1 cs37: computer architecture spring term, 2005 instructor: kate forbes-riley...

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CS37: Computer ArchitectureSpring Term, 2005

Instructor: Kate Forbes-Riley

[email protected]

mailto:[email protected]

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CS37: Lecture 7

• Floating point representation

• Floating point addition and multiplication

• Introduction to Machine Language and Assembly Language

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Review: GTE (≥)

1) GTE = ¬MSB (i.e, A – B ≥ 0)

2) GTE = (MSB ● Overflow) + (¬MSB ● ¬Overflow)

But this is just: ¬( MSB XOR Overflow)

(GTE = 1 only if neither or both =1)

MSB result

OverflowGTEout¬xor

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result 31

Review: 32-bit ALU w/ GTEIf GTEout = 1, A ≥ B, else GTEout = 0

GTEout is sent to GTEin of LSB; other GTEins = 0

if OP Code = 4, GTEins = Result

if A ≥ B, GTE Result = 00..01; else GTE Result = 00..00

Binvert Cin Op ALU 0

GTEin Cout


GTEin Cout GTEout


GTEin Cout


GTEin Cout

Binvert Op Code

…

result 0

result 1

result 30

A0B0

A1B1 0

A30B30 0

A31B31 0

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Review: Binary Integer Multiplication

1 0 0 multiplicand (m)

x 1 0 1 multiplier (n)

1 0 0 multiply and shift

0 0 0 intermediate

+ 1 0 0 products (Pb)

1 0 1 0 0 product: (p) m+n bits

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Review: Binary Multiplication (1st version PH3)

0 0 0 0 0 0 product = m + n bits0 0 0 1 0 0 m’cand in right m bits

1 0 1 test m’plier: LSB = 1 0 0 0 1 0 0 product = product + m’cand0 0 1 0 0 0 SLL 1 m’cand

0 1 0 SLR 1 m’plier 0 1 0 test m’plier: LSB = 0

0 1 0 0 0 0 SLL 1 m’cand 0 0 1 SLR 1 m’plier

0 0 1 test m’plier: LSB = 10 1 0 1 0 0 product = product + m’cand

1 0 0 0 0 0 SLL 1 m’cand 0 0 0 SLR 1 m’plier

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Floating Point

• Real Numbers: include fractions of whole #s 123,456.789 105 … 100.10-1 … 10-3

• Scientific notation: a single digit to the left of the decimal point 0.123456789 x 106

0.0123456789 x 107

0.00123456789 x 108

• Normalized Scientific notation: scientific notation with no leading zeros 1.23456789 x 105

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Floating Point with Binary Numbers

• Real Numbers: include fractions of whole #s 110111.101101

25 … 20.2-1 … 2-6

• Scientific notation: a single digit to the left of the binary point (with exponent in base 10) 0.110111101101 x 26

0.0110111101101 x 27

• Normalized Scientific notation: scientific notation with no leading zeros 1.10111101101 x 25

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• Normalized Scientific notation: always using the same form simplifies arithmetic and increases accuracy of stored reals b/c no leading zeros: leading digit is always 1

1.10111101101 x 25

Significand Base Exponent

Common to write significand in binary and base and exponent in decimal

Floating Point with Binary Numbers

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• Single precision = 32-bit representation (same procedure for double = 64 bit: 11 exp; 52 sig)

31 0

IEEE 754 floating-point standard

30 exponent 23 22 significand 031 sign

• 1 sign bit: 1 = negative; 0 = positive

• 8 bits for exponent: determines range of #s that can be represented

• 23 bits for significand: determines accuracy of #s that are represented

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IEEE 754 floating point: single precision

22 significand 0(fraction)

31 sign

• 23 bits for significand represents the fraction. The leading 1 is implicit!

• Because fraction is already in binary form, we can just put its (first/add) 23 bits into bits 0-22

• Similarly, we just put the sign in the sign bit

• E.g., 1.10111101101 x 25

30 exponent 23(biased)

22 significand 0 10111101101000000000000

31 0

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31 sign

• +/- Exponents aren’t represented with two’s complement!

• Exponent represented with “biased notation”.

This allows efficient sorting with integer HW:

00000001 … 11111111

negative exponents < positive exponents


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31 sign

• To compute biased exponent, add bias to exponent (both decimal) then convert to binary

• Bias for single precision = 127



Exp Biased Exp Binary 8-bits

-126 -126+127 = 1 20 00000001

-31 -31+127 = 96 26+25 01100000

5 5+127 = 132 27+22 10000100

126 126+127 = 253 27+26+25+24+23+22+21 11111110

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31 sign

• So biased exponent represents sign and magnitude of exponent in 8 bits.

