LC-3 Instruction Processing
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1 LC-3 Instruction Processing (Textbookʼs Chapter 4)# Next set of Slides:# Textbook Chapter #
2 Instruction Processing# It is impossible to do all of an instruction in one clock cycle.# Processors break an instruction into standard steps called phases.# CMPE12 Fall 2011 Joel Ferguson# 9-2#
3 Overview# Phases of instruction execution# Major components of LC-3 processor# Consider several instructions and how the LC-3 executes instructions using those components # CMPE12 Fall 2011 Joel Ferguson# 9-3#
4 Instruction Processing# Fetch instruction from memory Decode instruction Evaluate address Fetch operands from memory Execute operation Store result CMPE12 Fall 2011 Joel Ferguson# 9-4#
5 Phases of instruction processing# Six basic phases of instruction processing: #F D EA OP EX S# NOTE: # Not all phases are needed by every instruction# All instructions will go through F and D at least# Phases may take more than 1 clock cycle (if they involve memory)# CMPE12 Fall 2011 Joel Ferguson# 9-5#
6 FETCH# Load next instruction (at address stored in PC) from memory into Instruction Register (IR).# Copy contents of PC into MAR.# Send read signal to memory.# Copy contents of MDR into IR.# Then increment PC, so that it points to the next instruction in sequence.# PC becomes PC+1.# F D EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-6#
7 DECODE# First identify the opcode.# In LC-3, this is always the first four bits of instruction.# A 4-to-16 decoder asserts a control line corresponding to the desired opcode.# F D Depending on opcode, identify other operands from the remaining bits.# Example:# for LDR, last six bits is offset# for ADD, last three bits is second source operand# EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-7#
8 EVALUATE ADDRESS# For instructions that require memory access, compute address used for access.# F D Examples:# add offset to base register (as in LDR)# add offset to PC add offset to zero# set source registers addresses# EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-8#
9 FETCH OPERANDS# Obtain source operands needed to perform operation.# F D Examples:# load data from memory (LDR)# read data from register file (ADD)# EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-9#
10 EXECUTE# F Perform the operation, using the source operands.# Examples:# send operands to ALU and assert ADD signal# do nothing (e.g., for loads and stores)# D EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-10#
11 STORE RESULT# Write results to destination. (register or memory) Examples:# result of ADD is placed in destination register# result of memory load is placed in destination register# for store instruction, data is stored to memory# write address to MAR, data to MDR# assert WRITE signal to memory# F D EA OP EX S CMPE12 Fall 2011 Joel Ferguson# 9-11#
12 LC-3 Data Path# Filled arrow = info to be processed. Unfilled arrow = control signal. CMPE12 Fall 2011 Joel Ferguson# 9-12#
13 Global bus# Data Path Components (1)# special set of wires that carry a 16-bit signal to many components# inputs to the bus are tri-state devices, that only place a signal on the bus when they are enabled# only one device speaks on the bus at any given time# control unit decides which signal drives the bus# any number of components can read the bus# the control unit write-enables the destination device# CMPE12 Fall 2011 Joel Ferguson# 9-13#
