CS401 By Asad Naeem
Introduction:
Hi Everyone!✋
Here are CS401 Assembly Language Programming Mid Term Solved Subjective Question Answers for Sure 100% Success and Exam Preparation. All Questions are very important and repeated every time in exam.
All answers are according to CS401
Handout. Thus all students are request to prepare these Questions for better
CGPA and solid Concepts.
If any student prepare these questions sensibly then he no need of any other material and to prepare whole handout.
Benefits of These Notes:
1.
Time
Consuming:
Mostly students are jobs holder so these are very helpful for them. As I
mentioned above all are very important so you can save your time by preparing
these notes in a very short time, rather no need to prepare a complete Handout
of Assembly Language Programming CS401.
2. All are Past Exams Questions:
Sometime students confuse about syllabus and pattern about exams questions, so
these are very helpful to known about type of questions you will face in exams.
3.
Most
Repeated:
All are from Past Exams and most repeated. Sometime examiner change 1 or 2
words of question and its meaning remain the same. So prepare all by creating concepts.
4.
Completely
Theoretical:
Most advantage is that all questions are Theoretical and summary of complete handout. By preparing these notes you will
be able to answer the any question you face in exam.
5.
To
the point Answers:
All answers given are to the point and according to university and examiner necessities.
There is no additional and irrelevant material.
Therefore Prepare them by creating
your own concepts and write in your own words in exam. You will not forget these
concepts in your future. Then you can taught to others obviously.
Lets Start!
CS401 (Assembly Language Programming) MID TERM QUESTION ANSWER BY CH. ASAD NAEEM BC170200260 Asadnaeem280@gmail.com
Q 1: Affected flag of AND operation?
Affected Flag of AND are: CF (carry flag), OF (Over Flow Flag), PF (Parity Flag), SF (Sign Flag), ZF (zero flag) and AC (Auxiliary Carry).
Q 2: Relation between RET and CALL is dependent or
independent.
Technically they are
independent in their operation but both the instructions are commonly used as a
pair. RET works irrespective of the CALL and the CALL irrespective of the RET.
Q 3: What is the difference between LES (Load Extra
Segment) and LDS (Load Data Segment) instruction?
Or why we need the general register in LES and LED instruction?
There are
two special instructions that load a segment register and a general purpose
register. These instructions have two factors, one
is the general purpose register to be loaded and the other is the memory
location from which to load these registers.
Q 4: Write down the procedure to clear the selective
bit.
The source operand is
loaded with a mask containing at positions which keep their old value and zero.
The effect of applying this operation on the destination with mask in the
source is to clear the desired bits. That process is named as masking.
Q 5: Why REP prefix is generally not used with LODS
instruction?
LODS is used in a loop
and not with the REP prefix then the value formerly overloaded in the register
is overwritten, if the instruction is repeated and only the last value of the
block remains in the register.
Q 6: What is difference between REPE
(repeat while equal) and REPNE (repeat while not equal)?
REPE or REPZ repeat the next
string instruction while the zero flag is set and REPNE or REPNZ repeat the next
instruction while the zero flag is not set.
Q 7: Describe PUSH and POP with the help of an
example.
The operations of placing
items on the stack and removing them are called push and pop. Memory is like a
shelf as zero at the top and the maximum at the bottom. If a decrementing stack
begin at shelf five, the 1st item is located in shelf 5; the next entry
is located in shelf 4, same as the next will in shelf 3 and so on.
Q 8: Explain all characteristics of SCAS (scan string) instruction.
SCAS (Scan String) is
used to compare the source byte or word in register AL or AX with the
destination string element ES: DI and updates the flags. SCAS (Scan String) is frequently
used to place equality or in-equality in a string. SCAS (Scan String) is dissimilar
from the other instructions. It’s like a CMP instruction.
Q 9: Describe Local Variable?
Local variables are
created by stack. Global variables take permanent space but they not. Local
variables formed when the subroutine is called and wasted later. Therefore the
space used by them can be use again for the local variables of another
subroutine. Local variables only have meaning inside the subroutine and no
meaning outside it. The most suitable place to store these variables is the
stack.
Q 10: Explain divide overflow error.
If a large number is divided
by a very small number it is possible that the percentage is larger than the
space. In this case a stop is automatically generated and the program is
usually terminated as a result. That is termed as divided overflow error.
Q 11: For what purpose "INT4" is reserved.
For
Arithmetic Overflow INT4 is reserved, change of sign bit.
Q 12: Double working of Exchange instruction with help
of an example.
XCHG (Exchanges) instructions
exchange the contents of the destination (first) and source (second) operands. Two
general-purpose registers are called the operands.
Q 13: Two form of String Moves Instruction?
MOVS has the two forms
MOVSB and MOVSW.
Q 14: How value of IP (Instruction Pointer) Register is saved & updated when call Instruction
executed.
When Call instruction is
executed the stack pointer (SP) is decremented by 2 and value of IP is pushed on
stack.
Q 15: How many parameters a subroutine can receive
through stack.
There are 7 parameters a
subroutine can receive.
Q 16: How the string instruction is used in the block
of memory?
STOS (store string) transfers a byte
or word from register AL or AX to the string element addressed by ES: DI and
updates DI to point to the next location. STOS is often used to clear a block
of memory or fill it with a constant.
