1. Stored Program Architecture
Lecture 9

2. The von-Neumann Architecture
Stored Program Concept - Storing instructions in memory along with data
Memory
Processor - CPU
Input/Output devices

3. I/O input/output devices enable the user to interact with the computer
input devices – keyboards, mice, scanners, …
output devices – display screens, speakers, printers, …

4. 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.

5. 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.

6. 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

7. Bus Just like it sounds. What does a bus do?
Buses are circuits over which data travels
Buses connect parts of the the CPU and main memory
Takes time to get data from main memory, so it’s slower than registers.

8. 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)

10. 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

11. 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.

12. The above three steps are repeated until the end of the stored program.
A computer can execute millions of instructions per second.

13. 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

14. 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

15. Assembly language 0: LOAD R0 5 // R0 = M[5] 1: LOAD R1 6 // R1 = M[6]
2: ADD R2 R0 R1 // R2 = R0 + R1
3: STORE 7 R2 // M[7] = R2
4: HALT // HALT
5: 9 // 9 data value
6: 1 // 1 data value
7: 0 // will hold result

16. Machine Language (no spaces)
0: // R0 = M[5]
1: // R1 = M[6]
2: // R2 = R0 + R1
3: // M[7] = R2
4: // HALT
5: // 9 data value
6: // 1 data value
7: // will hold result

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

18. Program Execution
example: machine language program for adding two numbers in memory
program execution:
program counter is initialized: PC = 0
instruction at memory location 0 is fetched, PC is incremented to 1
instruction is decoded and CPU cycle is executed: load contents of memory location 5 into RO
next instruction is fetched at location 1, PC is incremented to 2
instruction is decoded and CPU cycle is executed: load contents of memory location 6 into R1
next instruction at memory location 2 is fetched, PC is incremented to 3
instruction is decoded and CPU cycle is executed: add contents of R0 and R1, store result in R2
next instruction at location 3 is fetched, PC is incremented to 4
instruction is decoded and CPU cycle is executed: copy contents of R2 into memory location 7
next instruction at location 4 is fetched, PC is incremented to 5
instruction is decoded and CPU cycle is executed: halt instruction  program ends