , 15H
In the above instruction the operand 1516
(source) is immediate operand which has to be move to the lower byte of AH
register i.e. AL(destination)
3) Memory
Operand Addressing :
To an
operand in the store in the memory and to fetch that address we need to calculate
the effective address(EA). This addressing mode consist of the different
representation of the EA. The memory operand addressing is further divided into
following parts:
1] Direct
Addressing Mode :
In this addressing
mode the 16-bit EA is taken directly from the displacement field of the
instruction. The displacement (unsigned 16-bit or signed-extended 8-bit number
) is stored in the location following the instruction.
E.g. Mov AL , [3000H] : here 3000H is the offset address
and by default the content of Data Segment will be segment address . The
combination of these two addresses having the address of particular memory
location that content some data which is moved to AL register.
2] Register
Indirect Addressing Mode :
In this
mode, the EA is specified in either a pointer register or an index register.
The pointer register can be either base register (BX) or base pointer(BP) and
the index register can be either source index (SI) or destination index (DI).
The physical address is computed by using DS and EA .
e.g. MOV [DI] , BX : The instruction copied the 16-bit
content of BX into the memory location offset by the value of EA specified in
the DI from the current contents in DS. Now , if we consider [DS]=7205H , [DI]=
0030H and BX = 8765H, the after execution of instruction , the content of BX is
copied to the memory locations 72080H and 72081H
Physical Address: [DS]x [10H]+EA = [DS]x [10H] + [DI]
3] Base
Addressing Mode :
In this addressing
mode , the effective address of the operand is obtained by adding a direct or
indirect displacement to the contents of either base register(BX) or base
pointer (BP). To find the actual physical address this displacement is also
added to the segment address and offset address.
e.g. MOV AX , [BP + 6] : If the content of BP =2300H and
SS=1000H then the content of memory locations 12306H and 12307H will move to
the 16-bit destination register AX
4] Index Addressing
Mode :
This addressing mode uses
only one of the two index registers (SI or DI) and the 8-bit or 16-bit
displacement is included in the operand field. To find the actual physical
address this displacement is also added with segment address and offset
address.
e.g. MOV AX , [SI + 16] : If the content of SI=1400H and DS
=4000H then the contents of memory locations 41416H and 41417H are move to the
16-bit destination Register AX.
5] Base Index
Addressing Mode :
This addressing Mode combines the features of the Base
Addressing Mode and Index Addressing Mode but does not allow the use of a
displacement . Two registers must be used as source and destination operand ,
one for BX/BP or another for SI/DI. This combination of two register can be
written as [BX][DI] or [BX+DI] both have the same meaning and gives the total
offset value with addition of these registers.
e.g. MOV [BP][SI] , AL : If the content of SS=0400H , BP= 4320H and SI=
4000H and AL = 34H is given the physical address will be calculated as
SS*10H+BP+SI so 04000H+4320H+4000H = 0C320H, the content of AL register will be
load at 0C320H memory location .
2]
Explain following instruction :
1) DAS
2) SBB 3) XALT 4) DAA
5)ADC
6) SUB 7) IMUL 8) LDA / LES
10) XCHG
11) IN (each for 3M/4M)
Ans
:
1) DAS
:
This
instruction is decimal adjust after subtraction is similar to DAA instruction
except that this adjusts decimal result after subtraction .This instruction
internally performs following steps :
Step 1: Check lower nibble of AL;
If it is greater than 9,
Subtract 6 to this nibble.
Else
No
change.
Step 2: Check upper nibble of AL:
If it is greater than 9,
Subtract 6 to this nibble
Else
No
change.
Example :
MOV AL ,10 ; AL = 10H
SUB AL , 3 ; AL = 10H – 3H will give result 0DH in AL.
DAS ; because lower nibble is greater than 9,
; subtracted by 6 , will give
result 7.
; upper nibble is zero , so on
change.
Finally
;
AL = 07 . This is our result.
2)SBB
:
The operation of SBB is
similar to SUB. However in subtraction to the source operand the processor also
subtract the content of carry flag.
It is primarily used for multiword subtract operation.
Destination = Destination – Source – Carry
Syntax
: SBB Destination , Source
Example : SBB BX
, CX
If the contents of
BX=1234H and contents of CX=0123H then after execution the result i.e. content
of BX=1111H carry flag will remain unset as no borrow is taken.