• Note range and reserved biased exponents (see PH3 pg 194):



Exponent Biased exp Significand Value

-126 to 127 1 to 254 any +/- reals

128 255 0 infinity (+/- )

128 255 nonzero NAN

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31 sign

• To compute the value of this FP representation:

Value = (-1)sign x (1+significand) x 2biased exp – bias

• Binary Val = -1.10000000000000000000000 x 23

(-1)1 x (1+.10000000000000000000000) x 2130 – 127

• Decimal Val = -12.0

-1 x (1 + 2-1) x 23 = -1 x (1.5) x 8 = -12.0



1 sign

10000010biased exp

10000000000000000000000 (significand fraction)

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31 sign

• To convert decimal to IEEE binary floating point

• If you’re lucky, it’s easy to manipulate:

0.75 = 3/4 = 3 x 2-2

= 0011. x 2-2 = (1.1 x 21) x 2-2 = 1.1 x 2-1

= (-1)0 x (1 + .10000…000) x 2((-1+127) – 127)

= (-1)0 x (1 + .10000…000) x 2(126 – 127)



31 0

22 significand 010000000000000000000000

30 exponent 2301111110

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31 sign

• If you’re unlucky, use brute force:

- 3.1415

1. convert integer 2. convert fraction:

(not sign): 11 Does 2-1 fit? Does 2-2 fit? Does 2-3 fit? Does 2-4 fit? …

• Here stop at 2-22 due to integer normalization



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31 sign

2. Converting fraction part: 11.00100100001…

.1415

- .1250 (1/8 = 2-3)

.0165

- .015625 (1/64 = 2-6)

.000875

- .000488 (1/2048 = 2-11)

…

Infinite # reals between 0…1; so some inaccuracy



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31 sign

3. Normalize with sign:

-11.00100100001…

-1.100100100001… x 21

4. Convert to IEEE 754 via equation:

= (-1)1 x (1 + .100100001…) x 2((1+127) – 127)

= (-1)1 x (1 + .100100001…) x 2(128 – 127)



1 sign

10000000(biased exp)

100100100001…(significand fraction)

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Convert -128.673828125 to IEEE FP standard

1. Convert integer part (w/o sign):

128 = 10000000

2. Convert fraction part:

= 10101100100000003. Normalize with sign:

-10000000.1010110010000000 =

-1.00000001010110010000000 x 27

2-1

.5

2-2

.25

2-3

.125

2-4

.0625

2-5

.03125

2-6

.015625

2-7

.0078125

2-8

.00390625

2-9

.001953125

1 0 1 0 1 1 0 0 1

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Convert -128.673828125 to IEEE FP standard

-1.00000001010110010000000 x 27

4. Convert to IEEE 754 equation:

(-1)1 x (1 + .00000001010110010000000) x 2((7+127) – 127)

(-1)1 x (1 + .00000001010110010000000) x 2(134 – 127)

5. Convert to IEEE 754 FP representation:

1 sign

10000110(biased exp)

00000001010110010000000(significand fraction)

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IEEE 754 Floating Point: Addition Algorithm

• Add 1.100 x 22 and 1.100 x 21

Step 1: Shift right (SRL) the significand of the smaller # to match exponent of larger #

1.100 x 21 0.110 x 22 lost accuracy: keep #bits

Step 2: Add the significands

1.100 x 22

+ 0.110 x 22

10.010 x 22

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IEEE 754 floating point: Addition Algorithm

Step 3: Normalize sum, checking for overflow (+ exponent too large for (8) bits) and underflow (- exponent too small for (8) bits)

10.010 x 22 1.0010 x 23

0 < 3 + 127 < 255 ok: no under/overflow

Step 4: Round the sum (variously, see pg 213), then normalize again if necessary

1.0010 x 23 1.001 x 23 lost accuracy: keep #bits

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IEEE 754 floating point: Multiplication Algorithm

• Multiply 1.100 x 22 by 1.100 x 21

Step 1: Step 2:

Add biased Multiply significands

exponents then 1.100

subtract bias x 1.100

(2+127) + (1+127) 0000

= 257 - 127 = 130 0000

1100

1100

10.010000

place binary point 3 + 3 digits from right

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IEEE 754 floating point: Multiplication Algorithm

Step 3: Normalize product, check over/underflow

10.010000 x 2130 1.0010000 x 2131

0 < 131 < 255 ok: no under/overflow

Step 4: Round (variously, see pg 213), then normalize again if necessary

1.0010000 x 2131 1.001 x 2131 lost accuracy

Step 5: Set sign of product (+ if operands’ signs are same; else -)

