Princess Sumaya Univ. Computer Engineering Dept. Chapter 2: Chapter 2: IT Students IT Students
Princess Sumaya Univ. Computer Engineering Dept. Chapter 2: IT Students.
Dec 14, 2015
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Princess Sumaya Univ.Computer Engineering Dept.
Chapter 2:Chapter 2:
IT StudentsIT Students
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.
22 / 9 / 9
Basic Computer OrganizationBasic Computer Organization
The 5 Classic Components of a Computer
The Basic Hardware/Software Interface
The Stored-Program Concept
IT StudentsIT Students
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.
33 / 9 / 9
RISC & CISCRISC & CISC
Reduced Instruction Set Computer
● The storage concept (32 bits):
♦ Registers (R1, R2 … R31)
♦ Memory
● Arithmetic operations:
♦ Simple
♦ Limited to registers
Complicated Instruction Set Computer
● IA-32
IT StudentsIT Students
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.
44 / 9 / 9
InstructionsInstructions
Assembly Language Instruction:
Perform: a = b + c
ADD a, b, c ; a b + c
What are ‘a’, ‘b’ and ‘c’?
Sequence of Instructions:
Perform:
a = (b + c) – (d + e)IT StudentsIT Students
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Memory OperandsMemory Operands
Data Transfer Instructions
● Memory Address (Location)
● Slower than registers
Load and Store Instructions
012••
Memory
IT StudentsIT Students
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66 / 9 / 9
Constant OperandsConstant Operands
Memory Operands
Perform a = b + 4
Immediate Operands
Memory
b
4
••21
22••
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Memory AccessMemory Access
Height and Width
1 KB
4 KB
Address & Data 16 Locations 1 K Locations 1 M Locations
Read & Write
Memory
4 3
0 1
••
21
22••
1024
8 bit
1 KB
IT StudentsIT Students
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88 / 9 / 9
Stored Program ArchitectureStored Program Architecture
Instructions(Program)
Operands(Data)
Memory1024 x 32
Opcode Operands
Binary Operand
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Memory OrganizationMemory Organization
Byte Access
Word Access
• •20 Byte
21 Byte • •
• •20 Byte Byte Byte Byte
24 Byte Byte Byte Byte • •
• •20 Byte Byte
22 Byte Byte • •
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1010 / 9 / 9
Instruction RepresentationInstruction Representation
Instruction Format
Machine Language & Machine Code
Sequence of instructions
Opcode Operands ADD R1, R2, 47
5 1 2 2F
0101 0001 0010 00101111
5 1 2 2 F
AssemblyLanguage
MachineLanguage
IT StudentsIT Students
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.Chapter 2Chapter 2
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.Chapter 2 Exercise 1Chapter 2 Exercise 1
Write an assembly language program to perform a+2b where ‘a’ and ‘b’ are stored in R1 and R2.
Write an assembly language program to swap two locations at memory addresses 10 and 20.
Rewrite the following program to run faster:SUB R1, R1, R1INC R1ADD R2, R3, R1
A memory has 4 bits of address and 4 bits of data. Determine the total number of bits inside it.
IT StudentsIT Students
Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Princess Sumaya University 22342 – Computer Org. & Assembly Lang. Computer Engineering Dept.Dept.Chapter 2 Exercise 2Chapter 2 Exercise 2
Determine the value of R1:LD R1, M [ 43 ]
Write the assemblyinstruction at 44.
Assuming 8-bit registers, determine the value of R1:Load Immediate R1, Binary ‘11110010’Arith Shift Right R1
Assuming i and j are integers, R1= i and R2 = j, write the C-language statement for: Compare R1, R2 Jump Equal Label_1 Clear R1Label_1: Clear R2
IT StudentsIT Students
40 00100011010001010000000100010000
44 10110001000110100001001000010001
OpcodeOpcode Operand 1Operand 1 Operand 2Operand 2 Operand 3Operand 3
Add (= B1) Reg (5 bits) Reg (5 bits) Reg (5 bits)
Add (= B2) Reg (5 bits) Reg (5 bits) Constant
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