3)XALT
:
This instruction
converts the content of the AL registers into a number stored in a memory
table. This instruction performs the direct table lookup techniques, which is
often used to convert one code to another .An XLAT instruction first adds the
content of AL to BX to form a memory address within the data segment . It
copies the content of this address into AL. This is the only instruction that
adds the 8-bit number to a 16-bit number. Syntax :
XLAT
Example : Suppose a translation table resides at
address 2000H (contents of BX registers ) and AL=05H and DS=4000H then the
working of XLAT instruction will check the memory location of the address DS *
10H + BX + AL i.e. 42005H which contains 8AH then the AL=8AH.
4)DAA
:
Sometimes we need to
add decimal numbers together and assumes results as decimal , but processor
solves numbers treating as a hexadecimal numbers so it is not always possible
to get correct decimal result after addition . For example we want to add 23
and 18, processor will treat these numbers as hexadecimal numbers and gives
answer 3BH , which is not correct answer form our respect , so DAA instruction
is able to correct this instruction adjust only lower byte of AX. This
instruction internally performs following steps :
Steps1: Check lower nibble of AL ;
If it is greater than 9,
Add 6 to this nibble . Auxiliary carry will be set.
Else
No change . Auxiliary carry will
be clear.
Step 2 : Check upper nibble of AL :
If it is greater than 9,
Add 6 to upper
nibble with Auxiliary carry
Else
Add upper nibble with Auxiliary
carry.
Syntax
: DAA
Example : ADD AL
, BL
DAA
5)
ADC :
The operation of
ADC is similar to ADD .However , in addition to the source operand the
processor also add contents of the carry flag . The ADC is commonly used to add
multi-bytes together , basically more
than storage capacity. Destination = Destination + Source + Carry
Syntax
: ADC Destination, Source
Example : Suppose
we want to perform 32-bit operation , but we have 16-bit register in 8086,so we
can store only 16-bit in any register. Consider AX store lower 16-bit number
AC59H . Now add the contents of AX register with 9678H , which will give result
42D1H and carry flag will be set
Now , our aim is to add upper 16-bit numbers (7654H + 24447H) and we
have to add carry too these numbers to get correct result. This operation can
not be performed by simple ADD instruction , so we need a new instruction which
will add the operands as well as carry , hence this instruction.
6) SUB :
This instruction is
used to subtract 8-bit(byte) or 16-bit (word) operands together. If source
operand is larger than the destination operand , the resulting borrow is
indicating by setting carry flag.
Syntax
: SUB Destination , Source
Example : SUB AX , BX
If AX=1234H and BX= 4321H the after the execution of instruction the AX
will be CF13H and carry flag will be set to indicate borrow was taken.
7) IMUL :
This instruction is
similar to MUL expert that operands are assumed to be a signed numbers. The
source operand specified in the instruction is multiplied by accumulator , if
source is 8-bit it multiplies with AL and result is placed into AX register. If
source is 16-bit then it multiplies with Ax register and result is placed into
DX : Ax registers
Syntax
: IMUL Source
Example : IMUL BL
If AL =80H and BL=20H, The content of AL will be treated as signed number which
is 2’s complement of (128)10 which is -128 * 32 then multiplication
of these numbers is -4096 , the 2’s complement of this number is F000H . So finally
AX will be F000H.
8) LDS / LES :
These
instruction are used to load two 16-bit registers from a 4-byte block of memory
. The first two bytes are copied into destination registers and the next two
bytes are copied into corresponding segment registers DS / ES . These
instruction are useful when we want to load segment and offset value by using
single instruction. Usually in string instruction we use the instruction to
load source string address (DS:SI) and destination string address (ES:DI) using
single instruction
Syntax
: LDS / LES Destination , Source
Example : LEA SI
, EA When executed it loads the SI
register with the offset address value. The value of this address is
represented by the effective address EA. The value of effective address can be
specified by any valid addressing mode. For instance , if the value in DI
equals 1000H and that in BX is 20H , then executing the instruction
LEA SI , [DI + BX + 5H] Will
load SI with the value (SI)= 1025H
9)
XCHG
:
This
instruction is used to swap the data . After executing this instruction
destination and source will swap .
Syntax
: XCHG Destination , Source
Example : XCHG
AL , BL If AL = 34H and BL= 76H then
after execution AL will be 76H and BL=34H.
10) IN :
Processor has a separate
I/O space from its memory space . among them 65536 I/O ports available for the
programmer to use.
These ports can be access
directly , using the ports address , for 65536 ports address in 0000H to FFFFH ,
the input port is actually a hardware device connected to the processor’s port
. The 8086 allows two different forms of the IN instruction. A fyll 16-bit
ports address must be loaded into DX register and instruction like IN Al , DX
or IN AX , DX may be used to read the input port . If the port number is in
between 00H to FFH then another form of IN instruction may be used as In AL ,
86H or IN AX , 80H.