Q 17: How many block processing instructions in 8088,
just name the number of processing instructions?
8088 contain only 5 block processing instructions. These 5 block instructions are STOS (store string), LODS (load string), CMPS (compare string), SCAS (scan string), and MOVS (move string) respectively.
Q 18: What are the instruction use by assembly
language for permanent diversion and temporarily diversion?
The instructions for permanent diversion in 8088 are the jump instructions, while the instruction for temporary diversion is the CALL instruction.
Q 19: How much string instructions are in assembly
language? Write down all instructions.
STOS –
Clearing the Screen
LODS –
String Printing
SCAS
-String Length
LES
(Load
Extra Segment) and LDS (Load Data Segment)
MOVS
-Screen Scrolling
CMPS – String Comparison
Q 20: Describe the purpose of these string instructions? (1) STOS (2) CMPS
1: STOS:-
STOS
transfers a byte or word from register AL or AX to the string element addressed
by ES: DI and updates DI to point to the next location. STOS is often used to
clear a block of memory or fill it with a constant.
2: CMPS:-
CMPS withdraws the source location DS: SI from the destination location ES: DI. Source and Destination are not changed. SI (Source Index) and DI (Destination Index) are updated accordingly. CMPS compares 2 blocks of memory for equality or inequality. It withdraws byte by byte or word by word.
Q 21: What is the syntax of PUSH instruction?
Push ax
Q 22: For what purpose DS and ES registers are
generally used in the context of video memory.
To access
the video memory both DS and ES are used. However usually we have DS for
accessing our data, and load ES with the part of video memory.
Q 23: At what time the instruction “push ax” is performed
in decrementing stack and how the value of SP will change.
The instruction “push ax” is executed in decrementing stack, Then SP is decremented by 2 and IP is pushed on the stack.
Q 24: What colors are repeated by 0th, 1st and 2nd
bits (from least significant side) in Attribute Byte:
Blue, green, Red
Q 25: How SCAS instructions can be used to detect Null
at the end of a string?
We use SCASB with REPNE and a zero in AL to find a zero byte in the string. We can load maximum 64K bytes In CX but actual strings will be much smaller. If SCAS and CMPS stopped due to equality or inequality then the index register have previously incremented. Hence when SCAS will stop DI thatwould be pointing the null character.
Q 26: From where does the contents
of SL and DS registers are loaded as a result of execution of the instruction”
LDS SI [BP +4]”?
“Lds si, [bp+4]” will load SI from BP+4 and DS from BP+6.
Q 27: How a 16-bit MUL operation is different from
32-bit MUL operation?
If the operands were 16bit the answer would be in 32bit and if the operands were 32bit the answer would be 64bit. For declaration of 16 bit we use “dw” and for 32-bit we use “dd”.
Q 28: Describe the working of the CALL
instruction with the reference of Stack.
During the CALL operation, the current value of the IP (instruction pointer) is automatically saved on the stack, and the destination of CALL is loaded in the IP (instruction pointer). Therefore Execution begin again from the destination of CALL.
Q 29: Explain the function of rotate right (ROR)
instruction.
In the rotate right operation every bit moves one position to the right and the bit dropped from the right is inserted at the left.
Q 30: Write all steps of algorithm
for printing number 352.
The steps
of our algorithm are below.
• Divide figure
or number by base (10 in case of decimal).
• The
remainder is its right most digits.
• Transform
the digit into ASCII representation (Add 0x30 to the remainder in case of
decimal).
• Save
this digit on stack.
• If the proportion
is non-zero then repeat the whole process to get the next digit, otherwise stop.
• Pop
digits one by one and print on screen left to right.
Q 31: What are the results after performing the instruction?
1. and ax, bx
2. or ax, bx
3. XOR ax, bx
Given that ax = 00110011 and bx = 00010001
1. 0x0011
2. 0x0033
3. 0x0022
Q 32: Explain the complete operation of Interrupt when
it is generated.
The result of an INT instruction is named as Interrupt (software interrupt) or it is created by an external hardware which passes the interrupt number by a different instrument. After completion of current executing instruction, the existing value of FLAGS pushed into the stack, then the existing code section is strapped, then the offset of the next instruction is strapped. After this it automatically clears the trap flag and the interrupt flag prohibit further interrupts until the current routine finishes.
Q 33: Which instruction is used to determine zero bits
in string?
REPE or REPZ repeat the next string instruction as zero flag is set and REPNE or REPNZ repeat the next instruction as zero flag is not set.
Q 34: Explain the use of TEST instruction.
For bit testing, test instruction is used. BX holds the mask and in every next iteration shifts to left, as our concerned bit is now the next bit.
Q 35: Describe MOVS and CMPS instructions?
1. MOVS
MOVS handovers
a byte or word from the source location DS: SI to the destination ES: DI and informs
SI and DI to point to the next sites. For moving a block of memory MOVS is used.
2. CMPS
CMPS withdraws the source location DS: SI from the destination location ES: DI. Source and Destination are not changed. SI and DI are updated consequently. CMPS compares 2 blocks of memory for equality or inequality of the block. It withdraws byte by byte or word by word.
Q 36: Define short jump?
