Answer Paper1

8/6/2019 Answer Paper1

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Ist internals solutions

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PES SCHOOL OF ENGINEERING

HOSUR ROAD (1KM BEFORE ELECTRONIC CITY), BANGALORE-100

Test I Solutions

Name of faculty: Mrs. AnuSophia/ Ms. Swapna

Q1.With a neat diagram, explain the architecture of 8086 microprocessor along

with the functions of each block and registers. 10marks

ES

CS

SS

Bus InterfaceUnit

Instructionqueue

Controllines

ControlUnit

GeneralRegisters

ALU

Flags

ExecutionUnit

AH(8) AL(8)BH(8) BL(8)

CH(8) CL(8)

DH(8) DL(8)

AXBX

CX

DX

SP

BPDISI

DS

IP

20

Adder

A1.The execution unit contains the data and address registers, the arithmetic and logic

unit and control unit. The bus interface unit contains bus interface logic, segment

registers, memory addressing logic and a 6 byte instruction object code queue. The

EU and BIU operate asynchronously.

EXECUTION UNIT AND BUS INTERFACE UNIT

i) BIU sends out the address, fetches instructions from memory, reads data from ports

and memory and writes data to port and memory. It handles all transfers of data and

address on the buses for EU.

ii) EU tells BIU where to fetch instructions or data from, decodes instructions and

executes instructions.

* EU consists of control circuitry which directs internal operation.

* Decodes translates instructions fetched from memory into a series of actions

which EU carries out.

*ALU:

performs add, subtract, AND,OR,XOR,increment,decrement,complement or shift

operations.

* It consists of pointer register and index register which is used to get the data

from memory or write data into the memory. The pointer registers are Base pointer,


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Source Pointer and Index Pointer. The index registers are Source Index and

Destination Index.

FLAG REGISTERS: A flag is a flip flop that indicates some condition produced by

execution of instruction or controls certain operations of EU.

* General purpose registers are 8 in number AX (AH & AL), BX (BH & BL),CX(CH & CL),DX(DH &DL) which can be used

For temporary storage of 8 bit or 16 bit data. The data can be accessed much

more quickly than it could be accessed in external memory.

* QUEUE: BIU stores the fetched 6 instructions in FIFO register called

queue.

* Segment Register: Segment register is used to hold the upper 16-bit bits of

starting address for each of the segments.

* Instruction pointer is a 16-bit register which contains offset address of nextinstruction to be executed.

* Code Segment register points to the base or start of the current code

segment.IP contains offset or distance from this base address to the next instruction

byte to be fetched.

* The stack pointer register in EU holds the 16-bit offset from the start of the

segment to the memory location where a word was most recently stored on stack.

* Base pointer consists of 16-bit source index register (SI) and 16-bit

destination index register (DI) which is used for temporary storage of data just like

general purpose register.

* DI and SI hold the 16-bit offset of a data word in one of the segments.

Q2) List out the different addressing modes in 8086 and differentiate between

the with an example.10 marks

1. REGISTER ADDRESSING MODE

Fastest addressing mode because processing the data between the register involves noreference to the memory. Size of operand should be the same.

Segment registers can also be used as operand with an exception that CS register

cannot be the destination operand.

Example: mov cx, ax cx=7418h, ax=3429h

After execution cx=ax=3429h


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2. IMMEDIATE ADDRESSING MODEIn immediate addressing mode an immediate number is specified in the instructions.

However immediate numbers cannot be used as destination operand. Immediate

operands can be accessed quickly because they are available directly from instruction

Queue.

Example: mov 4000h, dx = invalid since here immediate data is destination operand

mov ah,1000h = valid

3. DIRECT ADDRESSING MODE

In this addressing mode 16 bit offset address of memory location is directly specified

as a part of the instruction. The no. of bytes to be copied is determined by the size of

the destination.

Example: mov ax, [5000h]

4. INDIRECT ADDRESSING MODE

6 types of indirect addressing mode exists

a) Base register IAM: Offset address of the data to be accessed is available inbase register BX.

Example: MOV AX, [BX] ; Access the data at offset address specified in BXregister and copy into AX register.Bp register is used only when 8 bit or 16 bit

displacement.

b) Base register IAM with displacement: Part of the offset address of the data isstored in the register BX or BP and remaining part of the offset address is

specified as displacement.

