CS代考 Introduction to Computer Systems 15-213/18-243, spring 2009 1st Lecture, Ja

Introduction to Computer Systems 15-213/18-243, spring 2009 1st Lecture, Jan. 12th

Data representation II

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Acknowledgement: These slides are based on the textbook
(Computer Systems: A Programmer’s Perspective) and its slides.

Our focus today:
“How is real number stored in memory?”
“How is floating-point number stored in memory?”
#include

int main()
float a=3.1;
float b=4.2;
float c=a+b;
printf(“%f\n”,c);

Floating-point number  Real number
Example: Is (x + y) + z = x + (y + z)?
(1e20 + -1e20) + 3.14  3.14
1e20 + (-1e20 + 3.14)  ???
#include

int main() {
float a=(1e20 + -1e20) + 3.14;
float b=1e20 + (-1e20 + 3.14);
printf(“%f\n”,a);
printf(“%f\n”,b);

Today: Floating Point
Background: Fractional binary numbers
IEEE floating point standard: Definition
Example and properties
Rounding, addition, multiplication
Floating point in C

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition
Fractional Decimal Numbers
Integer number
Fractional number
3 1 5 9 610
3*104 + 1*103 + 5*102 + 9*101 + 6*100
3 1 . 5 9 610
3*101 + 1*100 + 5*10-1 + 9*10-2 + 6*10-3

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

bi bi-1 ••• b2 b1 b0 b-1 b-2 b-3 ••• b-j

Fractional Binary Numbers
Representation
Bits to right of “binary point” represent fractional powers of 2
Represents rational number:

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition
Latex source for equation: \sum_{k=-j}^i b_k \times 2^k

Fractional Binary Numbers
How to represent 11.62510 as
a fractional binary number?

0.625 = 1*(1/2) + 0*(1/4) + 1*(1/8)

Thus, we can represent it as 1011.1012

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Fractional Binary Numbers: Examples
Value Representation
5 3/4 101.112
2 7/8 010.1112
1 7/16 001.01112
Observations
Divide by 2 by shifting right (unsigned)
Multiply by 2 by shifting left
Numbers of form 0.111111…2 are less than 1.0
1/2 + 1/4 + 1/8 + … + 1/2i + … < 1.0 Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Representable Numbers Limitation #1 Can only exactly represent numbers of the form x/2k The following numbers cannot be exactly represented! Value Representation 1/3 0.0101010101[01]…2 1/5 0.001100110011[0011]…2 1/10 0.0001100110011[0011]…2 Limitation #2 Just one fixed binary point within the w bits? Limited range of numbers (very small values? very large?) Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition IEEE Floating Point IEEE Standard 754 Established in 1985 as uniform standard for floating point arithmetic Supported by all major CPUs Driven by numerical concerns Nice standards for rounding, overflow, underflow Hard to make fast in hardware Numerical analysts predominated over hardware designers in defining standard Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Scientific Notation Significand Decimal notation Scientific notation Normalized scientific notation: Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Numerical Form: (–1)s M 2E Sign bit s determines whether number is negative or positive Significand M normally a fractional value in range [1.0,2.0). Exponent E weights value by power of two MSB s is sign bit s exp field encodes E (but is not equal to E) frac field encodes M (but is not equal to M) Floating Point Representation s exp frac Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Precision options Single precision: 32 bits Double precision: 64 bits s exp frac 1 8-bits 23-bits s exp frac 1 11-bits 52-bits Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Overview: Floating Point Encodings Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition “Normalized” Values When: exp ≠ 000…0 and exp ≠ 111…1 Exponent coded as a biased value: E = Exp – Bias Exp: unsigned value of exp field Bias = 2k-1 - 1, where k is number of exponent bits Single precision: 127 (Exp: 1…254, E: -126…127) Double precision: 1023 (Exp: 1…2046, E: -1022…1023) Significand coded with implied leading 1: M = 1.xxx…x2  xxx…x: bits of frac field Minimum when frac=000…0 (M = 1.0) Maximum when frac=111…1 (M = 2.0 – ε) Get extra leading bit for “free” v = (–1)s M 2E s exp frac Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Normalized Encoding Example Value: float F = 15213.0; 1521310 = 111011011011012 = 1.11011011011012 x 213 Significand M = 1.11011011011012 frac = 110110110110100000000002 E = 13 Bias = 127 Exp = E + Bias = 140 = 100011002 0 10001100 11011011011010000000000 v = (–1)s M 2E E = Exp – Bias Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Normalized Encoding Example v = (–1)s M 2E E = Exp – Bias 0 10001100 11011011011010000000000 #include