If the offset is stored in a single byte as in 75F2 with the opcode 75 and operand F2, the jump is called a short jump. For example Unconditional jumps can be short, near, and far.
Q 37: Every character is displayed on the screen in
the form of a word. What individually byte of this word represents?
The second byte in the word chosen for one screen location holds the center and background colors for the character. Therefore the couple of the ASCII code in one byte and the attribute in the second byte creates the word that links to one location on the screen. The lower address contains the code however the higher one holds the attribute.
Q 38: IF DF=0 what
its represent and IF DF=1 what its represent?
It represent that the current operation has to be done from bottom to top of the block (D=0) or from top to bottom of the block (D=1).
Q 39: Explain how extended shifting is
performed?
Using our basic shifting and rotation instructions we can
effectively shift a 32bit number in memory word by word. We cannot shift the
whole number at once as our architecture is limited to word operations. We use
just two instructions and we name it extended shifting.
num1: dd 40000
Shl word [num1],
1
Rcl word
[num1+2], 1
The multiplicand is shifted left as a 32bit number using extended shifting operation.
Q 40: Explain MUL instruction in both
cases (i) if the source operand is byte (ii) if the source operand is a word?
MUL (multiply) performs an unsigned multiplication of the source operand and the accumulator. If the source operand is a byte, then it is multiplied by register AL and the double-length result is returned in AH (A High Byte) and AL (A Low Byte). If the source operand is a word, then it is multiplied by register AX, and the double-length result is returned in registers DX and AX.
Q 41: Why is it necessary to provide the
segment and offset address in case of FAR jump?
Far jump must be used a two byte segment and a two byte offset are given to it because It loads CS with the segment part and IP with the offset part.
Q 42: What’s your understanding about Incrementing
and Decrementing Stack?
A decrementing stack moves from higher addresses to lower addresses as elements are added in it while an incrementing stack moves from lower addresses to higher addresses as elements are added.
Q 43: What is the Difference between
CALL and RET?
The CALL instruction allows temporary change and then reusability of code. Therefore RET takes no arguments and transfers control back to the instruction.
Q 44: Tell the Formula to scroll up the
screen:
Rep movsw scroll
up
Scrollup: push bp
Mov bp, sp
Push ax
Push cx
Push si
Push di
Push es
Push ds
Mov ax, 80 ;
load chars per row in ax
Mul byte [bp+4] ;
calculate source position
Mov si, ax ;
load source position in si
Push si
; save position for later use
Shl si, 1
; convert to byte offset
Mov cx, 2000 ;
number of screen locations
Sub cx, ax ;
count words to move
Mov ax, 0xb800
Mov es, ax ;
point es to video base
Mov ds, ax ;
point ds to video base
Xor di, di
; point di to top left column
Cld
; set auto increment mode
Rep movsw ;
scroll up
Mov ax, 0x0720 ; space
in normal attribute
Pop cx ;
count of positions to clear
Rep stosw ;
clear the scrolled space
Pop ds
Pop es
Pop di
Pop si
Pop cx
Pop ax
Pop bp
Ret 2
Subroutine to calculate the length of a string
; Takes the segment and offset of a string as parameters
Strlen: push bp
Mov bp, sp
Push es
Push cx
Push di
Les di, [bp+4]
; point es: di to string
Mov cx, 0xffff ;
load maximum number in cx
Xor al, al
; load a zero in al
Repne scasb
; find zero in the string
Mov ax, 0xffff ;
load maximum number in ax
Sub ax, cx
; find change in cx
Dec ax
; exclude null from length
Pop di
Pop cx
Pop es
Pop bp
Ret 4
Q 46: What is difference between SHR (Shift
Logical Right) and SAR (Shift Arithmetic Right) instructions?
SHR (shift logical right operation) inserts a zero from the left
and transfers every bit one position to the right and copies the rightmost bit
in the carry flag.
The operation of SAR (shift arithmetic right) is to move every bit one place to the right with a copy of the most important bit.
Q 47: For what purpose "INT 1"
is reserved?
INT 1, Trap, Single step Interrupt
This interrupt is used in debugging with the trap flag. 8088 was the 1st processor that has this capability to support debugging.
Q 48: Define implied operand?
Implied operand means, it is continuously in a particular register say the accumulator, and it need not be mentioned in the instruction.
Q 49: Explain the process of ADC (add with carry)?
The ADC instruction is specially placed for extending the capability of ADD. Any size of number can be added using an appropriate arrangement of ADD and ADC. Entirely basic building blocks are providing for the assembly language programmer and the programmer can extend its capabilities as much as needed by using these fine instructions in suitable groupings. Further clarifying the operation of ADC, consider an instruction “ADC AX, BX.”
Q 50: Describe mismatch? Define with one example.
If we compose our
code in bytes but we state in word then there will be a logical error in the
code and it is called size mismatch error.
; A program add 3 numbers via byte variables
[Org 0x0100]
Mov al, [num1]; load first number in al
Mov bl, [num1+1]; load second number in bl
Add al, bl; gather sum in al
Mov bl, [num1+2]; load third number in bl
Add al, bl; gather sum in al
Mov [num1+3], al; store sum at num1+3
Mov ax, 0x4c00; terminate program
Int 0x21
num1: DW 5, 10, 15, 0
 |
| CS401 |
Q 51: Compose a
subroutine to clear screen.