Example:MOV AX, 23H [BX]; copies a word from data segment memory at

the offset address: [BX] +23H

MOV 50H [BP], 1234H

c) Indexed AM: Offset address of the data to be accessed is available in indexregisters SI or DI.

Example: MOV AX,[SI] ;Access the data at offset address specified in SI

register and copy into AX register

d) Indexed AM with displacement: Part of the offset address of the data is storedin the register SI or DI and remaining part of the offset address is called

effective offset address is specified as displacement.

Example:MOV AX, 23H [SI] ;copies a word from data segment memory at

the offset address :(SI)+23H

e) Base indexed AM: Part of the offset address of the data is stored in the registerBX or BP and SI or DI. Effective offset address is obtained by adding thecontents of BX/BP+SI/DI

Example: MOV AX,[BX+SI] ; access the data at offset address specified in

BX+SI Register and copy into AX register.

f) Base indexed AM with displacement: Effective offset address is obtained byadding the base register, index register and displacement specified in the

instruction.


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Example : MOV AX,1000H[BX+SI] ; Access the data at offset address

specified in BX+SI register+ displacement and copy into AX register.

Q3) Explain the following instructions with an example of each.10marks

1. AAA: ASCII adjustment after addition

Description: This instruction is used to make sure the result is correct unpacked

BCD.

Function:

1. Clear the higher order nibble of AL

2. If lower nibble of AL is greater than 9 or AF=1 then

a) add 6 to AL

b) add 1 to AH

c) set AF and CF to 1

Flags Affected: AF and CF

PF,SF,OF and ZF are left undefined.Example: AL: 0011 0101

BL: 0011 1001

ADD AL,BL Result AL=0110 1110=6EH which is incorrect

AAA Now AL=0000 0100 and CF=1 which is 14 correct answer

2. RCL: Rotate through carry left

Description: Instruction rotates all the bits in a specified word or byte some number

of bit positions to the left..

Functions: Operation is circular because the MSB of the operand is rotated into the

carry flag and the bit in the carry flag is rotated around into the LSB of the operand.

For multibit rotates, CF will contain the bit most recently rotated out of the

MSB.Destination operand can be a register or in a memory location. Count when

more than one is loaded into the CL register.

Flags affected: Affects only carry flag and overflow flag

Example: Cf=0, BH=10110011

RCL BH, 1; BH= 01100110

Cf=1, OF=1 because MSB Changed

3. DAS: Decimal Adjustment after Subtraction

Description: Used to make sure the result of adding two unpacked BCD nos is

adjusted to be legal BCD number.Function:1.If lower nibble of AL after an addition is greater than 9 or AF=1 then subtract 6 to

the lower nibble

2.If now the upper nibble of AL is greater than 9 or CF=1 Subtract 60H to AL.

Flags affected: AF,CF,PF,ZF OF is undefined

Example: AL=0100 0010=42h

BL=0010 1001=29h

SUB AL,BL AL=0001 1001=19H


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DAS subtract 6 to AL since 1001>9

AL=AL-6

AL=13 ( Since we take 2s complement of 29h carry gets complemented)

4.IMUL:Integer multiplication

Description:Multiplies a signed byte from some source operand to a signed byte inAL or a signed word to a signed word

in AX register.

If two signed bytes are multiplied the result is sored in Ax and if two signed word are

multiplied then the higher

order word is stored in Dx and lower order word is stored in AX.

Flags affected: OF and CF

AF,PF,SF and ZF are left undefined.

Example:IMUL BH: Signed byte in AL times signed byte in BH.

result in AX.

5. XLAT:

Description : Used to translate a byte from one code to another code. Replaces a byte

in al register with a byte pointed to by BX in a look up table in memory. Before the

execution of the instruction the look up table containing the values for the new code

must be put in memory and the offset of the starting address of the look up table must

be loaded in BX.

It puts the byte to be translated in AL . To point to the desired byte in the look up

table , the instruction adds the byte in al to the offset of the start of the table in BX.It

then copies the byte from the address pointed to by (bx+al) back into al register.

