1 instructions and addressing iosup/courses/2011_ti1400_4.ppt

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Instructions and Addressinghttp://www.pds.ewi.tudelft.nl/~iosup/Courses/2011_ti1400_4.ppt

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Lecture 1Making functions

time

A,B Y

A

BYADD

nand gates

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Lecture 1Making functions

Circuit Diagramhttp://xkcd.com/730/

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Lecture 2Programmable device

2,1 3ProgrammableDevice

input stream output stream

programREAD(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

• READ(X) means read next input value from input stream and store it internally as variable X

• WRITE(X) means put value in variable X on output stream• ADD(X,Y,Z) means assign value of X+Y to Z

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Lecture 3Von Neumann Architecture

READ(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

X: 1Y: 2Z: 3 • •

TEMP_A: TEMP_B: RESULT:

IR:

PC:

arithmeticunit

Central Processing Unit

CONTROL

Memory

Input

Output

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Problem: How to Represent and Use Data?

1. Representation2. Arithmetic3. Conversion

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Lecture 2Von Neumann Architecture

READ(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

X: 1Y: 2Z: 3 • •

TEMP_A: TEMP_B: RESULT:

IR:

PC:

arithmeticunit

Central Processing Unit

CONTROL

Memory

Input

Output

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Problem: How to Represent and Use Data Instructions?

1. Representation2. Use (Addressing Data)

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Instructions and Addressing

1. Memory Layout2. Types of Instructions3. Use of Accumulator/Registers4. Execution of Instructions5. Addressing

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Computer System

READ(X)READ(Y)ADD(X,Y,Z)WRITE(Z)

X: 1Y: 2Z: 3 • •

IR:

PC:

arithmeticunit

Central Processing Unit

controlunit

Main Memory

Input

Output

registers

Data and Instructions Instruction RegisterProgram Counter

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Main Memory Addressing

000000000 00000000 000000000000000

000000000 00000000 000000000000000

000000000 00000000 000000000000000

Bit

Byte(8 bits)

Word(16-64

bits)

Address0

Address31

0

8 16 24

1 2 3

0

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Q1: How many bits required to address?1. 1KB of memory

A. By Bit (13)B. By Byte (10)C. By Word of 32 bits

(8)

2. 1MB of memoryA. By Bit (23)B. By Byte (20)C. By Word of 32 bits

(18)

3. 1GB of memoryA. By BitB. By ByteC. By Word of 32 bits

4. 1TB of memoryA. By BitB. By ByteC. By Word of 32 bits

Q2: How much memory can be addressed by byte for a 32-bit architecture?Q3: Why use byte memory addressing?

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Instruction and wordlength (1/3)

instruction

instruction

instructioninstruction

012345678910

word

instructioninstructioninstruction instructioninstruction instruction

012345678910

wordaddress address

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Instruction and wordlength (2)

instructioninstructioninstruction instructioninstruction instruction

012345678910

wordaddress

instructioninstructioninstruction instructioninstruction instruction

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Instruction and wordlength (3)

012345678910

wordaddress

instructioninstr. instrinstruction instr. instrinstruction instr. instr

instructioninstr. instrinstruction instr. instrinstruction instr. instr

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32 bit word formats

opcode specifier operand specifiers

31 30 1 0

byte byte byte byte

a two’s complement number

4 ASCII characters

32 bit

a machine instruction

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Byte ordering (endianness)

0 1 2 34 5 6 78 9 - -- - - -- - - -

0

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Word Index

Big endiane.g., Motorola PowerPC 68k

3 2 1 07 6 5 4- - 9 8- - - -- - - -

0

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Little endiane.g., Intel

Jonathan Swift’s Gulliver’s Travells

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Q1 (the NUXI problem)

• What problems can occur in porting data between machines with big-endian and little-endian storage? (Hint: networked machines)

Q2 Why use the little endian byte ordering?

Q3 Why use the big endian byte ordering?

• (Hint: value 4 as single- and multi-byte)

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Instructions and Addressing

1. Memory Layout2. Types of Instructions3. Use of Accumulator/Registers4. Execution of Instructions5. Addressing 2.1. Data Copy Operations

2.2. Arithmetic And Logic Ops.2.3. Program flow control Ops.2.4. I/O operations

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Types of Instructions

• There are 4 types of instructions- Data Copy operations

• between memory and registers• between memory locations • between registers

- Arithmetic and Logic operations- Program flow control operations- I/O operations

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Symbolic notation

• Copy instructions [R1 ] M(LOC)

• Arithmetic operations M(C) M(A) + M(B)

• LOC, A, and B are memory addresses• M(address) means contents of memory

location at address.• [R] means contents of register R.

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Operand specification formats

• Three-address instructionsformat: INSTR source#1,source#2,destinationexample: Add A,B,Cmeans: M(C) M(B) + M(A)

• Problem: 3-address instructions means long instruction words.- If k bits are needed for memory addressing,

then 3k bits are needed for addressing operands. (k=32 for 32-bit platforms.)

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Operand specification formats

• Two address instructionsformat: INSTR source, destinationexample: Add A,Bmeans: M(B) M(B) + M(A)

• Problem: 2-address instructions mean somewhat long word or multiple FETCHes per instruction. - If k bits are needed for memory addressing,

then 2k bits for addressing operands. (k=32 for 32-bit platforms.)