; Clear screen using string instructions
[Org 0x0100]
Jmp start
; Subroutine to clear the screen
Clrscr: push es
Push ax
Push cx
Push di
Mov ax, 0xb800
Mov es, ax ;
point es to video base
Xor di, di ;
point di to top left column
Mov ax, 0x0720 ; space
char contain normal attribute
Mov cx, 2000 ;
number of screen locations
Cld
; auto increment style
Rep stosw ;
clear the whole screen
Pop di
Pop cx
Pop ax
Pop es
Ret
Start: call clrscr ;
call clrscr subroutine
Mov ax, 0x4c00 ;
terminate program
Int 0x21
Q 52: Replace the following invalid
instruction with the single valid instruction
(1)
Mov IP (2) Mov IP, L5
Mov IP
RET:-The instruction Ret will pop the IP address.
Mov IP, L5
CALL:-The instruction Ret will pop the IP address.
Q 53: Describe the selective bit testing and operation that can
be used with selective bit testing?
AND Operation can be used to check whether particular bits of a number are set or not. We change our multiplication algorithm by using selective bit testing instead of checking bits one by one. An AND Operation can be used with selective bit testing.
Q 54: What happen behind the screen when
a character displayed on the monitor screen? Or how the data is displayed on monitor screen?
When we press a
specific number like ‘A’. It’s shifted from the keyboard to the computer. Our
program read that number as the character ‘A’. When the same method repeated,
the Video Graphics Adapter (VGA) in our computer shows the shape of character.
ASCII code of corresponding character placed at a cell in the
VGA memory will cause the matching ASCII shape to be displayed on the corresponding
screen location.
When 0x40 is sent to the VGA card then it will turn into pixels ‘on’ and ‘off’ in such a way that a graphic representation of ‘A’ performs on the screen.
Q 55: What does “41” represent in the
instruction “mov word [es: 0], ox0741”?
The
41 is the lower byte is the ASCII code that display
on top left in white color on black screen. And display A.
Q 56: Briefly define the term”de facto
standard” by giving an example?
The lengthy ASCII code is just a de facto industry standard but it is not defined by an organization like the standard ASCII. ASCII has become the most common de facto standard and is used for Internet and global communication as well. For example Microsoft Word DOC, AutoCAD, IBM Personal Computer (PC) etc.
Q 57: Why do we need special
instructions like LES (loads extra segment register) and LDS (loads data
segment register)?
LES (loads extra segment register) loads ES while LDS (loads data segment register) loads DS. These instructions have 2 parameters, 1 is the general purpose register to be loaded, and the other is the memory location from where to load these registers. The major application of these instructions is when a subroutine receives a segment offset pair as an argument and loaded in a segment.
Q 58: Describe the working of the RET
with the reference of Stack.
RET (Return) transfers control back from a procedure to the instruction. RET pops the word at the top of the stack. The word at the top of stack is popped into the CS register,
Q 59: What is stack “n” defined with
example?
Stack
is a data structure that behaves like a LIFO manner. It contains many elements
and there is only one way for in and out of the container. When an element is
inserted it place on top of all other elements and when an element is removed then
the top element is removed first. For example,
Consider a test tube and put some balls in it. The second ball will come above the first and the third will come above the second. When a ball is taken out only the one at the top will be removed. The operation of placing an element on top of the stack is called pushing the element and the operation of removing an element from the top of the stack is called popping the element.
Q 60: Define address bus?
The
group of bits that the processor uses to inform the memory about which element
read or write is known as the address bus. Another important bus called the
data bus is used to move the data from the memory to processor. The address bus
is unidirectional and address continuously journeys from processor to memory.
Data moves from both, processor to memory and memory to processor, so the data bus is bidirectional.
Q 61: Explain the working of shift
logical left SHL?
The
shift logical left operation is the exact opposite of shift logical right. In this
operation the zero bits inserted from the right and every bit moves one position
to its left with the most significant bit dropping into the carry flag.
Shift arithmetic left is just another name for shift logical left.
Q 62: Write any two control
instructions.
The
instruction pointer and instructions run are two control instructions. These
are instructions that control the program execution.
Q 63: RET
instruction take how many arguments?
No arguments are taken by RET and also it transfers control back to the instruction.
Q 64: Write down the names of four
segment registers?
CS code Segment
DS data Segment
ES Extra Segment
SS Stack Segment
Q 65: What is direct addressing"?
Direct addressing: A
fixed offset is given in brackets and the memory is accessed to that offset.
Q 66: What is conditional jump"?
Conditional jump is the chief instruction, because it gives the processor decision making capability.
Q 67: Data clear algorithm in assembly
language code?
;
clear the screen
[Org
0x0100]
Mov ax, 0xb800; load
video base in ax
Mov es, ax; point es
to video base
Mov di, 0; shift di
to top left column
Nextchar: mov word [es:
di], 0x0720; clear next character on the screen
Add di, 2; shift to
next screen position
Cmp di, 4000; has the
whole screen cleared
Jne nextchar; if no
clear next position
Mov ax, 0x4c00;
terminate program
Int 0x21
Q 68: What is ‘Multiplication’ in assembly
language?