Flags affected : None

Example : MOV BX, OFFSET EBCDIC_TABLE ; point bx at start of EBCDIC table

In data segment.

XLATB ; replace ascii in al with ebcdic from table

Q4.a) What are the advantages of segmentation of memory? 5marks

Advantages of the segmented memory:

1. Allows the memory capacity to be 1MB although the actual address to be handled

is of 16 bit size.2. Allows the placing of code, data and stack portions of the same program in

different parts (segment) of memory, for data and code protection.

3. Permits a program and/or its data to be put into different areas of memory each

time program is executed.

4. Faster execution time & segments are distinct.

5. As we have different segments debugging is lot easier.


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Q4.b) With respect to the code segment register and instruction pointer register,

explain the calculation of the physical address of code byte with an

example.5marks

The instruction pointer register holds 16-bit address of the next code byte within the

code segment. The CS register points to the base or start of the current code segment.

The value contained in IP is referred to as an offset because this value must be addedto the segment base address in CS to produce the required 20=bit physical address

sent out to by BIU.The IP contains the distance or offset from this base address to the

next instruction byte to be fetched.

Physical address(20-bit)=(CS)*10H+IP

CS 3 4 8 A 0

IP + 4 2 1 4

---------------------

PHYSICAL ADDRESS 3 8 A B 4

Q5) Explain execution of following 8086 instruction. What will be the content ofdestination register? Also mention effect on flags and pointers after execution of

instruction.10marks

A5) i) DAA if AL = 8AH, CF=0,AF=0.

AL=10001010

lower byte of AL is 10 i.e.>9

add 6 of lower byte,AF is set

1000 1010

0110

---------

1001 0000 here higher byte not > 9

so destinstion register has 1001 0000 =90H

Carry flag and auxilliary flag is effected.

ii) AND AX,0FF0H if Ax =5678,CF=0

AX = 0101 0110 0111 1000

Adding 0000 1111 1111 0000

-------------------

0000 0110 0111 0000

now mask out upper 8 bitssince AX = 0000 0110 0111 0000

CF=0,OF=0.PF,SF & ZF are updated,AF is undefined.


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iii) DIV BH if AX=0050H,BX=1000H

AX=0050H

BH=10H

DIV AX/BH

AL=05H=5 decimal.

The quotient is stored in AL register while the remainder is stored in AHregister.AH=0H

iv)ROR AX,CX if AX =0420H,CX=0002H

The bits in the AX register is rotated towards right by the no. of times

specified in the CX register.

AX = 0000 0100 0010 0000

after rotation

AX = 0000 0001 0000 1000

v)PUSH CX if CX=4020H,SP=00FFFH

The value of CX register is pushed into the stack at a memory location which

is pointed to by the stack

pointer.The stack pointer points to the top of the stack where the new element

is pushed.

After the completion of the push operation the SP is decremented by 2 to point

to the next memory location

which now the top of the stack.Since CX is a word SP is decremented by 2

else for a byte it is decremented by 1.

SP = 0000 0000 1111 1111 1111

subtract 0000 0000 0000 0000 0010

------------------------

0000 0000 1111 1111 1101

SP = 00FFD.

Q6) What happens when the following assembler directives are

executed?10marks

1. X dw ?

* dw directive tells the assembler to define a variable of type word.

* dw-define word.

* it allocates memory for a word.

* ? Signifies that the memory is uninitialized.


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2. ARRAY db 10 dup(0)

* db directive tells the assembler that a variable of type byte is being defined.

* db-define byteA

* 10 byte location are allocated in memory and is given the name array.

* All the 10 memory locations are initialised to 0.

3. BCD_BIN endp

* BCD_BIN is a procedure.

* endp signifies the end of procedure BCD_BIN.

* proc along with endp is used to bracket a procedure.

4. ADD AX,word ptr[BX]

* The directive ptr tells the assembler to manipulate the byte or word pointed to by

BX.

* It adds a word in AX to a word word pointed to by BX.* The keyword word tells the assembler that the pointed to by BX is to be added.

5. ASSUME DS:DATA

* ASSUME directive is used to tell the assembler the name of the logical segment it

should use for the specified segment.

* DS:DATA tells the assembler that the logical segment DATA is to be used for

data.

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