Chapter Three - Jordan University of Science and Technologyyahya-t/cs351/ch3.pdf · — conventions...

1©2004 Morgan Kaufmann Publishers

Chapter Three


• Bits are just bits (no inherent meaning)— conventions define relationship between bits and numbers

• Binary numbers (base 2)0000 0001 0010 0011 0100 0101 0110 0111 1000 1001...decimal: 0...2n-1

• Of course it gets more complicated:numbers are finite (overflow)fractions and real numbersnegative numberse.g., no MIPS subi instruction; addi can add a negative number

• How do we represent negative numbers?i.e., which bit patterns will represent which numbers?

Numbers


• Sign Magnitude: One's Complement Two's Complement000 = +0 000 = +0 000 = +0001 = +1 001 = +1 001 = +1010 = +2 010 = +2 010 = +2011 = +3 011 = +3 011 = +3100 = -0 100 = -3 100 = -4101 = -1 101 = -2 101 = -3110 = -2 110 = -1 110 = -2111 = -3 111 = -0 111 = -1

• Issues: balance, number of zeros, ease of operations• Which one is best? Why?

Possible Representations


• 32 bit signed numbers:

0000 0000 0000 0000 0000 0000 0000 0000two = 0ten0000 0000 0000 0000 0000 0000 0000 0001two = + 1ten0000 0000 0000 0000 0000 0000 0000 0010two = + 2ten...0111 1111 1111 1111 1111 1111 1111 1110two = + 2,147,483,646ten0111 1111 1111 1111 1111 1111 1111 1111two = + 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0000two = – 2,147,483,648ten1000 0000 0000 0000 0000 0000 0000 0001two = – 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0010two = – 2,147,483,646ten...1111 1111 1111 1111 1111 1111 1111 1101two = – 3ten1111 1111 1111 1111 1111 1111 1111 1110two = – 2ten1111 1111 1111 1111 1111 1111 1111 1111two = – 1ten

maxint

minint

MIPS


• Negating a two's complement number: invert all bits and add 1

– remember: “negate” and “invert” are quite different!

• Converting n bit numbers into numbers with more than n bits:

– MIPS 16 bit immediate gets converted to 32 bits for arithmetic

– copy the most significant bit (the sign bit) into the other bits0010 -> 0000 0010

1010 -> 1111 1010

– "sign extension" (lbu vs. lb)

Two's Complement Operations


• Just like in grade school (carry/borrow 1s)0111 0111 0110

+ 0110 - 0110 - 0101

• Two's complement operations easy– subtraction using addition of negative numbers

0111+ 1010

• Overflow (result too large for finite computer word):– e.g., adding two n-bit numbers does not yield an n-bit number

0111+ 0001 note that overflow term is somewhat misleading,1000 it does not mean a carry “overflowed”

Addition & Subtraction


Arithmetic

• Where we've been:– Performance (seconds, cycles, instructions)

• What's up ahead:– Implementing the Architecture

32

32

32

operation

result

a

b

ALU


Constructing an ALU (Arithmetic Logic Unit)

• ALU is a device that performs the arithmetic operations like addition and subtraction, or logical operations like AND and OR.

• An ALU can be constructed from four hardware building blocks: AND gate, OR gate, Inverter, and Multiplexor. (What is the truth table of each?)

• The Multiplexor: If S == 0 then C = A else C = BSelects one of the inputs to be the output, based on a control input. S

CAB

01


• Let's build an ALU to support the andi and ori instructions– We'll just build a 1 bit ALU, and use 32 of

them because MIPS word is 32 bits wide

b

a

operation

result

op a b res

ALU (Cont.)


The 1-bit Logical Unit for AND and OR

a

b

0

1

Result

Operation


• Not easy to decide the “best” way to build something– Don't want too many inputs to a single gate– Don’t want to have to go through too many gates– for our purposes, ease of comprehension is important

• Let's look at a 1-bit ALU for addition:

• How could we build a 1-bit ALU for Add, AND, and OR?• How could we build a 32-bit ALU?

