Assignment on alp

Name ID

DULAL CHANDRA BARMAN 00805010

MD.JAHURUL ISLAM 00805034

MD.EMDADUL HUQ 00805038

MD.FARUQ AHMED 00805047

ASHOK KUMAR SARKER 00805049

FLAG REGISTER
FLAGS Register:

 Individual bits control the action or represent the status of the
processor

Control flags (TF, IF, DF):

 Determine how the processor responds to certain situations

Status flags (CF, PF, AF, ZF, SF, OF):

 Set to represent the result of certain operations

 Used to control conditional jump instructions

FLAG Register Bits:
Status Flags: Control Flags:

Bit Name Symb
ol Bit Name Symb
ol
0 Carry flag CF
8 Trap flag TF
2 Parity flag PF
4 Auxiliary carry flag AF 9 Interrupt flag IF

6 Zero flag ZF 10 Direction flag DF
7 Sign flag SF
11 Overflow flag OF

Status Flags:

The Carry Flag (CF):
 CF = 1 if there is a carry out from the msb (most
significant bit) on addition, or there is a borrow into the
msb on subtraction
 CF = 0 otherwise
 CF is also affected by shift and rotate instructions

The Parity Flag (PF):
 PF = 1 if the low byte of a result has an even number
of one bits (even parity)
 PF = 0 otherwise (odd parity)

The Auxiliary Carry Flag (AF):
 AF = 1 if there is a carry out from bit 3 on addition, or there is
a borrow into the bit 3 on subtraction
 AF = 0 otherwise
 AF is used in binary-coded decimal (BCD) operations

The Zero Flag (ZF):
 ZF = 1 for a zero result
 ZF = 0 for a non-zero result

The Sign Flag (SF):
 SF = 1 if the msb of a result is 1; it means the result is
negative if you are giving a signed interpretation
 SF = 0 if the msb is 0

The Overflow Flag (OF):
 OF = 1 if signed overflow occurred
 OF = 0 otherwise

Example 1. ADD AX,BX , where AX contains FFFFh , BX contains
FFFFh.
Solution: FFFFh
+ FFFFh
1FFFEh
The result stored in AX is FFFEh=1111 1111 1111 1110
SF = 1 because the msb is 1
PF = 0 because there are 7(odd number) of 1 bits in the low byte of
the result .
ZF = 0 because the result is nonzero.
CF = 1 because there is a carry out of the msb on addition.
OF = 0 because the sign of the stored result is the same as that of
the numbers being added (as a binary addition, there is a carry
into the msb and a carry out).

 Example 2. ADD AX, BX, where AX contains 7132h ,
BX contains 7000h
 Solution: 7132h
 + 7000h
 E132h
 The result stored in AX is E132h=1110 0001 0011 0010
 SF = 1 because the msb is 1
 PF = 0 because there are 3(odd number) of 1 bits in the
low byte of the result .
 ZF = 0 because the result is nonzero.
 CF = 0 because there is no a carry out of the msb on
addition.
 OF = 1

 Example-3: SUB AX, BX , where AX contains 8000h and BX
contains 0001h.
 Solution: 8000h
 -0001h
 7FFFh
 The result stored in AX is FFFEh=0111 1111 1111 1111
 SF = 0 because the msb is 0
 PF = 1 because there are 8(even number) of 1 bits in the low byte
of the result .
 ZF = 0 because the result is nonzero.
 CF = 0 because a smaller unsigned number is being subtracted
from a larger one.

 Now for OF. In a signal sense, we are subtracting a positive number
from a negative one, which is like adding two negatives. Because
the result is positive (the wrong sign), OF = 1.

OVERFLOW

 The overflow flag is set when the Most Significant
Bit (MSB) is set or cleared.
 For example, take the addition of 127 and 127.
The 8 bit signed binary number of 127 is
represented as 0111 1111.
 The MSB (the bit to the far left) is 0. When these
two 8 bit numbers are added the result is 254, or
1111 1110. Notice now that the MSB is now 1 and
not 0.
 Therefore, the overflow flag has been set. Here,
1111 1110 would be interpreted as a negative
number.

 The addition test shows whether the overflow
flag has been set.
 If two positive operands generate a negative
sum. For example, 0101 1100 and 0100 1010 is
1010 0110, or -90.
 The MSB changed from 0 to 1.
 And the two negative operands generate a
positive sum.
 For example 1100 0010 and 1000 0100 is
01101000 or 104.
 The MSB changed from 1 to 0, so the overflow
flag is being set.
 Overflow never occurs when the sign of two
addition operands are different.