MUL (multiply) performs an unsigned multiplication of the source operand and the accumulator. If the source operand is a ‘byte’, at that point it is multiplied by register AL and the double length result is returned in AH and AL. In that case if the source operand is a ‘word’, then it is multiplied by register AX, and the double length result is returned in registers DX and AX.
Q 69: Fill in the
Blanks of Given Questions.
1.
Stack is a _ data structure
_____
2.
Standard ASCII has 128 characters
3.
Which bit is referring to the red component of foreground color? (2bit)
Which bit is referring to the red component
of background color? (6bit)
4.
When a 32 bit number is divided by a 16 bit number, the quotient is of (16 bit)
5.
There are just 5 block processing instructions in 8088.
6. after the execution of instruction “RET 2” increment or decrement.
Q 70: Write down the purpose of “define
byte”?
To
state a part of our program as ‘holding data’ as an alternative of instructions
then we need a pair of elementary but special assembler directives. The first
directive is “define byte” written as “db.”
Q 71: Write down the purpose of FAR Jump?
Sometimes we may need to go from one code segment to another, and near and short jumps not available. Far jump must be used for that purpose and a 2 byte segment and a 2 byte offset are given to it.
Q 72: How the value of stack pointer
(sp) changes after every PUSH or POP instruction?
PUSH decrements SP (the stack pointer) by two and then transfers a word from the source operand to the top of stack. POP increments SP (the stack pointer) by two and then transfers a word from destination operand to the current top of stack.
Q 73: Explain AND, OR
and XOR operations with truth tables.
“AND” operation
AND performs the
logical bitwise and of the two operands (byte or word) and returns the
result to the destination operand.
“OR” operation
OR operation executes
the logical bitwise “inclusive or” of the 2 operands (byte or word) and returns
the result to the destination operand.
“XOR” operation
XOR (Exclusive or)
performs the logical bitwise “exclusive or” of the two operands and returns the
result to the destination operand.
Q 74: Why we need to clear the stack?
“Clear the stack” is performed when there are many calls to a subroutine then it repeated many times.
Q 75: Which two registers are changed
during CALL statement execution?
1.
IP only
2.
IP and SP only
3. SP only
4. BP and IP only
Answer:
IP and SP only.
Q 76: Which instruction is used to
determine the zero byte in string?
SCASB used with REPNE and a zero in AL to find a 0 byte in the string.
Q 77: The second byte in the word designated for one screen location holds the foreground and background colors for the character. Which colors are represented by 0th, 1st and 2nd bits (from least significant side)?
7 – Blinking of focus
character
6 – Red section of
background color
5 – Green section of
background color
4 –
Blue section of background color
3 – Intensity section
of focus color
2 – Red section of focus
color
1 – Green section of focus
color
0 – Blue section of focus color
Q 78: One character store in one word in
memory what is store in its upper byte and lower byte.
Upper byte (attributes) Lower byte (character code)
Q 79: In which situation branching technique used in INTEL architecture,
which instruction used to deliver temporary diversion and how?
The instruction for provisional change is the CALL instruction. The CALL instruction allows provisional change and therefore reusability of code. Now we can put the code for bubble sort at one place and reuse it again and again. That was not possible with permanent diversion.
Q 80: In the program HELLO WORLD desired location, what
functionality is achieved from the given code lines?
Mov al, [SI]
Mov [ES: DI], ax
Add DI, 2
Add SI, 1
Answer:
Mov al, [SI] ; load next character of string
Mov [ES: DI], ax ; display this character
on the screen
Add DI, 2 ; transfer
to next screen place
Add SI, 1 ; transfer to next character in string
Q 81: Which 2 instructions are used to checked that specific
bits are set or not? How these 2 instructions are vary from each other?
“AND instruction” can
be used to checked whether specific bits of a number are set or not.
“Shifting and JC” used to checked bits one by one. The “AND operation” abolishes the destination mask.
Q 82: Briefly explain how string instruction work on data?
String instructions work on a block of data that has a start and an end. The instructions can work from the start to the end and then from the end towards the start. In fact they can work in both directions; only string instructions allow moving data from memory to memory.
Q 83: Which instructions are used to call a subroutine?
Both CALL and RET instructions are used to call a subroutine.
Q 84: How 16-bit DIV operation different from 32-bit Div
operation?
DIV instruction presented
in the 8088 processor and there are two forms of the DIV instruction. The 1st
form divides a 32bit number in DX: AX by its 16bit operand and stores the 16bit
proportion in AX and the 16bit remainder in DX.
The 2nd form divides a 16bit number in AX by its 8bit operand and stores the 8bit proportion in AL and the 8bit remainder in AH.
Q 85: What is a REP prefix? Write
example.
A distinct prefix repeats the instruction in hardware that is called the REP prefix. REP repeats the string instruction CX times.
Q 86: Write any two logical bitwise operators.
“AND”
operation:
AND operation does
the logical bitwise “and” of the 2 operands.
“OR”
operation:
OR operation executes the logical bitwise “inclusive or” of the 2 operands.
Q
87: What this instruction
means? Mov [es: di], dx
Mov [es: di], dx ; print char on screen
Q 88: Which logical operations are used to see 0 and 1 bit?
AND Operation is used to see o and 1 bit.
Q 89: How push instruction work in an assembly language program?
