Computer Architecture

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FAMU-FSU College of Engineering

ComputerArchitectureEEL 4713/5764, Spring 2006

Dr. Michael Frank

Module #6 – MIPS ISA, part 1:Instructions and Addressing

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Course Instructional Objectives #2&3 As the syllabus says:

At the completion of this course, students should be able to…CIO #2 (ao). (AsmML) Derive machine code from assembly instructions.CIO #3 (aco). (CAsm) Derive assembler code from an equivalent C code representation

These CIOs relate to following Program Outcomes: Students graduating from the BSEE and BSCpE degree programs will have:

PO (a). (Apply) An ability to apply knowledge of mathematics, science and engineering;PO (c). (Design) An ability to design a system, component, or process to meet desired needs;PO (o). (Topics) EE: A knowledge of electrical engineering applications selected from the …digital

systems… areas; CpE: A knowledge of computer science and computer engineering topics including …computer architecture.

Under “assessment instruments,” the syllabus says:2. AsmML (a) Apply-3, (o) Topics-3 Students will solve exam problems in which they must hand-assemble MIPS (or similar)

assembly language instructions to their binary machine-language representations.3. CAsm (a) Apply-3, (c) Design-3, (o) Topics-3 Students will solve exam problems in which they must hand-compile short C language code

fragments to equivalent assembly language code for a MIPS or similar instruction-set architecture.

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Detailed Objectives for CIO#2CIO #2. AsmML: (ao) Derive machine code from assembly instructions. More specifically, given any assembly language instruction for a subset of the MIPS-32 instruction set (such as the MiniMIPS or MicroMIPS instructions discussed in the textbook), or a short sequence of such instructions, students should be able to hand-assemble the instructions to correct binary code, either in binary or hexadecimal format, and with all instruction fields clearly named and delineated. To accomplish this task, the student must be able to:

2.1. Identify the type (R/I/J) of a given source instruction.2.2. Identify the name, location, and size of the bit-fields in any given instruction type

(including op, rs, rt, rd, shamt, imm, jdest).2.3. Look up the opcode (and extended opcode, if necessary) for the given instruction.2.4. Select the appropriate bit field for encoding each instruction operand.2.5. Convert symbolic register names to 5-bit binary codes.2.6. Convert provided numerical constants to signed or unsigned binary formats.2.7 Resolve symbolic branch and jump destinations to the correct word offset relative to the

next PC or current segment, as appropriate.

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Detailed Objectives for CIO#3CIO #3. CAsm: (aco) Derive assembler code from an equivalent C code repres’n. More specifically, given any short code fragment or standalone function written in the ANSI standard C programming language, students should be able to hand-compile this source to correct working assembly language for the MIPS-32 instruction set, or a subset thereof such as MiniMIPS or MicroMIPS, while obeying standard MIPS register usage conventions. To accomplish this, students should be able to:

3.1. Identify which registers a function’s arguments can be found in, and which registers should be used to return function results.

3.2. Identify appropriate registers to use for storing local program variables.3.3. Generate correct caller/callee code to save and restore registers using the stack in

accordance with the standard MIPS calling conventions.3.4. Generate correct code for accessing data stored in arrays of words.3.5. Generate correct code for the evaluation of C integer arithmetic and logical expressions.3.6. Translate high-level control-flow constructs including if, if/else, for, and while statements

to equivalent assembly code using branch instructions.3.7. Correctly use assembler directives for providing static data.3.8. Correctly use pseudoinstructions for more concise and readable assembly.

June 2005 Computer Architecture, Instruction-Set Architecture Slide 5

Part IIInstruction-Set

Architecture


II Instruction Set Architecture

Topics in This Part

Chapter 5 Instructions and Addressing

Chapter 6 Procedures and Data

Chapter 7 Assembly Language Programs

Chapter 8 Instruction Set Variations

Introduce machine “words” and its “vocabulary,” learning:• A simple, yet realistic and useful instruction set• Machine language programs; how they are executed• RISC vs CISC instruction-set design philosophy


5 Instructions and Addressing

Topics in This Chapter

5.1 Abstract View of Hardware

5.2 Instruction Formats

5.3 Simple Arithmetic / Logic Instructions

5.4 Load and Store Instructions

5.5 Jump and Branch Instructions

5.6 Addressing Modes

First of two chapters on the instruction set of MiniMIPS:• Required for hardware concepts in later chapters• Not aiming for proficiency in assembler programming


5.1 Abstract View of Hardware

Figure 5.1 Memory and processing subsystems for MiniMIPS.

