Stored Program Architecture

Lecture 9

The von-Neumann Architecture

Stored Program Concept - Storing instructions in memory along with data

Memory Processor - CPU Input/Output devices

I/O

input/output devices enable the user to interact with the computer input devices – keyboards, mice,

scanners, … output devices – display screens,

speakers, printers, …

Main Memory

Main memory is a large number of memory locations. Each location is accessible via an address. It holds:

– Data– Active programs - instructions for the

CPU. Storing the instructions for the cpu in memory called the stored program concept (Von Neumann) . Program is stored in the machine language (binary) that corresponds to the circuitry of that processor.

Memory

Programmer writes a program in a high level language that is easy to understand, like Java, C++ or JavaScript.

Program is translated into machine language by a compiler or interpreter.

The binary code is stored in the main memory along with the data.

CPU – Central Processing Unit

ALU - arithmetic/logic unit circuitry for arithmetic and logic operations (e.g. addition, comparison, OR)

CU - control unit fetches instructions from memory decodes the instruction executes the instruction

registers – memory locations built into the CPU (to provide

fast access) hold data for immediate use by ALU

Bus

Just like it sounds. What does a bus do?

Buses are circuits over which data travelsBuses connect parts of the the CPU and

main memoryTakes time to get data from main memory,

so it’s slower than registers.

CPU

CPU acts as the brain of the computer it fetches data and instructions from

memory it executes the instructions it stores results back to memory

(Be patient for a few slides)

Control Unit

The control unit fetches instructions from memory decodes the instruction executes the instruction

The control unit has two important registers: PC- program counter - contains the address

in main memory of the next instruction IR- instruction register - holds the

instruction that is currently executing

Machine cycle

The control unit fetches from main memory the

instruction whose address is in the program counter. The instruction is stored in the instruction register and the program counter is incremented to the next instruction address.

decodes the instruction to understand what needs to be done.

executes the instruction by performing the operation on the data.

The above three steps are repeated until the end of the stored program.

A computer can execute millions of instructions per second.

Example

S’pose the control unit is executing an instruction to add two numbers: obtain first value from memory and

store in a register obtain second value from memory and

store in a register use the addition circuitry to add the two

numbers in the registers and place the result in a register

store the result in main memory

HW

Please read chap 14 and bring book to class for lab 9

The next few slides are preparation for the lab. They are a continuation of the Addition Example

Assembly language

0: LOAD R0 5 // R0 = M[5] 1: LOAD R1 6 // R1 = M[6]2: ADD R2 R0 R1 // R2 = R0 + R13: STORE 7 R2 // M[7] = R24: HALT // HALT 5: 9 // 9 data value6: 1 // 1 data value7: 0 // will hold result

Machine Language (no spaces)

0: 100000010 00 00101 // R0 = M[5]1: 100000010 01 00110 // R1 = M[6]2: 1010000100 10 00 01 // R2 = R0 + R13: 100000100 10 00111 // M[7] = R24: 1111111111111111 // HALT5: 0000000000001001 // 9 data value6: 0000000000000001 // 1 data value7: 0000000000000000 // will hold result

Simulator Machine Language

1st two instructions – select ALU operation and registers to operate onnext 3 instructions – control flow of data. Store should be location 5.last instruction – marks the end of instruction sequence

Program Execution

example: machine language program for adding two numbers in memory

program execution:1. program counter is initialized: PC = 02. instruction at memory location 0 is fetched, PC is incremented to 13. instruction is decoded and CPU cycle is executed: load contents of memory location 5 into RO4. next instruction is fetched at location 1, PC is incremented to 25. instruction is decoded and CPU cycle is executed: load contents of memory location 6 into R16. next instruction at memory location 2 is fetched, PC is incremented to 37. instruction is decoded and CPU cycle is executed: add contents of R0 and R1, store result in R28. next instruction at location 3 is fetched, PC is incremented to 49. instruction is decoded and CPU cycle is executed: copy contents of R2 into memory location 710. next instruction at location 4 is fetched, PC is incremented to 511. instruction is decoded and CPU cycle is executed: halt instruction program ends