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Chapter 2
Summary
Overview
• Numbers
– Decimal, Binary, Octal, Hexadecimal
– Their relationship
•
•
•
•
Sign-magnitude
One’s complement
Two’s complement
Arithmetic operation
Positive Numbers
• A computer represents positive integers in
binary
• Three methods for representing negative
numbers (signed numbers) are
– Sign-magnitude
– One’s complement
– Two’s complement
Sign Magnitude
• Representation the number’s sign and
magnitude (value)
• Sign -- Positive numbers  0 and negative
numbers  1
• e.g. 01000000 = 64 and 11000000 = -64
• Easy for human to understand but it requires
some special logic for arithmetic operations
(addition, subtraction)
One’s Complement
• The negative number is represented by flipping
the number’s bits
• E.g., 01001001 becomes 10110110
• E.g. consider 103 + -97
01100111 + (-01100001) = 01100111 + 10011110

= 100000101 (9 bits)
 00000101 + 1 = 00000110 (8 bits)
 103 – 97 = 6 = 00000110
 Try 113 + -42 = ?? and –75 + 13 = ??
Example:
Convert –5 into ones complement
representation (8 bit)
Solution:
• First, obtain +5 representation
in 8 bits  00000101
• Change every bit in the number
from 0 to 1 and vice-versa.
• –510 in ones complement is
111110102
Exercise:
Get the representation of ones
complement (6 bit) for the
following numbers:
i) +710
ii) –1010
Solution:
Solution:
(+7) = 0001112
(+10)10 = 0010102
So,
(-10)10 = 1101012
Twos complement
• Similar to ones complement, its
positive number is same as signand-magnitude
• Representation of its negative
number is obtained by adding 1 to
the ones complement of the number.
Example:
Convert –5 into twos complement
representation and give the answer
in 8 bits.
Solution:
 First, obtain +5 representation in 8
bits  000001012
 Obtain ones complement for –5
 111110102
 Add 1 to the ones complement number:
 111110102 + 12 = 111110112
 –5 in twos complement is 111110112
Exercise:
• Obtain representation of twos
complement (6 bit) for the
following numbers
i) +710
ii)–1010
Solution:
Solution:
(+7) = 0001112
(+10) 10 = 0010102
(same as sign-magnitude)
(-10) 10 = 1101012 + 12
= 1101102
So, twos compliment
for –10 is 1101102
Two’s Complement
• Rules
–
–
–
–
Just add all the bits
Throw away EAC (“end around carry”)
if a – b becomes a + (-b)
e.g.
111111 (-1)
10110 (-10)
+ 001000
(8)
+ 11101 (-3)
1000111
(7)
110011 (-13)
• try again –75 + 13 in 2’s complement
Chapter 3
Summary
Relationship Between Basic Operation
of Boolean and Basic Logic Gate
• The basic construction of a logical circuit is gates
• Gate is an electronic circuit that emits an output signal as a
result of a simple Boolean operation on its inputs
• Logical function is presented through the combination of
gates
• The basic gates used in digital logic is the same as the basic
Boolean algebra operations (e.g., AND, OR, NOT,…)
• The package Truth Tables and Boolean Algebra set out the basic
principles of logic.