1.001 x 2131

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Implementation of Floating Point Operations

• Block diagram of FP addition HW on pg 201

• Similar design to integer operations

• But need more HW (PH3 Figure 3.17, 3.18) 2 ALUs: one for exponent, one for significand Logic for Shifts Logic for Rounding

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Machine Languages: Instruction Sets

• An “instruction” is a specific sequence of binary numbers that tells the control to perform a single operation in a single step via the datapath

• An “instruction set” is the set of all instructions that a computer understands

• “Machine Language” is the numeric version of these instructions

• “Assembly Language” is a symbolic notation that the assembler translates into machine language (basically a 1-to-1 correspondence)

C Program swap(int v[], int k) {int tmp;tmp = v[k];v[k] = v[k+1];v[k+1] = tmp;}

MIPS AL Program swap: muli $2, $5, 4add $4, $2lw $15, 0($2)lw $16, 4($2)sw $16, 0($2)sw $15, 4($2)jr $31

MIPS ML Program 000000001010000100000000000110000000000010001110000110000010000110001100011000100000000000000000…

C compiler

MIPS assembler

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Processors and Assembly Languages

• Many different CPU architectures (machine languages, assembly languages, assemblers, instruction set types, etc.) PH3 CH 2 and X86 1.2 have some comparative discussion

• PH3 focuses mainly on the MIPS architecture

• We will learn the modern Intel 80x86 assembly language because it is the most prevalent today

• However, X86 is much more complex, due to backwards compatibility with earlier versions; thus, we will gloss over some details.

• We will use the X86 book on the website.

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Instruction Set Types

• MIPS uses RISC (Reduced Instruction Set Computer) Fixed instruction length: 32 bits Fewer instruction formats

• X86 uses CISC (Complex Instruction Set Computer) Instructions vary from 1 to 17 bytes; more

common operations are shorter More instruction formats

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Machine Language Programs

• An ML “program” consists of a specific sequence of instructions (machine code)

• In other words, ML programs are just sequences of (sequences of) binary numbers

• Therefore, both ML programs and data can be treated alike: both can be stored in memory to be read and written to: stored program concept

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Inside Computer: Stored Program Concept

Processor

Both the data and the programs for manipulating that data are stored in memory

Control

Datapath

REGISTERSALU

DEVICES

INPUT

OUTPUT

Main Memory (volatile)

Accounting program (machine code)

Editor program (machine code)

Monthly bills data

Master’s thesis data

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Inside Computer: Stored Program Concept

Processor

Control takes program as input; each instruction tells it to do operations on data in memory, via datapath and registers

Control

Datapath

REGISTERSALU

DEVICES

INPUT

OUTPUT

Main Memory

Accounting program (machine code)

Editor program (machine code)

Monthly bills data

Master’s thesis data

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Using Registers

• Basic repeated pattern occurs when performing a program: Load values from memory into registers Operate on values in registers Store values from registers into memory

• Why use registers as intermediary? Main memory is large (~1 GB) but slow Registers are much faster: they are closer

(on processor!) and smaller

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Using Registers

• Processor uses registers for “scratch paper”: Registers are the primary source and

destination for the results of operations These results are often used repeatedly, so

keeping them in registers is faster

• Some registers are “general purpose” (can be used for anything), others play specific roles

• MIPS has 32 32-bit registers (don’t worry about names right now); 24 are general purpose registers

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Using Registers

• X86 has 8 32-bit registers: ESP, EBP: often play specific roles EAX, ECX, EDX, EBX, ESI, EDI: truly general

purpose registers

• Don’t worry about the 16-bit registers discussed in the X86 book; most exist for backwards compatibility

• Why does X86 have so many fewer registers than MIPS? In part, because X86 has much more complex instructions

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How Instructions access Memory

• To access memory elements (e.g., load/store data), the instruction must know memory organization

• Memory is organized as a huge array, where the memory address is the index to the array

… …

2 10

1 00001

0 101

Memory address Data

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How Instructions access data in Memory

• In x86, memory is byte-addressable: address 0 is the first byte, address 1 is the second byte, etc. (different in MIPS)

• Note: in x86, 2 bytes = word; 4 bytes = double word

• Protected mode: each program has own virtual address space, managed by OS (details later)

… …

2 1 byte = 8 bits

1 1 byte = 8 bits

0 1 byte = 8 bits

Memory address Data

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Assignment

• Reading: PH Sections 3.6, skim 3.7 – 3.10 Skim PH3, Chapter 2 X86 book (as needed)

1 cs37: computer architecture spring term, 2005 instructor: kate forbes-riley...

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