Memory up to 2 words 30

Loc 0 Loc 4 Loc 8

Loc m 4

Loc m 8

4 B / location

m 2 32

$0

$1

$2

$31

Hi Lo

ALU

$0

$1

$2

$31 FP

arith

EPC

Cause

BadVaddr Status

EIU FPU

TMU

Execution & integer unit

Floating- point unit

Trap & memory unit

. . .

. . .

(Coproc. 1)

(Coproc. 0)

(Main proc.)

Integer mul/div

Chapter 10

Chapter 11

Chapter 12 Exception PC

PCSpim appearance


Data Types

MiniMIPS registers hold 32-bit (4-byte) words. Other common data sizes include byte, halfword, and doubleword.

Byte

Halfword

Word

Doubleword

Halfword = 2 bytes

Byte = 8 bits

Word = 4 bytes

Doubleword = 8 bytes


Register Conventions

Figure 5.2 Registers and data sizes in MiniMIPS.

Temporary values

More temporaries

Operands

Global pointer Stack pointer Frame pointer Return address

Saved

Saved Procedure arguments

Saved across

procedure calls

Procedure results

Reserved for assembler use

Reserved for OS (kernel)

$0 $1 $2 $3 $4 $5 $6 $7 $8 $9 $10 $11 $12 $13 $14 $15 $16 $17 $18 $19 $20 $21 $22 $23 $24 $25 $26 $27 $28 $29 $30 $31

0

$zero

$t0

$t2

$t4

$t6

$t1

$t3

$t5

$t7 $s0

$s2

$s4

$s6

$s1

$s3

$s5

$s7 $t8 $t9

$gp $sp $fp $ra

$at

$k0 $k1

$v0

$a0

$a2

$v1

$a1

$a3

A doubleword sits in consecutive registers or memory locations according to the big-endian order (most significant word comes first)

When loading a byte into a register, it goes in the low end Byte

Word

Doublew ord

Byte numbering: 0 1 2 3

3

2

1

0

A 4-byte word sits in consecutive memory addresses according to the big-endian order (most significant byte has the lowest address)

But in SPIM, theendian-ness depends on theunderlying HW


5.2 Instruction Formats

Figure 5.3 A typical instruction for MiniMIPS and steps in its execution.

Assembly language instruction:

Machine language instruction:

High-level language statement:

000000 10010 10001 11000 00000 100000

add $t8, $s2, $s1

a = b + c

ALU-type instruction

Register 18

Register 17

Register 24 Unused

Addition opcode

ALU

Instruction fetch

Register readout

Operation

Data read/store

Register writeback

Register file

Instruction

cache

Data cache (not used)

Register file

P C

$17 $18

$24

0 2 5 1 c 0 2 0


Add, Subtract, and Specification of Constants

MiniMIPS add & subtract instructions; e.g., compute: g = (b + c) (e + f)

add $t8,$s2,$s3 # put the sum b + c in $t8 add $t9,$s5,$s6 # put the sum e + f in $t9 sub $s7,$t8,$t9 # set g to ($t8) ($t9)

Decimal and hex constants

Decimal 25, 123456, 2873 Hexadecimal 0x59, 0x12b4c6, 0xffff0000

Machine instruction typically contains

an opcode one or more source operands possibly a destination operand


MiniMIPS Instruction Formats

Figure 5.4 MiniMIPS instructions come in only three formats: register (R), immediate (I), and jump (J).

5 bits 5 bits

31 25 20 15 0

Opcode Source register 1

Source register 2

op rs rt

R 6 bits 5 bits

rd

5 bits

sh

6 bits

10 5 fn

Destination register

Shift amount

Opcode extension

Immediate operand or address offset

31 25 20 15 0

Opcode Destination or data

Source or base

op rs rt operand / offset

I 5 bits 6 bits 16 bits 5 bits

0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0

31 0

Opcode

op jump target address

J Memory word address (byte address divided by 4)

26 bits

25

6 bits


5.3 Simple Arithmetic/Logic Instructions

Figure 5.5 The arithmetic instructions add and sub have a format that is common to all two-operand ALU instructions. For these, the fn field specifies the arithmetic/logic operation to be performed.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 x 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0

