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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 + -9701100111 + (-01100001) = 01100111 + 10011110 = 100000101 (9 bits)00000101 + 1 = 00000110 (8 bits)103 – 97 = 6 = 00000110Try 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:
(+7) = 0001112
Solution:
(+10)10 = 0010102
So,
(-10)10 = 1101012
Twos complement
• Similar to ones complement, its positive number is same as sign-and-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:
(+7) = 0001112(same as sign-magnitude)
Solution:
(+10) 10 = 0010102
(-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.
A B F0 0 00 1 01 0 01 1 1
A B F0 0 00 1 11 0 11 1 1
A B F0 0 00 1 11 0 11 1 1
A B F0 0 00 1 01 0 01 1 1
F
F
F
F
Name Graphic Symbol Boolean Algebra Truth Table
AB
AB
AB
AB
A F
AND
OR
NOT
NAND
NOR
F = A . B Or F = AB
F = A + B
_____ F = A + B
____ F = A . B Or F = AB
_ F = A
B F 0 1 1 0
the symbols, algebra signs and the truth table for the gates
1. Identity Elements 2. Inverse Elements 1 . A = A A . A = 0 0 + A = A A + A = 1 3. Idempotent Laws 4. Boundess Laws A + A = A A + 1 = 1 A . A = A A . 0 = 0 5. Distributive Laws 6. Order Exchange Laws A . (B + C) = A.B + A.C A . B = B . A A + (B . C) = (A+B) . (A+C) A + B = B + A 7. Absorption Laws 8. Associative Laws A + (A . B) = A A + (B + C) = (A + B) + C A . (A + B) = A A . (B . C) = (A . B) . C 9. Elimination Laws 10. De Morgan Theorem A + (A . B) = A + B (A + B) = A . B A . (A + B) = A . B (A . B) = A + B
Basic Theorems of Boolean Algebra
Relationship Between Boolean Function and Logic Circuit
A
B Q
Boolean function Q = AB + B = (NOT A AND B) OR B
Logic circuitA
AB
B= 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
AB
F
A.B = AB
CD C + D
= AB + C + D
G = A . (B + C + D)
A
B
CD
G = A . (B + C + D)
C + D
B + C + D
Truth Table
A
B Q
AAB
B= AB + B
Produce a truth table from the logic circuit
A B A AB Q
0 0 1 0 0
0 1 1 1 1
1 0 0 0 0
1 1 0 0 1
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
A B A B
A B A B
00 01
10 11
BA B
A
B
A
BA
1 0
1
0
0000 0001
0100
1100
1000
AB C D
A B
C D
A B
CD
A B
A B
C DC D
4 variables, A, B, C, D 24 = 16 squares
000 010 110 100
001 011 111 101
AB
C
A B
C
A BC A B A B
000 001
010 011
110 111
100 101
AB C
A B
C
A B
A B
A B
00 01 11 10
0
1
00
01
11
10
0 1
• List combinations in the order 00, 01, 11, 10
C
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
Truth Table
Karnaugh Map
1 1
1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C
How to create Karnaugh Map
1. Place 1 in the corresponding square
1 1
0 0 0 1 1 1 1 0AB
F = AB + AB
A BA B A B A B
Karnaugh Maps to Represent Boolean Functions
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
1 1
1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C
ABBC ABC F = BC AB ABC+ +
1
1 1 1 1
0 0 0 1 1 1 1 0BC
A
0
1
A
A
B CB C B C B C3 variables: 23-1 = 22 = 4 22-1 = 21 = 2 21-1 = 20 = 1
A
BC
F = A + BC
1 1
1
1
1 1 1
AB 01
01
00
00
CD
11
10
1011
4 variables, A, B, C, D 24-1 = 23 = 8 (maximum); 22 = 4;21 = 2; 20 = 1 (minimum);
CD + BD ABC+F =
Karnaugh Map
Boolean Function
Logic Circuit
Chapter 5
Summary
Address
Address
Address
Stack segment
Data segment
Code segment
Contains the beginning address of each segment
Segment register (in CPU)
memory(MM)
SS Register
DS Register
CS Register
MOV AL, 09HAX = ????
8 bit 8 bit
32 bits
AH AL
AX
EAX
MOV AX,0009HAX = ????
Value in AX = FFFFH, what is the value in AX after the following instruction is executed?
8 bit 8 bit
32 bits
CX
CH CL
ECX
MOV CX, 25HCX = ????
MOV CL, 0CX = ????
Value in CX = FFFFH, what is the value in CX after the following instruction is executed?
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?
