CS 265. Computer Architecture. Wei Lu, Ph.D., P.Eng.

Transcription

1 CS 265 Computer Architecture Wei Lu, Ph.D., P.Eng.

2 Part 3: von Neumann Architecture von Neumann Architecture Our goal: understand the basics of von Neumann architecture, including memory, control unit and arithmetic logic unit understand the basics of machine language instructions know how control unit works with machine instructions

3 Part 3: von Neumann Architecture Overview: why von Neumann architecture a rough idea about how von Neumann architecture works? main memory in von Neumann architecture arithmetic logic unit in von Neumann architecture control unit in von Neumann architecture in class practice: instruction code analysis based on von Neumann (aka IAS ) instruction set

4 Control Unit What is Control Unit? Machine Language Instructions in Control Unit How Control Unit Works

5 Typical machine instructions: an example Assuming variable: A stored in memory cell 001, B stored in memory cell 010, C stored in memory cell 011, and address 4 is assigned to register R register R A 010 B 011 C What the following instruction code does? LOAD 001 STORE 010

6 Typical machine instructions: an example LOAD A 001 A 010 B 010 B register R 011 C LOAD C 100 A 101

7 Typical machine instructions: an example Store A 001 A 010 B 010 A register R 011 C 100 A STORE C 100 A STORE X Load content of R to memory location X

8 Structure of a control unit PC (Program Counter): stores the address of next instruction to fetch IR (Instruction Register): stores the instruction fetched from memory Instruction Decoder: Decodes instruction and activates necessary circuitry PC IR +1 Instruction Decoder

9 Structure of a control unit: an example An Example about PC Register and IR register Suppose address 4 is assigned to PC register and address 5 is assigned to IR register PC Register IR Register Binary code of instruction e.g. LOAD X means instruction LOAD X

10 How control unit works with instructions Instruction Processing Fetch instruction from memory Decode instruction Evaluate address Fetch operands from memory Execute operation Store result

11 How control unit works: an example Example: Instruction Processing of ADD X,Y ADD X,Y means add content of memory locations X and Y, and store back in memory location Y. Instruction ADD X,Y stored in memory as: ADD X Y

12 How control unit works: an example suppose address 4 is assigned to MAR (memory address register), address 5 is assigned to MDR (memory data register), address 6 is assigned to IR (instruction register) and address 63 is assigned to PC (program counter) register MAR MDR Location of instruction program ADD X,Y IR PC

13 Instruction Processing: FETCH Load next instruction ADD (at address stored MAR MDR IR PC How control unit works: an example in PC) from memory into Instruction Register (IR) Load contents of PC into MAR. 2. Copy the content of memory cell with specified address into MDR 3. Read the content of MDR to IR 4. increment PC to PC+1 to point the next instruction in sequence.

14 How control unit works: an example Instruction Processing: DECODE First identify the opcode in IR, e.g. the first 8 bits means ADD under the defined instruction set ADD X Y Depending on opcode, identify other operands from the remaining bits, e.g. the address X is 0011 the address Y is 0100 F D EA OP EX S

15 How control unit works: an example Instruction Processing: Evaluate Address For instructions that require memory access, compute address used for access F D Y EA OP EX S

16 How control unit works: an example Instruction Processing: FETCH OPERANDS Obtain source operands needed to perform ADD operation e.g. read the content of X and Y to MDR F D EA X Y MDR OP EX S

17 How control unit works: an example Instruction Processing: EXECUTE Perform the operation using the source operands e.g. Obtain the content of MDR and then send them to ALU for calculation F D EA OP EX S

18 How control unit works: an example Instruction Processing: STORE Write results to destination (register or memory) e.g. result of ADD X,Y will be placed in destination Y F D EA X Y OP EX S

19 von Neumann Architecture

20 Thank you for your attendance Any questions?