Different Implementations

S u m

C a r r y I n

C a r r y O u t

a

b

• A full adder or a (3,2) adder has three inputs (a, b, & Cin) and two outputs (Sum & Cout).

• A half adder or a (2,2) adder has only 2 inputs (a & b) and 2 outputs (Sum & Cout).


Input and Output Specification for 1-bit Adder

1111101011011011000101110100101010000000

SumCarryOutCarryInbaOutputsInputs


Values of the Inputs when CarryOut is a 1

• CarryOut = (b . CarryIn) + (a . CarryIn) + (a . b) + (a . b . CarryIn)• If a . b . CarryIn is true, then all of the other terms must be true,

so we leave out the last term.• cout = a b + a cin + b cin

111011101110

CarryInbaInputs


Adder Hardware for the Carry Out Signal

• The above Hardware was constructed from the following equation: CarryOut = (b . CarryIn) + (a . CarryIn) + (a . b)

CarryIn

a

b

CarryOut


The Sum bit

• The Sum bit is set to 1 when exactly one input is 1 or when all three inputs are 1.(check Truth Table on slide 7).

• Sum = (a . b’ . CarryIn’) + (a’ . b . CarryIn’) + (a’ . b’ . CarryIn) + (a . b . CarryIn)

• Sum = a XOR b XOR cin

• Draw the logic circuit? (Left as exercise).


A 1-bit ALU that Performs AND, OR, & Addition

b

0

2

R e s u lt

O p e ra t io n

a

1

C a r ry In

C a r ry O u t


A 32 bit ALU Constructed from 32 1-bit ALUs.

R e s u l t 3 1a 3 1

b 3 1

R e s u l t 0

C a r r y I n

a 0

b 0

R e s u l t 1a 1

b 1

R e s u l t 2a 2

b 2

O p e r a t i o n

A L U 0

C a r r y I n

C a r r y O u t

A L U 1

C a r r y I n

C a r r y O u t

A L U 2

C a r r y I n

C a r r y O u t

A L U 3 1

C a r r y I n

Ripple carry: CarryOut of the less significant bit is connected to the CarryIn of the more significant bit.


• Two's complement approach: just negate b and add.• How do we negate?

• A very clever solution:

What about subtraction (a – b) ?

0

2

Result

Operation

a

1

CarryIn

CarryOut

0

1

Binvert

b


• No overflow when adding a positive and a negative number• No overflow when signs are the same for subtraction• Overflow occurs when the value affects the sign:

– overflow when adding two positives yields a negative – or, adding two negatives gives a positive– or, subtract a negative from a positive and get a negative– or, subtract a positive from a negative and get a positive

• Consider the operations A + B, and A – B– Can overflow occur if B is 0 ?– Can overflow occur if A is 0 ?

Detecting Overflow


• An exception (interrupt) occurs– Control jumps to predefined address for exception– Interrupted address is saved for possible resumption

• Details based on software system / language– example: flight control vs. homework assignment

• Don't always want to detect overflow— new MIPS instructions: addu, addiu, subu

note: addiu still sign-extends!note: sltu, sltiu for unsigned comparisons

Effects of Overflow


• More complicated than addition– accomplished via shifting and addition

• More time and more area• Let's look at 3 versions based on grade school algorithm

1000 (multiplicand)__x_1001 (multiplier)

10000000

00001000

1001000 (Product)• If we ignore the sign bits, the length of the multiplication of

an n-bit multiplicand and an m-bit multiplier is a product that is n + m bits long.

• That is, n + m bits are required to represent all possible products.

Multiplication


Multiplication

• Each step of multiplication is simple:1. Just place a copy of the multiplicand

(1 x multiplicand) in the proper place if the multiplier digit is 1, or

2. Place 0 (0 x multiplicand) in the proper place if the digit is 0.

• Negative numbers: convert and multiply– there are better techniques, we won’t look at

them.


Multiplication: Implementation

DatapathControl

MultiplicandShift left

64 bits

64-bit ALU

ProductWrite

64 bits

Control test

MultiplierShift right

32 bits

32nd repetition?