31 25 20 15 0

ALU instruction

Source register 1

Source register 2

op rs rt

R rd sh

10 5 fn

Destination register

Unused add = 32 sub = 34

Add and subtract already discussed; logical instructions are similar

add $t0,$s0,$s1 # set $t0 to ($s0)+($s1) sub $t0,$s0,$s1 # set $t0 to ($s0)-($s1) and $t0,$s0,$s1 # set $t0 to ($s0)($s1) or $t0,$s0,$s1 # set $t0 to ($s0)($s1) xor $t0,$s0,$s1 # set $t0 to ($s0)($s1) nor $t0,$s0,$s1 # set $t0 to (($s0)($s1))


Arithmetic/Logic with One Immediate Operand

Figure 5.6 Instructions such as addi allow us to perform an arithmetic or logic operation for which one operand is a small constant.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0

31 25 20 15 0

addi = 8 Destination Source Immediate operand


I 1

An operand in the range [32 768, 32 767], or [0x0000, 0xffff], can be specified in the immediate field.

addi $t0,$s0,61 # set $t0 to ($s0)+61 andi $t0,$s0,61 # set $t0 to ($s0)61 ori $t0,$s0,61 # set $t0 to ($s0)61 xori $t0,$s0,0x00ff # set $t0 to ($s0) 0x00ff

For arithmetic instructions, the immediate operand is sign-extended


5.4 Load and Store Instructions

Figure 5.7 MiniMIPS lw and sw instructions and their memory addressing convention that allows for simple access to array elements

via a base address and an offset (offset = 4i leads us to the ith word).

0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 x 1 0 0 0 0 0 0

31 25 20 15 0

lw = 35 sw = 43

Base register

Data register

Offset relative to base


I 1 1 0 0 1 1 1 1 1

A[0] A[1] A[2]

A[i]

Address in base register

Offset = 4i

.

.

.

Memory

Element i of array A

Note on base and offset: The memory address is the sum of (rs) and an immediate value. Calling one of these the base and the other the offset is quite arbitrary. It would make perfect sense to interpret the address A($s3) as having the base A and the offset ($s3). However, a 16-bit base confines us to a small portion of memory space.


lw, sw, and lui Instructions

Figure 5.8 The lui instruction allows us to load an arbitrary 16-bit value into the upper half of a register while setting its lower half to 0s.

0

0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0

31 25 20 15 0

lui = 15 Destination Unused Immediate operand


I

Content of $s0 after the instruction is executed

lw $t0,40($s3) # load mem[40+($s3)] in $t0 sw $t0,A($s3) # store ($t0) in mem[A+($s3)]

# “($s3)” means “content of

$s3” lui $s0,61 # The immediate value 61 is

# loaded in upper half of $s0 # with lower 16b set to 0s


Initializing a RegisterExample 5.2

Show how each of these bit patterns can be loaded into $s0:

0010 0001 0001 0000 0000 0000 0011 1101 1111 1111 1111 1111 1111 1111 1111 1111

Solution

The first bit pattern has the hex representation: 0x2110003d

lui $s0,0x2110 # put the upper half in $s0 ori $s0,0x003d # put the lower half in $s0

Same can be done, with immediate values changed to 0xffff for the second bit pattern. But, the following is simpler and faster:

nor $s0,$zero,$zero # because (0 0) = 1


5.5 Jump and Branch Instructions

Unconditional jump and jump through register instructions

j verify # go to mem loc named “verify” jr $ra # go to address that is in $ra;

# $ra may hold a return address

Figure 5.9 The jump instruction j of MiniMIPS is a J-type instruction which is shown along with how its effective target address is obtained. The jump register (jr) instruction is R-type, with its specified register often being $ra.

0

0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0

31 0

j = 2

op jump target address

J

Effective target address (32 bits)

25

From PC

0 0

x x x x

0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0

31 25 20 15 0

ALU instruction

Source register

Unused

op rs rt

R rd sh

10 5 fn

Unused Unused jr = 8


Conditional Branch Instructions

Figure 5.10 (part 1) Conditional branch instructions of MiniMIPS.

