1 / 17

Chap 4 & 5

Chap 4 & 5. Chap 4 Homework – due Monday October 27 Chap 5 Homework – due Wednesday October 29 Project 2 Designs (Working Schematics) – due Wednesday October 29 Project 2 Reports – due Wednesday November 5. LC-3 Computer LC-3 Instructions.

javier
Download Presentation

Chap 4 & 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chap 4 & 5 Chap 4 Homework – due Monday October 27 Chap 5 Homework – due Wednesday October 29 Project 2 Designs (Working Schematics) – due Wednesday October 29 Project 2 Reports – due Wednesday November 5 LC-3 Computer LC-3 Instructions Project 1Feedback – I would like to see more clarification/discussion of what you did and observed, especially linking the timing traces to the circuits. Note: Alt PrintScreen is a good way to get a 1 page schematic with timing traces

  2. LC-3 Memory Map (64K of 16 bit words) 256 words 256 words 23.5 K words 39.5 K words 512 words

  3. The LC-3 Computeravon Neumannmachine • The Instruction Cycle • Fetch: Next Instruction from Memory • (PC)  (points to) next instruction • PC (PC) + 1 • Decode: Fetched Instruction • Evaluate: Instr &Address (es) • (find where the data is) • Fetch: Operand (s) • (get data as specified) • Execute: Operation • Store: Result • (if specified) PSW Memory PSW (Program Status Word): Bits: 15 10 9 8 2 1 0 | S| |Priority| | N| Z| P|

  4. Computer Machine Instruction Formats • What is IN an instruction? • Operation code – what to do • Input Operand(s) – where to get input operands (memory, registers) • Output Operand(s) – Where to put results (memory, registers) • What are the major instruction types? • Data Movement (load, store, etc.) • Operate (add, sub, mult, OR, AND, etc.) • Control (branch, jump to subroutine, etc.)

  5. LC-3 Instructions (Fig 5.3 – Appendix a) • Addressing Modes • Register • (Operand is in one of the 8 registers) • PC-relative • (Operand is “offset” from where the PC points) • Base + Offset (Base relative) • (Operand is “offset” from the contents of a register) • Immediate • (Operand is in the instruction) • Indirect • (The “Operand” points to the real address of Operand • – rather than being the operand)

  6. TRAP Instruction • Calls a service routine, identified by 8-bit “trap vector.” • When service routine is done, PC is set to the instruction following TRAP (using RET instruction).

  7. TRAPS See page 543.

  8. Example LC-3 Program • Write a program to add 12 integers and store the result in a Register.

  9. Compute the Sum of 12 Integers Program • Program begins at location x3000. • Integers begin at location x3100. R1  x3100R3  0 (Sum)R2  12(count) R2=0? R4  M[R1] R3  R3+R4R1  R1+1 R2  R2-1 NO YES R1: “Array” index pointer (Begin with location 3100) R3: Accumulator for the sum of integers R2: Loop counter (Count down from 12) R4: Temporary register to store next integer

  10. Sum integers from x3100 – x310B R1: “Array” index pointer (Begin with location 3100) R3: Accumulator for the sum of integers R2: Loop counter (Count down from 12) R4: Temporary register to store next integer

  11. The Sum program in “binary” 0011000000000000 ;start x3000 x3000 1110001011111111 ;R1=x3100 x3001 0101011011100000 ;R3=0 x3002 0101010010100000 ;R2=0 x3003 0001010010101100 ;R2=R2+12 x3004 0000010000000101 ;If z goto x300A x3005 0110100001000000 ;Load next value into R4 x3006 0001011011000100 ;R3=R3+R4 x3007 0001001001100001 ;R1=R1+1 x3008 0001010010111111 ;R2=R2-1 x3009 0000111111111010 ;goto x3004 x300A 1111000000100101 ;halt

  12. The Sum program in “hex” 3000 ;start x3000 x3000 E2FF ;R1=x3100 x3001 56E0 ;R3=0 x3002 54A0 ;R2=0 x3003 14AC ;R2=R2+12 x3004 0405 ;If z goto x300A x3005 6840 ;Load next value into R4 x3006 16C4 ;R3=R3+R4 x3007 1261 ;R1=R1+1 x3008 14BF ;R2=R2-1 x3009 0FFA ;goto x3004 x300A F025 ;halt

  13. The Sum program Data in “hex” 3100 ; Begin data at x3100 x3100 0001 ; Loc x3100 x3101 0002 x3102 0004 x3103 0008 x3104 FFFF x3105 1C10 x3106 11B1 x3107 0019 x3108 0F07 x3109 0004 x310A 0A00 x310B 400F ; Loc x310B

  14. LC3 Edit • Enter (or Load) the program into LC3 Edit - Store it as prog.bin for a binary file, or Store it as prog.hex for a hex file - Create a prog.obj file with the Editor • Enter (or Load) the data into LC3 Edit - Store it as data.bin for a binary file, or Store it as data.hex for a hex file - Create a data.obj file with the Editor

  15. LC-3 Simulator • Open LC-3 Simulator - Load prog.obj - Load data.obj - Set breakpoint(s) - Step through program

  16. Simulator Screen

  17. LC3 Edit

More Related