Data Manipulation CSC 2001. Overview Computer Architecture Machine Language Computer Architecture...

Data ManipulationData Manipulation

CSC 2001CSC 2001

OverviewOverview

Computer ArchitectureMachine Language

Computer ArchitectureMachine Language

DisclaimerDisclaimer

There is A LOT more detail to computer architecture & machine language than what I’ll go over today.

That detail is of great interest, but beyond the scope of this class.

Basically, this is a simplified, high-level view. We can explore more outside of class if you are interested.

There is A LOT more detail to computer architecture & machine language than what I’ll go over today.

That detail is of great interest, but beyond the scope of this class.

Basically, this is a simplified, high-level view. We can explore more outside of class if you are interested.

General approachGeneral approach

Build up a basic understanding of architecture and begin to look at how it gets used to manipulate data.

Build up a basic understanding of architecture and begin to look at how it gets used to manipulate data.

Computer ArchitectureComputer Architecture

Inputdevice Output

device

Auxiliarystoragedevice

Central Processing Unit

Control Unit

Arithmetic/logicunit

Memory unit

Computer ArchitectureComputer Architecture

Central Processing Unit

Control Unit

Arithmetic/logicunit

Memory unit

Bus

Central Processing Unit (CPU)

Central Processing Unit (CPU)

Central Processing Unit

Control Unit

Arithmetic/logicunit

• Pentium (Intel)• PowerPC (Motorola & IBM)• SPARC

CPUCPU

CPUCPU

~55 milliontransistors

Control UnitControl Unit

Central Processing Unit

Control Unit

Arithmetic/logicunit

Memory unit

Bus

Control UnitControl Unit

Control Unit

Circuits for coordinatingthe machine’s activities

Arithmetic/Logic UnitArithmetic/Logic Unit

Central Processing Unit

Control Unit

Arithmetic/logicunit

Memory unit

Bus

Arithmetic/Logic UnitArithmetic/Logic Unit

Arithmetic/logicunit

Circuits for performingdata manipulation.

MemoryMemory

Main memoryRegistersCache memory

Main memoryRegistersCache memory

RegistersRegisters

The CPU circuitry has to have a place to hold data that it is working with.

Very fast accessUse:

Control unit transfers data from memory to registers.

Informs the ALU where the data is.Activates the ALUTells it which register to store the result in

The CPU circuitry has to have a place to hold data that it is working with.

Very fast accessUse:

Control unit transfers data from memory to registers.

Informs the ALU where the data is.Activates the ALUTells it which register to store the result in

Example: Adding two numbers

Example: Adding two numbers

Load a register with one value from main memory (control unit/main memory/bus)

Load another register with a value from main memory (control unit/main memory/bus)

Tell the ALU to add two specific registers and where to store the result (control unit/ALU)

Do the addition and store the result (ALU)

Copy the result to main memory (control unit/main memory/bus)

Load a register with one value from main memory (control unit/main memory/bus)

Load another register with a value from main memory (control unit/main memory/bus)

Tell the ALU to add two specific registers and where to store the result (control unit/ALU)

Do the addition and store the result (ALU)

Copy the result to main memory (control unit/main memory/bus)

Cache memoryCache memory

Purpose:Store pertinent part of main memory

on CPU itself for faster access.Level 1 (L1/primary):

typically small, but on CPULevel 2 (L2/secondary/backside):

small compared to main memory, larger than L1 cache, not on CPU

Purpose:Store pertinent part of main memory

on CPU itself for faster access.Level 1 (L1/primary):

typically small, but on CPULevel 2 (L2/secondary/backside):

small compared to main memory, larger than L1 cache, not on CPU

BusBus

Central Processing Unit

Control Unit

Arithmetic/logicunit

Memory unit

Bus

BusBus

Central Processing Unit

Memory unit

Bus

Physically transfersdata between CPUand main memory

Early programmingEarly programming

Hardwiring (rewiring) the control unit.

Hardwiring (rewiring) the control unit.

BreakthroughBreakthrough

Program can be encoded and stored like data.

Generalize control unit to read and execute instructions.

A computer’s set of instructions and the encoding system comprise a machine language.

Program can be encoded and stored like data.

Generalize control unit to read and execute instructions.

A computer’s set of instructions and the encoding system comprise a machine language.

Machine LanguageMachine Language

Instruction set:CPU dependentUsually relatively smallTwo philosophies:

Reduced Instruction Set Computer (RISC)minimal set of instructions; efficient and fastexample?

Complex Instruction Set Computer (CISC) larger set; easy to program (at the machine level);

one CISC instruction = several RISC instructionsexample?

Instruction set:CPU dependentUsually relatively smallTwo philosophies:

Reduced Instruction Set Computer (RISC)minimal set of instructions; efficient and fastexample?

Complex Instruction Set Computer (CISC) larger set; easy to program (at the machine level);

one CISC instruction = several RISC instructionsexample?

Instruction categoriesInstruction categories

Data transferArithmetic/logicControl

Data transferArithmetic/logicControl

Data transferData transfer

Movement of dataLOAD or STORE

Movement of dataLOAD or STORE

Arithmetic/logicArithmetic/logic

AND, OR, XOR, SHIFT, ROTATEAND, OR, XOR, SHIFT, ROTATE

ControlControl

directs the execution of programJUMP to another part of the

program

directs the execution of programJUMP to another part of the

program

ExampleExample

Appendix CReview machine architectureReview syntax of languageReview semantics of language

ExampleFigure 2.2/Figure 2.7

Appendix CReview machine architectureReview syntax of languageReview semantics of language

ExampleFigure 2.2/Figure 2.7

Another exampleAnother example

Store the ASCII letter A in memory location 56.

Store the ASCII letter B in memory location 112.

Swap the bit patterns in memory locations 56 and 112

Halt

Store the ASCII letter A in memory location 56.

Store the ASCII letter B in memory location 112.

Swap the bit patterns in memory locations 56 and 112

Halt

First stepsFirst steps

ASCII A?01000001 (Hex: 0100 0001 = 41)

ASCII B?01000010 (Hex: 0100 0010 = 42)

Hex 56?38 (3*16 + 8 = 48 + 8 = 56)

Hex 112?70 (7*16 + 0 = 112)

ASCII A?01000001 (Hex: 0100 0001 = 41)

ASCII B?01000010 (Hex: 0100 0010 = 42)

Hex 56?38 (3*16 + 8 = 48 + 8 = 56)

Hex 112?70 (7*16 + 0 = 112)

ProgramProgram

Store the ASCII letter A in memory location 56.21413138

Store the ASCII letter B in memory location 112.21423170

Store the ASCII letter A in memory location 56.21413138

Store the ASCII letter B in memory location 112.21423170

Program (continued)Program (continued)

Swap the bit patterns in memory locations 56 and 1121138127031703238

HaltC000

Swap the bit patterns in memory locations 56 and 1121138127031703238

HaltC000

Final programFinal program

21413138214231701138127031703238C000

21413138214231701138127031703238C000

Simple program completelystorable as data.

Practice problemsPractice problems

If you find this difficult, do some of the practice problems on page 82 (answers start on page 518).

Do enough to either convince yourself you understand or find the places you are having problems!!

If you find this difficult, do some of the practice problems on page 82 (answers start on page 518).

Do enough to either convince yourself you understand or find the places you are having problems!!