Conditional branches use PC-relative addressing

bltz $s1,L # branch on ($s1)< 0 beq $s1,$s2,L # branch on ($s1)=($s2) bne $s1,$s2,L # branch on ($s1)($s2)

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0

31 25 20 15 0

bltz = 1 Zero Source Relative branch distance in words


I 0

1 1 0 0 x 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0

31 25 20 15 0

beq = 4 bne = 5

Source 2 Source 1 Relative branch distance in words


I 1


Comparison Instructions for Conditional Branching

Figure 5.10 (part 2) Comparison instructions of MiniMIPS.

slt $s1,$s2,$s3 # if ($s2)<($s3), set $s1 to 1 # else set $s1 to 0;

# often followed by beq/bne slti $s1,$s2,61 # if ($s2)<61, set $s1 to 1 # else set $s1 to 0

1 1 1 0 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0

31 25 20 15 0

ALU instruction

Source 1 register

Source 2 register

op rs rt

R rd sh

10 5 fn

Destination Unused slt = 42

1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0

31 25 20 15 0

slti = 10 Destination Source Immediate operand


I 1


Examples for Conditional Branching

If the branch target is too far to be reachable with a 16-bit offset (rare occurrence), the assembler automatically replaces the branch instruction beq $s0,$s1,L1 with:

bne $s1,$s2,L2 # skip jump if (s1)(s2)

j L1 # goto L1 if (s1)=(s2) L2: ...

Forming if-then constructs; e.g., if (i == j) x = x + y

bne $s1,$s2,endif # branch on ij add $t1,$t1,$t2 # execute the “then”

partendif: ...

If the condition were (i < j), we would change the first line to:

slt $t0,$s1,$s2 # set $t0 to 1 if i<j beq $t0,$0,endif # branch if ($t0)=0;

# i.e., i not< j or ij


Example 5.3

Compiling if-then-else Statements

Show a sequence of MiniMIPS instructions corresponding to:

if (i<=j){x = x+1; z = 1;} else {y = y–1; z = 2*z}

Solution

Similar to the “if-then” statement, but we need instructions for the“else” part and a way of skipping the “else” part after the “then” part.

slt $t0,$s2,$s1 # j<i? (inverse condition) bne $t0,$zero,else # if j<i goto else part addi $t1,$t1,1 # begin then part: x = x+1 addi $t3,$zero,1 # z = 1 j endif # skip the else part

else: addi $t2,$t2,-1 # begin else part: y = y–1 add $t3,$t3,$t3 # z = z+z

endif:...


5.6 Addressing Modes

Figure 5.11 Schematic representation of addressing modes in MiniMIPS.

Addressing Instruction Other elements involved Operand

Implied

Immediate

Register

Base

PC-relative

Pseudodirect

Some place in the machine

Extend, if required

Reg file Reg spec Reg data

Memory Add

Reg file

Mem addr

Constant offset

Reg base Reg data

Mem data

Add

PC

Constant offset

Memory

Mem addr Mem

data

Memory Mem data

PC Mem addr


Example 5.5

Finding the Maximum Value in a List of Integers

List A is stored in memory beginning at the address given in $s1. List length is given in $s2. Find the largest integer in the list and copy it into $t0.

Solution

Scan the list, holding the largest element identified thus far in $t0.

lw $t0,0($s1) # initialize maximum to A[0]addi $t1,$zero,0 # initialize index i to 0

loop: add $t1,$t1,1 # increment index i by 1beq $t1,$s2,done # if all elements examined,

quitadd $t2,$t1,$t1 # compute 2i in $t2add $t2,$t2,$t2 # compute 4i in $t2 add $t2,$t2,$s1 # form address of A[i] in $t2 lw $t3,0($t2) # load value of A[i] into $t3slt $t4,$t0,$t3 # maximum < A[i]?beq $t4,$zero,loop # if not, repeat with no

changeaddi $t0,$t3,0 # if so, A[i] is the new

maximum j loop # change completed; now repeat

done: ... # continuation of the program


The 20 MiniMIPS Instructions

Covered So Far

Instruction UsageLoad upper immediate lui rt,imm

Add add rd,rs,rt

Subtract sub rd,rs,rt

Set less than slt rd,rs,rt

Add immediate addi rt,rs,imm

Set less than immediate slti rd,rs,imm

AND and rd,rs,rt

OR or rd,rs,rt

XOR xor rd,rs,rt

NOR nor rd,rs,rt

AND immediate andi rt,rs,imm

OR immediate ori rt,rs,imm

XOR immediate xori rt,rs,imm

Load word lw rt,imm(rs)

Store word sw rt,imm(rs)

Jump j L

Jump register jr rs

Branch less than 0 bltz rs,L

Branch equal beq rs,rt,L

Branch not equal bne rs,rt,L

Copy

Control transfer

Logic

Arithmetic

Memory access

op15

0008

100000

1213143543

20145

fn

323442

36373839

8 Table 5.1

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