6.1) Assembly Language Program Format
6.2) Features of Assembly Language
6.3) Data Definition
CHAPTER 6
ASSEMBLY LANGUAGE PROGRAM FORMAT AND DATA
DEFINITION
PAGE directive to establish 60 lines and 132 columns per page
TITLE directive to identify the program’s name as A04ASM1
; symbol is for comment
STACK to define the stack segment
DATASEG to define the data segmentCODESEG to
define the code segment
ASSUME directive is used to tell the assembler the starting address of segments with the segment registers
Initialize the address of data segment in DS
Procedure MAIN
END directive to tell the assembler that this is the end of the source program
Request to end the program and return to the OS
Ending Program ExecutionAfter 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, 132TITLE A04DEFIN (EXE) Define data directives
.MODEL SMALL
.DATA;DB – Define Bytes:;-----------------------BYTE1 DB ? ; UninitializedBYTE2 DB 48 ; Decimal constantBYTE3 DB 30H ; Hex constantBYTE4 DB 01111010B ; Binary constantBYTE5 DB 10 DUP (0) ; Ten zerosBYTE6 DB ‘PC FAIR’ ; Character stringBYTE7 DB ‘12345’ ; Number as charactersBYTE8 DB 01, ‘Jan’, 02, ‘Feb’ ; Table of months
;DW – Define Words:;-------------------------WORD1 DW 0FFF0H ; Hex constantWORD2 DW 01111010B ; Binary constantWORD3 DW BYTE8 ; Address constantWORD4 DW 2, 4, 6, 7, 9 ; Table of 5 constantsWORD5 DW 6 DUP (0) ; Six zeros
;DQ – Define Doublewords:;---------------------------------DWORD1 DD ? ; UninitializedDWORD2 DD 41562 ; Decimal valueDWORD3 DD 24, 48 ; Two constantsDWORD4 DD BYTE3 – BYTE2 ; Difference between addressesEND
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 DB ? ; Uninitialized item, size of 1 baitDATAY DB 25 ; 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 DUP(?) ; Ten words, uninitializedDB 5 DUP(12) ; Five bytes containing
0C0C0C0C0CDB 3 DUP(5 DUP(4)) ; Fifteen 4s
CHAPTER 7 ASSEMBLY LANGUAGE
INSTRUCTIONS
7.1 Introduction
7.2 Data Transfer Instructions
7.3 Arithmetic Instructions
7.4 Bit Shifting Instructions
7.5 Looping Instructions
7.6 Unconditional Transfer Instructions
7.7 Conditional Jump Instructions
7.8 Other Instructions
Data Transfer InstructionsSome examples on MOV:
BYTEFLD DB ? ; define byteWORDFLD DW ? ; define word
…MOV EDX , ECX ; register to registerMOV BYTEFLD , DH ; register to memoryMOV WORDFLD , 1234 ; immediate to memoryMOV CH , BYTEFLD ; memory to registerMOV CX , 40H ; immediate to register
MOV AX , DS ; segment register to register
Example of the ADD and SUB instructions:
BYTE1 DB 24H ;Data elements WORD1 DW 4000H
. . .MOV CL , BYTE1 ; byte processing MOV DL , 40HADD CL , DL ; register to register SUB CL , 20H ; Immediate from register ADD BYTE1 , BL ; register to memory MOV CX , WORD1 ; word processing MOV DX , 2000HSUB CX , DX ; register from registerSUB CX , 124H ; Immediate from memoryADD WORD1 , DX ; register to memory
7.3.1 Addition and Subtraction Of Binary Data
MUL is used for unsigned data Examples on the usage of the MUL instructions using the data as defined below:
BYTE1 DB 80HBYTE2 DB 40HWORD1 DW 8000HWORD2 DW 2000HDWORD1 DD 00018402HDWORD2 DD 00012501H
(a) MOV AL, BYTE1 ; AL (multiplicand)=80HMUL BYTE2 ; byte x byte, product in AX
; 80H x 40H, AX= 2000H
(b) MOV AX, WORD1 ; AX (multiplicand)=8000HMUL WORD2 ; 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 DB 80H ; Byte valueBYTE2 DB 16HWORD1 DW 2000H ; Word valueWORD2 DW 0010HWORD3 DW 1000H
(a) MOV AX, WORD1 ; AX=2000H IDIV BYTE1 ; 2000H(+ve)/80H (-ve),
; quotient=C0H (-ve), remainder=00H; AL=C0H, AH=00H
(b) MOV DX, WORD2 ;DX=0010H MOV AX, WORD3 ;AX=1000H,dividend in DX:AX (WORD2:WORD3)
;DX:AX = 0010 1000H (+ve)
IDIV WORD1 ;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
1 0 1 1 0 1 1 1BH
C
0 1 0 1 1 0 1 1 1
SHR BH, 01
0 0 0 1 0 1 1 0 1
SHR BH, CL
0 0 0 0 0 1 0 1 1
SHR BH, 02
0
00
00
Example of the SAR instructionThe 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
1 0 1 1 0 1 1 1BH
1 1 0 1 1 0 1 1 1
SAR BH, 01
1 1 1 1 0 1 1 0 1
SAR BH, CL
1 1 1 1 1 1 0 1 1
SAR BH, 02
1
11
11
0 0 0 0 0 1 0 1 BH
0 0 0 0 0 1 0 1 0
0 0 0 1 0 1 0 0 0
0 1 0 1 0 0 0 0 0
00
0
00
A few examples on ROR:
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 0 1 1 0 1 1 1BH
1 1 0 1 1 0 1 1 1
0 1 1 1 1 0 1 1 0011
1
011
1
2
3
1
2
3
Below are instances of the ROL instruction :
1 0 1 1 0 1 1 1 BH1
2
3
1
1 0 1 1 0 1 1 1 11
1
2
3
1 0 1 1 1 1 0 1 1
011
011
LOOP Instruction – example using DEBUG
-A 1004A66:0100 MOV CX,5 ;LOOP COUNTER=54A66:0103 MOV AX,04A66:0106 ADD AX,CX4A66:0108 LOOP 106 ;LOOP TO LOCATION 0106