Syntax
: IN accumulator , port
Example : If we
want to read input 80 H then we may give instruction like
IN AL , 08 H or
MOV DX , 80H
IN
AL , DX
3]
Explain Instruction format of 8086.
Ans
:
•A machine language
instruction format has one or more number of fields associated with it.
•The first field is called as operation
code field or op-code field, which indicates the
type of operation to
be performed by the CPU
•The instruction format also contains
other fields known as operand fields
•The CPU executes the instruction using the information
which reside in these fields
•There are six general formats of
instructions in 8086 instruction set.
•The length of an instruction may vary
from 1 byte to 6 bytes. The instruction formats
are described as follows
1]
One Byte Instruction:
•This format is only one byte long and may
have the implied data or register operands.
•The least significant 3-bits of the opcode are used for
specifying the register operand, if any
•Otherwise, all the 8 bits form an
opcode and the operands are implied 2]
Register to Register:
•This format is 2 bytes long
•The first byte of the code specifies the
operation code and width of the operand specified by ‘w’ bit.
•The second byte of the code shows the
register operands and R/M field, as shown below:
•The register represented by the REG field
is one of the operands.
•The R/M field specifies another register
or memory location i.e. the other operand.
3] Register to/from memory with no
displacement:
•This format is also 2 bytes long and
similar to the Register to Register format except for the MOD field as shown.
•The MOD field shows the mode of
addressing. The MOD, R/M, REG and the ‘W’ fields are decided in Table 2.2.
4] Register to/from Memory with Displacement: •This type of instruction format contains 1 or 2 additional
bytes for displacement along with 2 byte format of the register to/from memory
without displacement. The format is as shown below.
5] Immediate Operand to Register:
•In this format, the first byte as well as the 3-bits from
the second byte which are used for REG field in case of register to register
format are used for opcode.
•It also contains one or two bytes of immediate data. The
complete instruction format is as shown below.
3] Immediate Operand to Memory with 16-bit
displacement:
•This type of instruction format requires
5 or 6 bytes for coding.
•The first 2 bytes contain the information
regarding OPCODE, MOD and R/M fields.
The remaining 4 bytes contain 2 bytes of displacement and 2
bytes of data as shown.
4]
State and Explain Arithmetic group of Instruction.
Ans
:The operation such as Addition , Subtraction ,
Multiplication and Division are the arithmetic . To perform these operation the
8086 is provided with the arithmetic group of instruction which consist of
following instruction :
Addition
:
1] ADD
: This instruction is use to perform addition of two 8-bit or 16-bit registers or immediate operand
e.g. ADD AX , BX
2] ADC
: This instruction is use to perform addition of two 8-bit or 16-bit registers or immediate operand and the
content of carry flag are added to the result if it is set.
e.g. ADC AX , BX
3] INC
: The instruction increases the operand by one
e.g. INC AX
4] AAA:
After performing the addition of the ascii values the result is adjust by ascii
to get the required output. This instruction is used after ADD instruction e.g.
ADD AX , BX
AAA
5] DAA
: The instruction is performed on the packed BCD numbers after performing
operation the result is adjust with the decimal. It is used after ADD
instruction.
e.g. ADD AL , BL
DAA
Subtraction
:
1] SUB
: This instruction is used to subtract two operand representing in an form of
addressing modes .e.g. SUB AX , BX
2] SBB
: After performing the subtraction of
the if the borrow is taken then the
carry flag is set . So, the
content of carry flag are also subtracted from the result.
e.g. SBB AL , BL
3] DEC
: The decrease the operand by 1 .
e.g. DEC AL
4] AAS
: After performing the subtraction of the ascii values the result is adjust by
ascii to get the required output. This instruction is used after SUB
instruction.
e.g. SUB AL , BL
AAS
5] DAS
: The instruction is performed on the packed BCD numbers after performing
operation the result is adjust with the decimal. It is used after SUB
instruction.
e.g. SUB AL , BL
DAS
6] NEG
: The operand is convert to its 2’s complement
e.g. NEG AL
Multiplication
/ Division :
1] MUL
: It multiplies the two unsigned integer
values, one of the operand always resides in the AX register.
e.g. MUL BL
2] IMUL
: It multiplies the two signed integer values, one of the operand always
resides in the AX register.
e.g. IMUL BL
3] AAM
: It adjust the AX register for multiplication
4] DIV
: 1] MUL : It divide the two unsigned
integer values, one of the operand always resides in the AX register.
e.g. DIV BL
5] IDIV
: It divides the two signed integer
values, one of the operand always resides in the AX register.
e.g. IDIV BL
6] AAD
: It adjust the AX register for division.