Name
Graphic Symbol
AND
A
B
F
OR
A
B
F
NOT
NAND
NOR
A
A
B
A
B
Boolean Algebra
F =A. B
Or
F = AB
F =A+ B
_
F =A
F
____
F=A.B
Or
F = AB
F
F
_____
F=A+B
Truth Table
A
0
0
1
1
A
0
0
1
1
B
0
1
0
1
B
0
1
0
1
B
0
1
F
0
0
0
1
F
0
1
1
1
F
1
0
A B
0 0
0 1
1 0
1 1
F
0
1
1
1
A B
0 0
0 1
1 0
1 1
F
0
0
0
1
the symbols, algebra signs and the truth table for the gates
Basic Theorems of Boolean Algebra
1. Identity Elements
1.A=A
0+A=A
2. Inverse Elements
A.A=0
A+A=1
3. Idempotent Laws
A+A=A
A.A=A
4. Boundess Laws
A+1=1
A.0=0
5. Distributive Laws
A . (B + C) = A.B + A.C
A + (B . C) = (A+B) . (A+C)
6. Order Exchange Laws
A.B=B.A
A+B=B+A
7. Absorption Laws
A + (A . B) = A
A . (A + B) = A
9. Elimination Laws

A + (A . B) = A + B

A . (A + B) = A . B
8. Associative Laws
A + (B + C) = (A + B) + C
A . (B . C) = (A . B) . C
10. De Morgan Theorem
  
(A + B) = A . B
  
(A . B) = A + B
Relationship Between Boolean
Function and Logic Circuit
Boolean function  Q = AB + B
= (NOT A AND B) OR B
Logic circuit
A
A
AB
B
B
Q
= AB + B
Relationship Between Boolean
Function and Logic Circuit
• Any Boolean function can be implemented in electronic
form as a network of gates called logic circuit
A
B
A.B = AB
F = AB + C + D
C
D
C+D
G = A . (B + C + D)
A
G = A . (B + C + D)
B
C
D
B+C+D
C+D
Truth Table
A
A
AB
B
B
Produce a truth table from the logic circuit
A
B
A
AB
Q
0
0
0
1
1
1
0
1
0
1
1
0
0
0
0
1
1
0
0
1
Q
= AB + B
Karnaugh Map
• A graphical way of depicting the content of a truth table
where the adjacent expressions differ by only one variable
• For the purposes simplification, the Karnaugh map is a
convenient way of representing a Boolean function of a
small number (up to four) of variables
• The map is an array of 2n squares, representing all possible
combination of values of n binary variables
• Example: 2 variables, A and B
B
B
B
A
AB
AB
A
AB
AB
A
B
A
0
1
1
0
00
01
10
11
4 variables, A, B, C, D  24 = 16 squares
AB
CD
CD
CD
AB
0000
0001
AB
0100
AB
1100
AB
1000
CD
CD
• List combinations in
the order 00, 01, 11, 10
00
01
11
C
AB
AB
AB
AB
0 C
000
010
110
100
1 C
001
011
111
101
C
0
C
1
C
00
AB
000
001
01
AB
010
011
11
AB
110
111
10
AB
100
101
AB
AB
10
How to create Karnaugh Map
Truth Table
A
B
C
F
0
0
0
1
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
0
1
1
1
0
1. Place 1 in the corresponding
square
Karnaugh Map
BC
BC
00
A 0
1
A
A 1
1
BC
01
BC
11
1
1
BC
10
Karnaugh Maps to Represent Boolean Functions
AB
00
AB
01
AB
1
11
AB
10
AB
1
F = AB + AB
2. Group the adjacent squares:
Begin grouping square with 2n-1 for n variables
• e.g. 3 variables, A, B, and C
23-1 = 22 = 4
= 21 = 2
= 20 = 1
BC
BC
00
A 0
1
A
1
A 1
BC
BC
01
BC
11
BC
10
1
1
AB
ABC
F = BC + AB + ABC
BC
A
BC
00
BC
11
BC
10
1
A 0
A 1
BC
01
1
1
A
F = A + BC
BC
1
1
3 variables:
23-1 = 22 = 4
22-1 = 21 = 2
21-1 = 20 = 1
4 variables, A, B, C, D  24-1 = 23 = 8 (maximum); 22 = 4;
21 = 2; 20 = 1 (minimum);
AB
CD
00
00
1
01
1
11
1
10
1
F=
01
11
10
1
1
CD + BD + ABC
1
Karnaugh Map
Boolean Function
Logic Circuit
Chapter 5
Summary
Contains the beginning
address of each segment
SS Register
Address
Stack segment
DS Register
Address
Data segment
CS Register
Address
Code segment
Segment register
(in CPU)
memory
(MM)
32 bits
8 bit
8 bit
AH
AL
AX
EAX
Value in AX = FFFFH, what is the value in AX after the
following instruction is executed?
MOV AL, 09H
AX = ????
MOV AX,0009H
AX = ????
32 bits
8 bit
8 bit
CH
CL
CX
ECX
Value in CX = FFFFH, what is the value in CX after the
following instruction is executed?
MOV CX, 25H
CX = ????
MOV CL, 0
CX = ????
How many bytes/ bits for the following instructions?
MOV AX, 0123
MOV AH, 09
What are the values in CS register and IP register?
What is the value in IP after MOV CL,42 is executed?