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Two address instruction Additional Problem: Operand OverrideAdd A,B

• Two operand instructions destroy contents of the B location

• Need other instruction to avoid that:Move B,C

• We then have Move B,C

Add A,C

meaning: M(C) M(B); M(C) M(C) + M(A);

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One address instructions

• One address - have implicit source (often called Accumulator)Load AAdd BStore Cmeaning[Accu ] M(A);[Accu ] [Accu ] + M(B);M(C) [Accu ]

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Instructions and Addressing

1. Memory Layout2. Types of Instructions3. Use of Accumulator/Registers4. Execution of Instructions5. Addressing

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Registers

• Many computers have a number of General-Purpose registers inside the CPU

• Access to registers is faster than to memory locations

• Used to store temporary data during processing

• Registers require less bits of address than main memory

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Register addressing

• Let Ri denote register• General operation ADD A,B,C can be

broken down toMove A,R0Add B,R0Store R0,Cmeaning [R0] M(A); [R0] [R0] + M(B); M(C) [R0]

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Instruction Formats (Summary)

opcode specifier operand specifiers

general format

one operand addressing

opcode operand operand

two operand addressing

opcode operand

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Accumulator architecture

AccumulatorPC

CPU

MainMemory

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Question

How many instructions can be defined when the opcode field is 5 bit ?

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Example

instruction

accu

4 bits 12 bits

opcode operandm

15 0

sign bit

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Instruction setCode Opcode Meaning

01000101011001111000100110101011

LoadStoreAddSubMulDivAndOr

[Accu] M(m)M(m) [Accu][Accu] [Accu] + M(m)[Accu] [Accu] - M(m)[Accu] [Accu] * M(m)[Accu] [Accu] / M(m)[Accu] [Accu] & M(m)[Accu] [Accu] v M(m)

000000010010

JMPJSRRTS

Goto mCall Subroutine at mReturn from Subroutine

1100110111101111

JZJNZJPJN

Goto m if [Accu]= 0Goto m if [Accu] =/ 0Goto m if [Accu] > 0Goto m if [Accu] < 0

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Multiple register architecture

R0CPU

MainMemory

R1

R2

R3

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Instructions and Addressing

1. Memory Layout2. Types of Instructions3. Use of Accumulator/Registers4. Execution of Instructions5. Addressing

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Straight-line sequencing

…...…...

…….

Move A,R0Add B,R0Move R0,C • •

ii+4i+8

AB

C

address

Program for M(C)=M(B)+M(A)

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Straight-line sequencing

Add Nn,R0Move R0,S

Move N1,R0Add N2,R0Add N3,R0 • •

ii+4i+8

i+4n-4i+4n

address

Program for addition of n numbers

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Branching

......

n

Clear R0Move N,R1

Decr R1Branch>0 LMove R0,S

L

SNN1

NnProgram for addition of n numbers

Determine address of“next” number and add itto R0

programloop

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Common branch conditions

• N (negative) set to 1 of result is negative• Z (zero) set to 1 of result is zero• V (overflow) set to 1 of result overflows• C (carry) set to 1 of carry-out results

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Question

Why is the carry condition important?

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Instructions and Addressing

1. Memory Layout2. Types of Instructions3. Use of Accumulator/Registers4. Execution of Instructions5. Addressing 5.1. Addressing Modes

5.2. Immediate Addressing5.3. Direct Addressing5.4. Indirect Addressing5.5. Index Addressing

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Addressing modes

• Addressing modes determine how the address of operands is determined

• Typical 4 addressing modes- immediate addressing- direct addressing- indirect addressing- index addressing

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Immediate addressing (1)

opcode specifier operand

instruction

ADD # -1JNZ 10 • •

10

example: simple counting loop

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Immediate addressing (2)• Advantages:

- no additional calculations needed to obtain operand

- fast

• Disadvantages:- Operand value must be known- Operand value cannot be changed- Limited no of bits available

• Notation: MOVE #200,R0• Meaning: [R0] 200

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Direct addressing(1)

opcode specifier mem or reg address

instruction

memoryor

registers

ADD 13JNZ 10 • # -1

10

13

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Direct addressing(2)• Advantages:

- Operand separate from instruction- Can be changed- Full word length available

• Disadvantages:- More memory accesses- More storage occupation

• Notation: ADD R1,R2• Meaning: [R2] [R2] +[R1]• Also called “Absolute addressing” (Ham.)

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Indirect addressing (1)

ADD (12)JNZ 10 13 # -1

10

13

opcode specifier mem or reg address

instruction

op. address

operand

memoryor

registers

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Indirect addressing (2)

• Advantages:- Actual address of operand is not in instruction- Can be changed

• Disadvantages:- Even more memory or register references- More memory occupation

• Notation: ADD (R1),R2• Meaning: [R2] [R2] + M([R1])

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Example indirect addressing

programloop L

N

N1

Clear R0

Move N,R1

Move #N1,R2

Add (R2),R0

Add #4,R2

Decr R1

Branch>0 L

Move R0,S

n

S

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Index addressing (1)

opcode Reg index

instruction

operand

memoryor

registers

operand

+

registers

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Index addressing (2)

• Advantages:- Allows specification of fixed offset to operand

address

• Disadvantages:- Extra addition to operand address

• Notation: ADD X(R1),R3 (X=number)• Meaning: [R3] [R3] + M([R1] + X)

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Example index addressing

NE

Program with index addressing

programloop

sexagesalary

nEmpoyee IDsexagesalaryEmpoyee ID

L

Move #E,R0

Move N,R1

Clear R2

Add 8(R0),R2

Add #16,R0

Decrement R1

Branch>0 L

Div R1,R2

Move R2,Sum

Move N,R1

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Additional modes

• Some computers have auto-increment (decrement instructions)

• Example: (R0)+• Meaning .. M(R0)..; [R0] [R0]+1• Example: -(R0)• Meaning [R0] [R0]-1; .. M(R0)..