The PUSH instruction increases the stack pointer and stores the value of the specified byte operand inside the RAM address. By this instruction flags are not affected.
Q 91: What is the distinctive
feature RET n?
RET n discarded the arguments rather than popping them.
Q 92: What is divide overflow error,
mov ax, bl, DIV bl......?
Overflow error
If a large
number is divided by a very small number it is possible that the percentage is
larger than the space. In this case a stop is automatically generated and the
program is usually terminated as a result a divide overflow error occur.
Mov ax
These instructions are used to move data
from one place to another. These places can be registers, memory, or even
inside secondary devices.
BL:
The BL and BLX instructions copy the address of the next
instruction into LR.
DIV bl:
“DIV BL” has 8bit
operand, therefore the indirect dividend is 16bit that would store in the AX register.
“DIV BX” takes 16bit operand therefore the indirect dividend would 32bit.
Q 93: When Z
flag is set?
If the last mathematical or logical instruction has produced a 0 in its destination then the Z flag (Zero flag) is set.
Q 94: What is difference between JNA and JBE?
JNA Ã Jump if not above ZF = 1
JBE Ã Jump if below or equal CF = 1
This jump is taken after a CMP, if the unspecified destination is smaller than or equal to the unidentified source.
Q 95: Define Multiplication of 1101, 0101 with complete
steps.
1101 = 13
0101 = 5
-----
1101
0000x
1101xx
0000xxx
------------
1000001 = 65
Q 96: What will
happen if we do not clear the stack after subroutine call?
If the stack is not cleared or is partially cleared the stack will finally become full, SP will reach 0, and thereafter coil will produce unexpected results that’s known as SO (stack overflow). Thus clearance of everything that is placed on the stack is very important.
Q 97: What is the purpose of JAE? Explain with
example.
Answer:
Instruction Jump Condition Test
JAE Jump if Above or Equal CF=0
Q 98: Write 2 other standers for computer other then ASCII.
Gray Codes, EBCDIC
Q 99: Assume that there are
4 parameters in a stack write a single instruction to clear the stack?
1:
CALLAR 2: CALLEE 3: REMOVE 4: RET
Q 100: Write a code to left shift 32 bit of
number in a memory label num1?
Shl bl, 1 ;
shift multiplicand left
Q
101: How can we convert a two dimensional video memory into one
dimensional?
For two dimensional coordinate system of the screen to the one dimensional memory, we need to multiply the row number by 80 since there are 80 columns per row and add the column number to it and again multiply by two since there are 2 bytes for each character.
Q 102: How local variables
are created in the stack?
PUSH and POP instructions are used to create local variables.
Q 103: Briefly explain
the working of following registers AX, BX, IP, CS.
AX Register:
To return a value from a subroutine is to use the AX register.
That is why AX is not saved and restored in the subroutine.
IP (Instruction Pointer) Register:
This is the special register containing the address of the next
instruction to be executed. No mathematics or memory access can be done through
this register. It is out of our direct control and is automatically used. It
is dangerous and required special care.
CS Register:
The code segment register is the special registers related to
the Intel segmented memory model.
1 window that is pointy to the CS register holds the currently performing
code. So CS holds the zero or the base of code.
BX Register.
BX registers points to our data and we can add them using the instruction “mov ax, [bx]” without fixing the address of our data in the instructions.
Q 103: Why we use callee
not caller explain.
If RET n is not accessible, stack clearing through callee is a difficult procedure. It will save the return address in a register, then eliminate the parameters, and then place back the return address.
Q
104: What are intra segment call and inter segment call?
Near calls are called intra segment calls, while far calls are called inter-segment calls.
Q 105: Illustrate the reason of introducing string instruction
in 8088?
String instructions are complex instruction that perform a number of tasks against one instruction. DI and SI are used to access memory. SI and DI are called source index and destination index because of string instructions. If DF (Direction Flag) is cleared, this is called the auto increment mode of string instruction, and if DF (Direction Flag) is set, this is called the auto decrement mode.
Q 106: Identify only two
prefix which are meaningful with CMPS instruction. Justify the answer.
REPE and REPNE are two prefix that are meaningful with CMPS
instruction. CMPS used with a REPE or a REPNE
prefix as the blocks are same or because they are different and use zero
flag.
Q 107: Describe working of
"NOT “operation?
NOT invert the bits (forms the one’s complement) of the byte or
word operand known as single operand instruction not a logical operation.
“Not ax” and “not byte [mem]” are examples.
Q 108: What reasoning the
default segment association of BP with SS?
The stack segment has a very critical and fine use and there is a reason default segment association of BP with SS. DS (Default segment) link with the BP register is the SS (stack segment). The reason is to preview inside the stack using the BP register and read the parameters without removing them and without touching the stack pointer.
Q 109: Show the process of
applying selective bit inversion on the lower nibble of following bits
10011011?
XOR would use as a masking operation to invert selective bits. The operand should be applied with a mask of 0F in the source to invert the lower nibble 10011011. The upper nibble will hold its value and the lower nibble will be set as a result.
Q 110: Write the code
to print 'B' on the screen with black back ground and white foreground?
Mov word [es: 1], 0x0742
Q 111: Describe the
two forms of DIV instruction?
DIV instruction
presented in the 8088 processor and there are two forms of the DIV instruction.