(Signed) Overflow
 Can only occur when adding numbers of the same sign (subtracting
with different signs)

 Detected when carry into MSB is not equal to carry out of MSB

 Easily detected because this implies the result has a different sign
than the sign of the operands

 Programs can ignore the Flags

 Example 4. ADD AX,BX , where AX and BX both contains 7FFFh .

Solution:
Hex Binary
7FFFh 0111 1111 1111 1111
7FFFh 0111 1111 1111 1111
FFFEh 1111 1111 1111 1110
The signed and unsigned interpretation of 7FFFh is 32767. Thus for
both signed and unsigned
addition,7FFFh+7FFFh=32767+32767=65534.This is out of range
for signed numbers, the signed interpretation of the stored answer
FFFEh is -2,so signed overflow occurred .However the unsigned
interpretation of FFFEh is 65534,which is the right answer, so there
is no unsigned overflow.

Unsigned overflow:
 On addition, unsigned overflow occurs when there is a carry out of
the msb.
 On subtraction unsigned overflow occurs when there is a borrow
into the msb.

 Example 5. ADD AX, BX, where AX contains FF12h, BX contains
1ACBh.
Solution:
FF12h
+ 1ACBh
119DDh
The result is stored in AX is 19DDh=0001 11 111 1101
FF12h=-1 and 1ACBh=1, and FF12h+1ACBh=-1+1=0, so sign overflow
did
not occur. Here a 1 is carried out of the msb, so unsigned overflow
occurred.

The importance of flag registers:

Flag register has it's importance by following resion.
1. Sign flag: use to store the sign of any number under processing i.e
when arithmetical logic has negative sign it is in set condition. and
rest in reset.
2. Zero flag: when ALU operation is zero it is in reset condition.
3. Auxiliary carry flag: it hold the carry during internal processing of
addition ,sub , mul and div.
4. Parity flag: hold 1 for even parity,0 for odd parity.
5. Carry flag: hold carry after end of processing of operation.

FLAGS Register & Flow
control Instruction

Flags Register
Individual bits control the action or
represent the status of the processor

Control flags (TF, IF, DF): Determine how
the processor responds
to certain situations
Status flags (CF, PF, AF, ZF, SF, OF):
• Set to represent the result of certain
operations
• Used to control conditional jump
instructions

Flow Control Instructions:

In this assignment we will practice how to
control the flow of an assembly language
program using the compare instruction, the
different jump instructions and the loop
instructions.
Objectives:
1- Jump Instructions.
2- Compare Instruction.
3- Loop Instructions.

Jump Instructions:

The jump instructions are used to transfer
the flow of the program to the indicated
Operator.
Here we discus about various categories.
Conditional Jumps:
There are Three categories of Conditional
Jumps:
a)Singed Jumps
b)Unsigned Jumps:
c) Single-Flag Jumps:

The Unconditional Jump:

The unconditional jump may be
used to make infinite loops. Though
the use of such instructions is not
recommended in high level
languages, due to the availability of
program control structures, in
assembly however, most of the time
one must use the unconditional
jump.

The JMP Instructions

The JMP instruction causes an
unconditional transfer of
control(unconditional jump).
The syntax is

IF-THEN:

• The IF-THEN structure may be expressed
in pseudocode as follow

IF (condition is true)
THEN
Execute true-branch statements
END_IF

IF-THEN-ELSE
IF condition is true
THEN
Execute true-branch statements

ELSE
Execute false-branch statements

END-IF

Loop
Loops represent the final basic control structure
(sequences, decisions, and loops) which make
up a typical program. Program loops consist of
three components: an optional initialization
component, a loop termination test, and the
body of the loop.

FOR LOOP
FOR LOOP_COUNT Initialize
count
times DO
Statements Statement

END_FOR
Count=count-1

false
count=
o

true

EXAMPLE
Write a count-controlled loop to display A to Z 26
character
The code is
MOV CX, 26
MOV AH, 2
MOV DL, ’A’
TOP:
INT 21H
LOOP TOP

While Loops
WHILE condition
Do
Statements
END_WHILE

Example
I = 0; mov I, 0
WHILE (I<100) WhileLp: cmp I, 100
I = I + 1; jge WhileDone
inc I
jmp WhileLp

REPEAT LOOP

REAPET Statement

Statements
UNTLIE condition

count=
false
o

true

Example
MOV AH,1
LOOP1:
INT 21H
CMP al, '.'
JE EndLoop
JNE LOOP1
EndLoop:

Assignment on alp

Recommended

More Related Content

What's hot (20)

Viewers also liked (11)

Similar to Assignment on alp (20)

Recently uploaded (20)

Assignment on alp