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CS 230: Computer Organization and Assembly Language

CS 230: Computer Organization and Assembly Language. Aviral Shrivastava. Department of Computer Science and Engineering School of Computing and Informatics Arizona State University. Slides courtesy: Prof. Yann Hang Lee, ASU, Prof. Mary Jane Irwin, PSU, Ande Carle, UCB. Announcements.

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CS 230: Computer Organization and Assembly Language

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  1. CS 230: Computer Organization and Assembly Language Aviral Shrivastava Department of Computer Science and Engineering School of Computing and Informatics Arizona State University Slides courtesy: Prof. Yann Hang Lee, ASU, Prof. Mary Jane Irwin, PSU, Ande Carle, UCB

  2. Announcements • Project 3 • MIPS Assembler • Midterm • Thursday, Oct 2009 • MIPS ISA and Programming • Function calls and register conventions • Assembling MIPS Instructions • 2’s complement number system • FP number system • Finals • Tuesday, Dec 08, 2009

  3. Computer Organization • We have leaned the ISA of the processor till now • Given an algorithm, express it in terms of the processor ISA software Instruction Set Architecture hardware

  4. Below the Program • High-level language program (in C) • swap (int v[], int k) • . . . • Assembly language program (for MIPS) • swap: sll $2, $5, 2 • add $2, $4, $2 • lw $15, 0($2) • lw $16, 4($2) • sw $16, 0($2) • sw $15, 4($2) • jr $31 • Machine (object) code (for MIPS) • 000000 00000 00101 0001000010000000 • 000000 00100 00010 0001000000100000 • 100011 00010 01111 0000000000000000 • 100011 00010 10000 0000000000000100 • 101011 00010 10000 0000000000000000 • 101011 00010 01111 0000000000000100 • 000000 11111 00000 0000000000001000 C - Compiler Assembler

  5. Datapath & Control Fetch PC = PC+4 Exec Decode • We're ready to look at an implementation of the MIPS • Simplified to contain only: • memory-reference instructions: lw, sw • arithmetic-logical instructions: add, sub, and, or, slt • control flow instructions: beq, j • Generic implementation: • use the program counter (PC) to supply the instruction address and fetch the instruction from memory (and update the PC) • decode the instruction (and read registers) • execute the instruction • All instructions (except j) use the ALU after reading the registers • Why? memory-reference? arithmetic? control flow?

  6. Abstract Implementation View • Two types of functional units: • elements that operate on data values (combinational) • elements that contain state (sequential) • Single cycle operation • Split memory (Harvard) model - one memory for instructions and one for data Write Data Instruction Memory Address Read Data Register File Reg Addr Data Memory Read Data PC Address Instruction ALU Reg Addr Read Data Write Data Reg Addr

  7. 32-bit ALU 32 32 • Supports all the Arithmetic/Logic operations operation a ALU 32 result b

  8. ALU Design op overflow add/subt A0 result0 B0 + less A1 result1 B1 zero + . . . 0 less . . . A31 result31 + B31 0 less set

  9. Fetching Instructions • Fetching instructions involves • reading the instruction from the Instruction Memory • updating the PC to hold the address of the next instruction • PC is updated every cycle, so it does not need an explicit write control signal • Instruction Memory is read every cycle, so it doesn’t need an explicit read control signal Add 4 Instruction Memory Read Address PC Instruction

  10. Decoding Instructions Read Addr 1 Read Data 1 Register File Read Addr 2 Write Addr Read Data 2 Write Data • Decoding instructions involves • sending the fetched instruction’s opcode and function field bits to the control unit Control Unit Instruction • reading two values from the Register File • Register File addresses are contained in the instruction

  11. Executing R Format Operations 31 25 20 15 10 5 0 R-type: op rs rt rd shamt funct • R format operations (add, sub, slt, and, or) • perform the indicated (by op and funct) operation on values in rs and rt • store the result back into the Register File (into location rd) • Note that Register File is not written every cycle (e.g. sw), so we need an explicit write control signal for the Register File RegWrite ALU control Read Addr 1 Read Data 1 Register File Read Addr 2 overflow Instruction zero ALU Write Addr Read Data 2 Write Data