14 LC-3 Global Bus# 16 bits wide# Output drivers control source# Input signals control destination# Not all I/O signals to global bus are shown# CMPE12 Fall 2011 Joel Ferguson# 9-14#
15 Data Path Components (1A)# Memory# Control and data registers for memory and I/O devices# memory: MAR, MDR (also control signal for read/write)# CMPE12 Fall 2011 Joel Ferguson# 9-15#
16 LC-3 Memory and I/O systems# Memory Components: Memory Mem. Address Reg. Mem. Data Reg. Mem Enable, R, W Input and Output (has same signals as memory but not shown) CMPE12 Fall 2011 Joel Ferguson# 9-16#
17 ALU# Data Path Components (2) Accepts inputs from register file and from sign-extended bits from IR (immediate field).# Output goes to bus, and is used by condition code logic, register file, memory# Register File# Two read addresses (SR1, SR2), one write address (DR)# Input from bus# result of ALU operation or memory read# Two 16-bit outputs# used by ALU, PC, memory address# data for store instructions passes through ALU# CMPE12 Fall 2011 Joel Ferguson# 9-17#
18 LC-3 Register files and ALU# Register File Registers R0-R7 16 bits Data in 2 x 16 bits Data out 10 control bits ALU and SR2MUX Data: 3 16-bit words Control: 2 bits for ALU and 1 bit for MUX CMPE12 Fall 2011 Joel Ferguson# 9-18#
19 Data Path Components (3) PC and PCMUX# There are three inputs to PC, controlled by PCMUX# 1. PC+1 FETCH stage# 2. Address adder BR, JMP 3. bus TRAP (discussed later)# MAR and MARMUX# There are two inputs to MAR, controlled by MARMUX# 1. Address adder LD/ST, LDR/STR 2. Zero-extended IR[7:0] -- TRAP (discussed later)# CMPE12 Fall 2011 Joel Ferguson# 9-19#
20 LC-3 PC and MARMUX# PC; PCMUX; incrementer PC gets data from 1 of 8 sources. MAR and MARMUX MAR gets data from how many sources? CMPE12 Fall 2011 Joel Ferguson# 9-20#
21 Data Path Components (4) Condition Code Logic# Looks at value on bus and generates N, Z, P signals# Registers set only when control unit enables them (LD.CC)# only certain instructions set the condition codes (ADD, AND, NOT, LD, LDI, LDR, LEA)# Control Unit Finite State Machine# On each machine cycle, changes control signals for next phase of instruction processing# who drives the bus? (GatePC, GateALU, )! which registers are write enabled? (LD.IR, LD.REG, )! which operation should ALU perform? (ALUK)! # Logic includes decoder for opcode, etc.# CMPE12 Fall 2011 Joel Ferguson# 9-21#
22 LC-3 C.Code Registers and FSM# Condition Code Logic Sets codes when LD.CC allows. Control Unit Is an FSM Controls everything based on instruction and CC. CMPE12 Fall 2011 Joel Ferguson# 9-22#
23 Data Path Components (5) Effective Address Mux and Adder# Computes addresses for PC and MAR# Used for most instructions that affect the PC or access memory (BRA, JMP, LD, ST, JSR, etc.)# CMPE12 Fall 2011 Joel Ferguson# 9-23#
24 LC-3 Effective Address Computation Units: For almost all PC and MAR computations. CMPE12 Fall 2011 Joel Ferguson# 9-24#
25 Example 1:# x30a2 add R2,R0,R1 R0 = x0230 R1 = x1111 R2 = x2222 x30a2# 1) Fetch(from mem) 1 st step# x1401# CMPE12 Fall 2011 Joel Ferguson# 9-25#