The 1st form divides a 32bit number in DX: AX by its 16bit
operand and stores the 16bit proportion in AX and the 16bit remainder in DX.
The 2nd form divides a 16bit number in AX by its 8bit operand and stores the 8bit proportion in AL and the 8bit remainder in AH.
Q 112: which interrupts are
generated as a result of (a) dividing a number by zero (b) if the trap flag is
set
INT
0, Division by zero?
It’s Mean that the proportion did not fit in the DS (destination
register)
INT 1, Trap, Single step Interrupt:
This interrupt is used in debugging with the trap flag. Whenever trap flag would set, the Single Step Interrupt would generated after every instruction.
Q 113: Define legal and illegal instructions? Give
reason why these are illegal instructions such as mov ax, bx “mov
[num1], [num2]” mov bx, [num1].
Answer:
The assembler only keeps writing illegal instructions which the
processor cannot execute that is also the difference between a syntax error and
a logic error.
“mov [num1], 5”
is legal instruction but the processor can’t recognize the data movement scope
in this process. The variable num1 can be cured as a byte or word and also 5
can be treated as a byte or word.
Mov ax, bx these
instructions are used to move data from one place to another.
“mov [num1], [num2]” This instruction is illegal
because memory to memory data movement is not acceptable.
Mov [num1+6], [num1]; ILLEGAL
However this form is illegal and not allowed on the Intel architecture.
Q 114: How Assembly language strings
declare in ASCII code.
Add
dl, 0x30 ;
convert digit into ASCII value
converting a digit into its ASCII
representation by adding 0x30 and subtracting 0x20 from it.
Q 115: why address bus is a unidirectional bus also
mention the flow of address bus.
The address bus is unidirectional and
address continuously travels from processor to memory.
Because memory is a dumb device and cannot expect which element
the processor need at a particular time.
Q 116: Define RET n?
There is a form of the RET instruction called “RET n” where n is a numeric argument. After performing the operation of RET, it further increments the stack pointer by this number.
Q 117: Explain the correlation between interrupt numbers and interrupt
handler?
The correlation process from the interrupt number to the interrupt handler uses a table called interrupt vector table. Its location is fixed to physical memory address. Each entry of the table is four bytes.
Q 118: How video devices are viewed by processor?
The video device is seen by the processor because a memory area is containing the ASCII codes that are currently displayed on the screen and a set of I/O ports controlling things like the resolution, the cursor height, and the cursor position.
Q 119: Why interrupts are considered to be asynchronous and
unpredictable?
Unpredictable means the time in which interrupt cannot be predicted, while asynchronous means that the interrupt has nothing to do with the currently executing instruction and the current state of the processor.
Q 120: Describe ASCII stander code and extended ASCII code?
ASCII stands for American Standard Code for Information
Interchange. The name shows that this is a code that allows the interchange of
information; All ASCII based computers use the same code.
The extended ASCII number is just a de facto industry standard but then it is not defined by an organization like the standard ASCII.
Q 121: What is the Function of ISR interrupts service
routine?
The routine that performs in reaction of an INT instruction is called the interrupt service routine (ISR) or the interrupt handler.
Q 122: How many operands in ADC?
ADC has three operands.
Q 123: Which instruction is most use in screen rolling function and how
screen rolling works.
The instruction MOVS is used for Screen Scrolling. Scrolling is the process when all the lines on the screen move one or more lines towards the top and towards the bottom. REP with MOVS will operate the full processor power to organize the scrolling in less period.
Q 124: What 32 bit word is defined?
32 bit word defines the double.
Q 125: What will be the result if 2 16bit operands multiplied?
The outcome will be 32bit.
Q 126: What we write for software int?
For software int we write INT.
Q 127: which instruction and its limitation is used to swap the
values?
A new instruction XCHG has been introduced. The instruction swaps source and destination operands. It has reduced the code size by one instruction.
Q 128: Which instructions are used for insert and retrieve data from
stack.
When CALL operation occur the current value of the instruction pointer is automatically saved on the stack, When the RET instruction is executed, it recovers the value of the instruction pointer from the stack.
Q 129: What does the following instruction do? Xor
di, di
Xor di, di ; point di to top left column
Q 130: Explain the function of rotate left (ROL) instruction.
In the operation of (ROL) rotate left instruction, the most important bit is copied to the
carry flag and is injected from the right, causing every bit move to the left position.
Q 131: What is the function of segment override prefix and what changes it brings to the opcode?
To dominate the relationship for one instruction of one of the registers BX, BP, SI or DI, we use the segment override prefix. Opcode has not altered.
Q
132: What are the different types of jumps available? Define position
relative addressing.
Near Jump: We can jump anywhere
within a segment by using a near jump.
Short Jump: This is the limitation of a byte in char representation. It used
with segment.
Far Jump: Far capability makes
intra segment control possible.
Q 133: What is purpose of DF direction flag?
This flag tells that the current operation has to be done from bottom to top of the block (D=0) or from top to bottom of the block (D=1).
Q 134: Write instruction of following:
MOV AX 10
MOV AX 0X10
Answer:
Mov
ax, 10 ; space used by one task
Mov ax, 0x10 ; load all seg regs to 0x10
Q 135: Write the command to
access the debugger and clarify individually portion of debugger.