7] CBW
: It convert the Byte value to the Word value i.e. 8-bit to 16-bit 8] CWD : It convert the word to the
double word i.e. 16-bit to 32-bit
e.g. MOV AL , 0A1H
CBW
CWD
5]
Assume that (AX)=001016 , (BX)=010016 and (DS)=100016,
What happens if XLAT instruction is executed?
Ans
: Given,
(AX) = 001016
(BX) = 010016 (DS) = 100016
Hence lookup table resides at DS x 10 i.e. 10000H
+ BX 0100H
10100H
So the lookup table is at the address 10100H . The value of
the AL=10H.The data from the memory location will transfer to the AL register.
6] Assume that (AX)=110016 , (BX)=0ABC16. What
is the result of executing instruction ADD AX,BX?
Ans
:
Given,
(AX)= 110016
(BX)= 0ABC16
Instruction :
ADD AX, BX
The content of register AX and BX are added and the result
is stored in the AX register
(AX) = 110016 = 0001000100000002
ADD
(BX) = 0ABC16 =0000101010111100 2
(AX) =1BBC 16= 00011011101111002
Hence after executing the instruction the result stored in
the AX is 1BBC16.
7] Write a program to add 8-bit data at offset 0400H in 3000H segments
to another 8-bit data available at 0500H offset in the same segment and store
the result at 0700H in the same segment.
Ans
:
MOV AX , 3000H ; Move the data 3000H into AX register
MOV DS ,
AX ; The content of AX
i.e. 3000H is move to DS
The
above two instruction establish the Data Segment and set segment address to
3000H.
MOV AL , [0400H] ; The 8-bit content of memory location
having
Address 3000H : 0400H are move to 8-bit AL
Register
ADD AL , [0500H] ; The
content of memory location having address
3000H : 0500H are added with the content of AL
register
and result is store in AL register
MOV [0700H] , AL ; the 8-bit content of AL register are moved to the memory location having address 3000H : 0700H
8] Assume that (AX) : ABCD16 and (BX) : 432116.
What is the result of executing ADD AX
,BX ?
Ans
:
Given,
(AX)= ABCD16
(BX)= 432116
Instruction :
ADD AX, BX
The content of register AX and BX are added and the result
is stored in the AX register
(AX) = ABCD16 = 10101011110011012
ADD
(BX) = 432116 = 01000011001000012
(AX) = DDDD16= 11101110111011102
Hence after executing the instruction the result stored in
the AX is DDDD16.
9) Two byte sized register are stored at the symbolic offset address
NUM1 and NUM2 respectively. Write an instruction sequence to generate their sum
and store it at NUM3. The SUM is to be formed by adding the values at NUM1 and
that at NUM2. Assume that all storage location are in the current data segment.
(6 M/7 M) Ans :
Let us assume that the memory locations NUM1 ,NUM2 and NUM3 are
in the same data segment
MOV AX, @data ; Establish data segment
MOV
DS , AX
MOV AL, [NUM1] ; Get the second BCD number
MOV
AL, [NUM2] ; Adding the values
DAA ; Apply BCD adjustment
MOV
[NUM3] , AL ; Save the result
10) Memory location 00490H through 00493H contain respectively 0A,9C,
82 and 78. What does AX contain after each instruction? (Assume that SI
contains 0490H and BP contain 0002H)
i) MOV
Ax, [SI] ii) MOV Ax, [SI +1]
iii) MOV Ax, [SI] [BP]. (6 M)
Given , (SI) = 0490H
= 0490H x 10H
= 04900H (BP)= 0002H i] MOV AX ,
[SI]
In this instruction the content of memory location whose
address is contain by SI is moved to the AX register. The content of memory
location 00490H is 0AH and 00491H is 9CH .Hence, the content of 00490H and
00491H is moved to the register AX as AX is a 16-bit. After performing
instruction the content of AX will be:
(AX)=(9C0A)16
ii] MOV
AX , [SI + 1]
In this instruction the content
of memory location whose address is contain by (SI + 1) is moved to the AX
register. The memory address contain by (SI + 1) is 00491H .The content of
memory location 00491H is 9CH and 00492H is B2H. Hence, the content of 00491H
and 00492H is moved to the register AX. After performing instruction the
content of AX will be:
(AX)=(B29C)16
iii] MOV
AX , [SI] [BP]
In this instruction the content
of memory location whose address is contain by (SI + BP) is moved to the AX
register. The memory address contain by (SI + BP) is 00492H .The content of
memory location 00492H is B2H and 00493H is 78H. Hence, the content of 00492H
and 00493H is moved to the register AX. After performing instruction the
content of AX will be:
(AX)=(78B2)16