26 Example 1:# x30a2 add R2,R0,R1 R0 = x0230 R1 = x1111 R2 = x2222 x1401# 1) Fetch 2 nd step # CMPE12 Fall 2011 Joel Ferguson# 9-26#
27 Example 1:# x30a2 add R2,R0,R1 R0 = x0230 R1 = x1111 R2 = x2222 x1401# 2) Decode# CMPE12 Fall 2011 Joel Ferguson# 9-27#
28 Example 1:# x1341# x30a2 add R2,R0,R1 R0 = x0230 R1 = x1111 R2 = x2222 x1111# x0230# 5) Execute# x1341# CMPE12 Fall 2011 Joel Ferguson# 9-28#
29 Example 1:# x30a2 add R2,R0,R1 R0 = x0230 R1 = x1111 R2 = x1341 x1111# x0230# x1341# 6) Store Results# CMPE12 Fall 2011 Joel Ferguson# 9-29#
30 Example 2:# X3117 LDR R3,R5,xB R3 = x3451 R5 = x8800 x3117# 1) Fetch 1 st step # x674b# CMPE12 Fall 2011 Joel Ferguson# 9-30#
31 Example 2:# X3117 LDR R3,R5,xB R3 = x3451 R5 = x8800 1) Fetch 2 nd step # x674b# CMPE12 Fall 2011 Joel Ferguson# 9-31#
32 Example 2:# X3117 LDR R3,R5,xB R3 = x3451 R5 = x8800 x674b# 2) Decode# CMPE12 Fall 2011 Joel Ferguson# 9-32#
33 Example 2:# X3117 LDR R3,R5,xB R3 = x3451 R5 = x8800 x000b# x880b# x8800# x0b# 4) Evaluate Address# x880b# CMPE12 Fall 2011 Joel Ferguson# 9-33#
34 Example 2:# X3117 LDR R3,R5,xB R3 = x3451 R5 = x8800 4) Fetch Operands# x880b# X????# CMPE12 Fall 2011 Joel Ferguson# 9-34#
35 Example 2:# x????# X3117 LDR R3,R5,xB R3 = mem[x880b] R5 = x8800 x????# 6) Store Results# CMPE12 Fall 2011 Joel Ferguson# 9-35#
36 Example 3:# x3040 BRZ EndLoop EndLoop = x3030 x3040 Therefore mem[x3040] = x05ef 1) Fetch 1 st step # x05ef# CMPE12 Fall 2011 Joel Ferguson# 9-36#
37 Example 3:# x3041# x3040 BRZ EndLoop EndLoop = x3030 1) Fetch 2 nd step # x05ef# CMPE12 Fall 2011 Joel Ferguson# 9-37#
38 Example 3:# x3040 BRZ EndLoop x05ef# 2) Decode# CMPE12 Fall 2011 Joel Ferguson# 9-38#
39 Example 3:# x3030# x3040 BRZ EndLoop x3041# xffef (16bits)# x1ef (9bits)# 3) Evaluate Address# CMPE12 Fall 2011 Joel Ferguson# 9-39#
40 Example 3:# x3030 x3040 BRZ EndLoop x3041 xffef (16bits) (no Operand Fetch)# (no Execute)# 6) Store Results# CMPE12 Fall 2011 Joel Ferguson# 9-40#
41 Example 4:# X3317 STR R3,R5,xB R3 = x3451 R5 = x8800 x3317# 1) Fetch 1 st step # x774b# CMPE12 Fall 2011 Joel Ferguson# 9-41#
42 Example 4:# X3117 STR R3,R5,xB R3 = x3451 R5 = x8800 1) Fetch 2 nd step # x774b# CMPE12 Fall 2011 Joel Ferguson# 9-42#
43 Example 4:# X3117 STR R3,R5,xB R3 = x3451 R5 = x8800 x774b# 2) Decode# CMPE12 Fall 2011 Joel Ferguson# 9-43#
44 Example 4:# X3117 STR R3,R5,xB R3 = x3451 R5 = x8800 x000b# x880b# x8800# b001011# 4) Evaluate Address# x774b# x880b# CMPE12 Fall 2011 Joel Ferguson# 9-44#
45 Example 4:# X3117 STR R3,R5,xB R3 = x3451 R5 = x8800 x3451 ALU told to AND;# AND(x3451,x3451) # #= x3451# 4) Fetch Operands# x3451# x880b# CMPE12 Fall 2011 Joel Ferguson# 9-45#
46 Example 4:# X3117 STR R3,R5,xB R3 = x3451 R5 = x8800 mem[x880b]<-x3451 No Execute phase# 6) Store Results# x3451# CMPE12 Fall 2011 Joel Ferguson# 9-46#
47 Recommended and Homework exercises for Nov. 16:# Recommended: Ex 4.8, 4.10# Recommended: Ex 4.13 and 4.16 (a little bit more advanced)# To Turn in: For the instruction LD R1, NAME, show the Evaluate Address phase of the instruction by showing the inputs and outputs of the Effective Address Computation Units. Instruction at x3012 NAME at x3001# CMPE12 Fall 2011 Joel Ferguson# 9-47#
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