The debugger displays the values of registers, flags, stack, our code and 1 or 2 areas of the system memory as data. Inside the debugger we can see that the registers before and after the CALL operation are precisely identical. The debugger provides a close vision of the processor.
Q 136: How most significant bit is affected when we execute
shift logical right and shift arithmetic right instruction?
The shift logical right operation insets a 0 from the left and
moves every bit one position to the right and copies the rightmost bit in the
carry flag.
The operation of shift arithmetic right is to shift every bit one place to the right with a copy of the most important bit.
Q 137: How Stack data structure is helpful in assembly
programming?
Stack data structure use PUSH and POP to
store and recover data.
Sometimes we want to save the
information existing in the registers. This information may be lost when
program is run. So we save this information in Stack by using PUSH, at the end
we can recover this information using POP.
Q 138: Which logical operation can be used to see whether a bit
is 1 or 0?
We can use logical operation AND to see
if bit is 1 or 0.
Q 139: Suppose AL contains 11001011b and CF= 1. Provide the original
contents of AL after individually instructions executed. Assume the former early
situations for individually part of this question.
1.
SHL AL,1
2.
SHR AL,1
3.
ROL AL,CL ;if CL comprises
2
4.
SAR AL,CL ;if CL comprises
2
5. RCR AL,CL ;if CL comprises 3
Answer:
SHL
AL, 1
1001011b1
SHR
AL, 1
111001011
ROL AL, CL ; if CL contains 2
1001011b1
SAR AL, CL ; if CL contains 2
121001011b
RCR
AL, CL ; if CL contains 3
311001011
Q 140: Following part of code belong to the program "scrolling up
the screen". Write against each instruction what it does.
Mov ax, 80
Mul byte [bp+4]
Mov
si, ax
Push si
Shl si, 1
Answer:
Mov ax, 80 ; each line consists of 80 words, so we move
to next line
Mul byte [bp+4] ; we move 4 words in front
Mov si, ax ; contents of accumulator register ax are
moved to source index SI
Push si ; a copy of contents of source index SI
are moved to stack
Shl si, 1 ; contents of source index are
moved to left by 1 unit.
Q 141: Write down an assembly language program that displays zero (0)
on the whole DOS screen.
.model small
.stack 100H
.data
.code
Call proc
Mov cx, 1
Mov dx, 0
L1:
Mov bx, 0
Add dx, bx
Loop L1
Endp
Q 142: Write an assembly program
which shows two arrays using bubble sort method.
Mov ax, [ARR+si]
Mov dx, [ARR+di]
Start:
Mov ax, code
Mov ds, ax
Jmp start2
ARR: dw 1,2,4,3,6,5,9
start2:
Mov ch, 0h
Q 143: Enlist the flag that will be affected by DIV instruction.
The CF, OF, SF, ZF, AF, and PF flags are undefined.
Q
144: Write a code for adding two 32 bit number using extends addition.
Mov edx, 0
Mov eax, 0FFFFFFFFh
Add eax, 0FFFFFFFFh
ADC edx, 0 ; EDX += CF, add "leftover" carry
Q
145: Write assembly code for sorting two arrays with bubble sort.
Mov ax, [ARR+si]
Mov dx, [ARR+di]
Start:
Mov ax, code
Mov ds, ax
Jmp start2
ARR: dw 1,2,4,3,6,5,9
start2:
Mov ch, 0h
Q 146: Calculate the physical memory address generated by the
followingsegment offset pairs.
1. 0000:6727
Physical Address = Segment x10 + Offset
Physical Address = 0000 x10 + 6727
Physical Address = 00000 + 6727
Physical Address = 6727h
2. FFFF: 4336
Physical
Address = Segment x10 + Offset
Physical
Address = FFFF x10 + 4336
Physical
Address = FFFF0 + 4336
Physical
Address = 104326h
To make 20bit physical address we have to drop the carry bit
so
Physical
Address = 04326h
3. AB01: FFFF
Physical
Address = Segment x10 + Offset
Physical
Address = AB01 x10 + FFFF
Physical
Address = AB010 + FFFF
Physical Address = BB00Fh
Q 147: Replace the following invalid
instructions with a single instructionthat has the same effect.
A. pop ip
Ret
B. mov ip, L5
Jmp L5
C. sub sp, 2
Mov [ss: sp], ax
Push ax
D. mov ax, [ss: sp]
Add sp, 2
Pop ax
E. add sp, 6
Mov ip, [ss: sp-6]
Call
Q 148: Suppose we have AX = 0x123, write
an assembly language instruction which clear the lowest nibble of the value
placed in AX.
AND AX, 000Fh
This will clear all
but the low-nibble of AX,
Q 149: Write
a program which does extended multiplication between 32bit operands.
Mov
eax, 12345h
Mov
ebx, 1000h
Mul ebx; EDX: EAX = 0000000012345000h, CF=0
Q 150: How many BYTES will be move by
each of the following block of codes?
a) MOV cx, 384 REP movsb
b) MOV cx, 384 REP movsw
In block “A” move 384 bytes (B). In block B move
384 words (W)
Which amounts to 2 x 384 = 768 bytes
Q
151: What shifting operation we use for
16-bit multiplication by 2 and why?
See on Google or Handout.
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