CHAPTER 6
ASSEMBLY LANGUAGE
PROGRAM FORMAT AND DATA
DEFINITION
6.1) Assembly Language Program Format
6.2) Features of Assembly Language
6.3) Data Definition
PAGE directive to
establish 60 lines
and 132 columns
per page
TITLE directive
to identify the
program’s name
as A04ASM1
CODESEG to
define the code
segment
; symbol
is for
comment
STACK to
define the
stack
segment
DATASEG
to define
the data
segment
ASSUME
directive is used
to tell the
assembler the
starting address
of segments with
the segment
registers
Procedure MAIN
END directive to
tell the
assembler that
this is the end of
the source
program
Initialize
the
address
of data
segment
in DS
Request to
end the
program
and return
to the OS
Ending Program Execution
After executing an assembly language program, the programmer
must tell the system to terminate the executing program with the
help of DOS interrupt services.
INT 21H is the commonly used interrupt service. It used the
function code in the AH register to determine the next course of
action.
INT 21H can also be used to control input from the keyboard,
control the screen, disk I/O and output to the printer.
INT 21H with function code 4CH is used to terminate program
execution. The function code 4CH must be priory entered into
AH. Example:
Data Definition
Assembler offers a few directives that enable programmers to define data
according to its type and length. Format for data definition:
[name]
Data names are optional because in assembly language programming, data is
not necessarily reference by its name.
Dn Directive
Next slide are the common directives to define data and also directives used in
MASM 6.0
The following are some examples of numeric and character data definition
Page
60,
132
TITLE
A04DEFIN (EXE) Define data directives
.MODEL SMALL
.DATA
;DB – Define Bytes:
;----------------------BYTE1
DB
?
; Uninitialized
BYTE2
DB
48
; Decimal constant
BYTE3
DB
30H
; Hex constant
BYTE4
DB
01111010B
; Binary constant
BYTE5
DB
10
DUP
(0)
; Ten zeros
BYTE6
DB
‘PC FAIR’
; Character string
BYTE7
DB
‘12345’
; Number as characters
BYTE8
DB
01, ‘Jan’, 02, ‘Feb’ ; Table of months
;DW – Define Words:
;------------------------WORD1 DW
0FFF0H
WORD2 DW
01111010B
WORD3 DW
BYTE8
WORD4 DW
2, 4, 6, 7, 9
WORD5 DW
6 DUP (0)
;DQ – Define Doublewords:
;--------------------------------DWORD1
DD
?
DWORD2
DD
41562
DWORD3
DD
24, 48
DWORD4
DD
BYTE3 – BYTE2
END
; Hex constant
; Binary constant
; Address constant
; Table of 5 constants
; Six zeros
; Uninitialized
; Decimal value
; Two constants
; Difference between addresses
DB or BYTE
-to define item with the size of one byte. The range of its value is
stated in the table before.
DW or WORD
-to define item with the size of one word or 2 bytes. The range of
its value is stated in the table before. Assembler will change numeric
constants into binary object code (presented in hexadecimal) and
kept in the memory in reverse bytes. For instance, if the real value is
3039H it will be kept as 3930H in the data segment
DD or DWORD
- to define item with the size of 4 bytes. Its range of values is stated
in the table above. As usual data is kept in reverse byte or reverse
sequence. For example, if data is 00BC614EH it will be kept as
4E61BC00H.
Expressions
Expressions in operand may specify an uninitialized value or a constant
value. Example:
DATAX
DATAYDB
DB
25
?
; Uninitialized item, size of 1 bait
; Initialized item, DATAY with value 25
Uninitialized item is used to store a value which size is defined. The value of a
data can be used and edited to suit the program’s needs. Expressions can
contain a few constants that is separated by the sign ‘,’ and the quantity is
limited to the row length.
Example:
DATAZ
DB
21, 22, 23, 24, 25, 26, …
The assembler defines the above constant byte by byte , from left to right.
DATAZ or DATAZ+0 contains the value 21, DATAZ+1 contains 22,
DATAZ+2 contains 23 and so forth. Example of instruction MOV AL,
DATAZ+3 will enter the value 24 into the CL register
Expressions also allows duplication of constants
using the format below:
Example:
DW
10
DB
5
0C0C0C0C0C
DB
3
DUP(?)
DUP(12)
; Ten words, uninitialized
; Five bytes containing
DUP(5 DUP(4)) ; Fifteen 4s
CHAPTER 7
ASSEMBLY LANGUAGE
INSTRUCTIONS
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
Introduction
Data Transfer Instructions
Arithmetic Instructions
Bit Shifting Instructions
Looping Instructions
Unconditional Transfer Instructions
Conditional Jump Instructions
Other Instructions
Data Transfer Instructions
Some examples on MOV:
BYTEFLD
WORDFLD
DB
DW
…
MOV
MOV
MOV
MOV
MOV
EDX , ECX
BYTEFLD , DH
WORDFLD , 1234
CH , BYTEFLD
CX , 40H
MOV
AX , DS
?