1a. Add multiplicand to product andplace the result in Product register

Multiplier0 = 01. TestMultiplier0

Start

Multiplier0 = 1

2. Shift the Multiplicand register left 1 bit

3. Shift the Multiplier register right 1 bit

No: < 32 repetitions

Yes: 32 repetitions

Done


First Version of the Multiplication Hardware

• The Multiplicand register, ALU, and Product register are all 64 bits wide, with only the Multiplier register containing 32 bits.

• We will need to move the multiplicand left one digit each step as it may be added to the intermediate products. So, over 32 steps a 32-bit multiplicand would move 32 bits to the left. Hence we need a 64-bit Multiplicand register, initialized with the 32-bit multiplication in the right half and 0 in the left half.

• The multiplier is shifted in the opposite direction at each step.• The Product register initialized to 0.• Control decides when to shift the Multiplicand and Multiplier

registers and when to write new values into the Product register.


Final Version

Multiplicand

32 bits

32-bit ALU

ProductWrite

64 bits

Controltest

Shift right

32nd repetition?

Product0 = 01. TestProduct0

Start

Product0 = 1

3. Shift the Product register right 1 bit

No: < 32 repetitions

Yes: 32 repetitions

Done

What goes here?

•Multiplier starts in right half of product


Multiply in MIPS

• MIPS provides a separate pair of 32-bit registers to contain the 64-bit product, called Hi and Lo.

• To produce a properly signed or unsigned product, MIPS has two instructions: multiply (mult) and multiply unsigned (multu).


Floating Point (a brief look)

• We need a way to represent– numbers with fractions, e.g., 3.1416– very small numbers, e.g., .000000001– very large numbers, e.g., 3.15576 × 109

• Representation:– sign, exponent, significand: (–1)sign × significand × 2exponent

– more bits for significand gives more accuracy– more bits for exponent increases range

• IEEE 754 floating point standard: – single precision: 8 bit exponent, 23 bit significand– double precision: 11 bit exponent, 52 bit significand


IEEE 754 floating-point standard

• Leading “1” bit of significand is implicit

• Exponent is “biased” to make sorting easier– all 0s is smallest exponent all 1s is largest– bias of 127 for single precision and 1023 for double precision– summary: (–1)sign × (1+significand) × 2exponent – bias

• Example:

– decimal: -.75 = - ( ½ + ¼ )– binary: -.11 = -1.1 x 2-1

– floating point: exponent = 126 = 01111110

– IEEE single precision: 10111111010000000000000000000000


Floating point addition

•

Still normalized?

4. Round the significand to the appropriatenumber of bits

YesOverflow orunderflow?

Start

No

Yes

Done

1. Compare the exponents of the two numbers.Shift the smaller number to the right until itsexponent would match the larger exponent

2. Add the significands

3. Normalize the sum, either shifting right andincrementing the exponent or shifting left

and decrementing the exponent

No Exception

Small ALU

Exponentdifference

Control

ExponentSign Fraction

Big ALU


0 1 0 1 0 1

Shift right

0 1 0 1

Increment ordecrement

Shift left or right

Rounding hardware



Floating Point Complexities

• Operations are somewhat more complicated (see text)• In addition to overflow we can have “underflow”• Accuracy can be a big problem

– IEEE 754 keeps two extra bits, guard and round– four rounding modes– positive divided by zero yields “infinity”– zero divide by zero yields “not a number”– other complexities

• Implementing the standard can be tricky• Not using the standard can be even worse

– see text for description of 80x86 and Pentium bug!


Chapter Three Summary

• Computer arithmetic is constrained by limited precision• Bit patterns have no inherent meaning but standards do exist

– two’s complement– IEEE 754 floating point

• Computer instructions determine “meaning” of the bit patterns• Performance and accuracy are important so there are many

complexities in real machines • Algorithm choice is important and may lead to hardware

optimizations for both space and time (e.g., multiplication)

• You may want to look back (Section 3.10 is great reading!)

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