  12. Load and Store Operations 31 25 20 15 0 I-Type: address offset op rs rt • Load and store operations • compute a memory address by adding the base register (in rs) to the 16-bit signed offset field in the instruction • base register was read from the Register File during decode • offset value in the low order 16 bits of the instruction must be sign extended to create a 32-bit signed value • store value, read from the Register File during decode, must be written to the Data Memory • load value, read from the Data Memory, must be stored in the Register File

  13. Load and Store Operations RegWrite ALU control MemWrite overflow zero Read Addr 1 Read Data 1 Address Register File Read Addr 2 Instruction Data Memory Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32

  14. Executing Branch Operations 31 25 20 15 0 I-Type: address offset op rs rt • Branch operations have to • compare the operands read from the Register File during decode (rs and rt values) for equality (zero ALU output) • compute the branch target address by adding the updated PC to the sign extended16-bit signed offset field in the instruction • “base register” is the updated PC • offset value in the low order 16 bits of the instruction must be sign extended to create a 32-bit signed value and then shifted left 2 bits to turn it into a word address

  15. Executing Branch Operations Branch target address Add Add 4 Shift left 2 ALU control PC zero (to branch control logic) Read Addr 1 Read Data 1 Register File Read Addr 2 Instruction ALU Write Addr Read Data 2 Write Data Sign Extend 16 32

  16. Executing Jump Operations 31 25 0 J-Type: jump target address op • Jump operations have to • replace the lower 28 bits of the PC with the lower 26 bits of the fetched instruction shifted left by 2 bits Add 4 4 Jump address Instruction Memory Shift left 2 28 Read Address Instruction PC 26

  17. Adding the pieces together Add RegWrite ALU control MemWrite 4 ovf zero Read Addr 1 Instruction Memory Read Data 1 Address Register File Read Addr 2 Data Memory Read Address PC Instruction Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32

  18. Multiplexor Insertion Add RegWrite ALUSrc ALU control MemWrite MemtoReg 4 ovf zero Read Addr 1 Instruction Memory Read Data 1 Address Register File Read Addr 2 Data Memory Read Address PC Instruction Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32

  19. Clock Distribution System Clock clock cycle MemtoReg RegWrite MemWrite Add ALUSrc ALU control 4 ovf zero Read Addr 1 Instruction Memory Read Data 1 Address Register File Read Addr 2 Data Memory Read Address PC Instruction Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32

  20. Adding the Branch Portion Add Add 4 Shift left 2 PCSrc RegWrite ALUSrc ALU control MemWrite MemtoReg ovf zero Read Addr 1 Instruction Memory Read Data 1 Address Register File Read Addr 2 Data Memory Read Address PC Instruction Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32

  21. Adding the Jump Portion RegWrite ALUSrc ALU control MemWrite MemtoReg ovf zero Read Addr 1 Read Data 1 Address Register File Read Addr 2 Data Memory Read Data ALU Write Addr Read Data 2 Write Data Write Data MemRead Sign Extend 16 32 Jump 32 26 1 Shift left 2 28 PC+4[31-28] 0 Add Add 4 Shift left 2 PCSrc Instruction Memory Read Address PC Instruction

  22. MIPS Machine (with Controls) Instr[25-0] 1 Shift left 2 28 32 26 0 PC+4[31-28] 0 Add Add 1 4 Shift left 2 PCSrc Jump ALUOp Branch MemRead Instr[31-26] Control Unit MemtoReg MemWrite ALUSrc RegWrite RegDst ovf Instr[25-21] Read Addr 1 Instruction Memory Read Data 1 Address Register File Instr[20-16] zero Read Addr 2 Data Memory Read Address PC Instr[31-0] 0 Read Data 1 ALU Write Addr Read Data 2 0 1 Write Data 0 Instr[15 -11] Write Data 1 Instr[15-0] Sign Extend ALU control 16 32 Instr[5-0]

  23. Yoda says… Use your feelings, Obi-Wan, and find him you will

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