?
; define byte
; define word
;
;
;
;
;
register to register
register to memory
immediate to memory
memory to register
immediate to register
; segment register to register
7.3.1 Addition and Subtraction Of
Binary Data
Example of the ADD and SUB instructions:
BYTE1
WORD1
MOV
MOV
ADD
SUB
ADD
MOV
MOV
SUB
SUB
ADD
DB
DW
24H
4000H
. . .
CL , BYTE1
DL , 40H
CL , DL
CL , 20H
BYTE1 , BL
CX , WORD1
DX , 2000H
CX , DX
CX , 124H
WORD1 , DX
;Data elements
; byte processing
;
;
;
;
register to register
Immediate from register
register to memory
word processing
; register from register
; Immediate from memory
; register to memory
MUL is used for unsigned data
Examples on the usage of the MUL instructions using the
data as defined below:
BYTE1
BYTE2
WORD1
WORD2
DWORD1
DWORD2
DB
DB
DW
DW
DD
DD
80H
40H
8000H
2000H
00018402H
00012501H
(a)
MOV
MUL
AL, BYTE1
BYTE2
; AL (multiplicand)=80H
; byte x byte, product in AX
; 80H x 40H, AX= 2000H
(b)
MOV
MUL
AX, WORD1
WORD2
;
;
;
;
AX (multiplicand)=8000H
word x word, product in DX:AX
80000H x 2000H,
DX:AX= 1000 0000H
The following are a few examples of the IDIV instruction using the data
definition below:
BYTE1
BYTE2
WORD1
WORD2
WORD3
DB
DB
DW
DW
DW
80H
16H
2000H
0010H
1000H
(a) MOV AX, WORD1
IDIV BYTE1
;
;
;
;
(b) MOV DX, WORD2
MOV AX, WORD3
;DX=0010H
IDIV WORD1
; Byte value
; Word value
AX=2000H
2000H(+ve)/80H (-ve),
quotient=C0H (-ve), remainder=00H
AL=C0H, AH=00H
;AX=1000H,dividend in DX:AX (WORD2:WORD3)
;DX:AX = 0010 1000H (+ve)
;00101000H (+ve)/2000H (+ve)
;remainder:quotient in DX:AX
;1000H:0080H
Example of the SHR instruction
As in the example above, the SHR instruction will enter the value 0 to the
leftmost bit after the shift. Carry flag will contain the last bit shifted out after
the shift
C
BH
1
0
1
1
0
1
1
1
0
1
0
1
1
0
1
1
1
0
0
0
1
0
1
1
0
1
SHR BH, 01
0
SHR BH, CL
00
SHR BH, 02
00
0
0
0
0
0
1
0
1
1
Example
of
the
SAR
instruction
The SAR instruction is used on signed number. SAR will enter the sign
bit (whether 0 (+ve) or 1 (–ve)) into the leftmost bit after every shift.
Examples of SAR instruction
BH
1
0
1
1
0
1
1
1
1
1
0
1
1
0
1
1
1
1
1
1
1
0
1
1
0
1
SAR BH, 01
1
SAR BH, CL
11
SAR BH, 02
11
1
1
1
1
1
1
0
1
1
0
0
0
0
0
1
0
1
BH
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
0
1
0
1
0
0
0
0
0
00
00
A few examples on ROR:
1
2
3
The difference between ROR and RCR is only the way of operation.
In RCR, every bit that is rotated will enter the carry flag before
entering the leftmost bit
1
BH
1
0
1
1
0
1
1
1
1
2
1
1
0
1
1
0
1
1
1
011
3
0
011
1
1
1
1
0
1
1
0
Below are instances of the ROL instruction :
1
2
3
1
2
1
0
1
1
0
1
1
1
0
1
1
0
1
1
1
1
BH
1
1
1
011
3
1
0
1
1
1
1
0
1
1
011
LOOP Instruction – example using
DEBUG
-A 100
4A66:0100
4A66:0103
4A66:0106
4A66:0108
MOV CX,5
MOV AX,0
ADD AX,CX
LOOP 106
;LOOP COUNTER=5
;LOOP TO LOCATION 0106