typedef unsigned char *pointer;

void show_bytes(pointer start, size_t len){
for (i = 0; i < len; i++) printf("%p\t0x%.2x\n",start+i, start[i]); printf("\n"); int main() { float a=15213; show_bytes((pointer)&a,sizeof(float)); 0x7ffcc3ce673c 0x00 0x7ffcc3ce673d 0xb4 0x7ffcc3ce673e 0x6d 0x7ffcc3ce673f 0x46 Result on x86 (little Endian): Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Denormalized Values Condition: exp = 000…0 Exponent value: E = 1 – Bias (instead of E = 0 – Bias) Significand coded with implied leading 0: M = 0.xxx…x2 xxx…x: bits of frac exp = 000…0, frac = 000…0 Represents zero value exp = 000…0, frac ≠ 000…0 Numbers closest to 0.0 v = (–1)s M 2E E = 1 – Bias s exp frac Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Special Values Condition: exp = 111…1 Case: exp = 111…1, frac = 000…0 Represents value  (infinity) Operation that overflows E.g., the result of 1.0/0.0 Case: exp = 111…1, frac ≠ 000…0 Not-a-Number (NaN) Represents case when no numeric value can be determined E.g., the results of sqrt(–1),  − ,   0 s exp frac Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Visualization: Floating Point Encodings Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Visualization: Floating Point Encodings +Normalized −Normalized Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Interesting Numbers Description e xp frac Numeric Value Zero 00…00 00…00 0.0 Smallest Pos. Denorm. 00…00 00…01 2– {23,52} x 2– {126,1022} Single ≈ 1.4 x 10–45 Double ≈ 4.9 x 10–324 Largest Denormalized 00…00 11…11 (1.0 – ε) x 2– {126,1022} Single ≈ 1.18 x 10–38 Double ≈ 2.2 x 10–308 Smallest Pos. Normalized 00…01 00…00 1.0 x 2– {126,1022} Just larger than largest denormalized One 01…11 00…00 1.0 Largest Normalized 11…10 11…11 (2.0 – ε) x 2{127,1023} Single ≈ 3.4 x 1038 Double ≈ 1.8 x 10308 {single,double} Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Tiny Floating Point Example Example: 8-bit Floating Point Representation the sign bit is in the most significant bit the next four bits are the exponent, with a bias of 7 the last three bits are the frac Let’s use the form like the IEEE Format normalized, denormalized representation of 0, NaN, infinity s exp frac 1 4-bits 3-bits Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition s exp frac E Value 0 0000 000 -6 0 0 0000 001 -6 1/8*1/64 = 1/512 0 0000 010 -6 2/8*1/64 = 2/512 0 0000 110 -6 6/8*1/64 = 6/512 0 0000 111 -6 7/8*1/64 = 7/512 0 0001 000 -6 8/8*1/64 = 8/512 0 0001 001 -6 9/8*1/64 = 9/512 0 0110 110 -1 14/8*1/2 = 14/16 0 0110 111 -1 15/8*1/2 = 15/16 0 0111 000 0 8/8*1 = 1 0 0111 001 0 9/8*1 = 9/8 0 0111 010 0 10/8*1 = 10/8 0 1110 110 7 14/8*128 = 224 0 1110 111 7 15/8*128 = 240 0 1111 000 n/a inf Dynamic Range (Positive Only) closest to zero largest denorm smallest norm closest to 1 below closest to 1 above largest norm Denormalized Normalized v = (–1)s M 2E n: E = Exp – Bias d: E = 1 – Bias Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Creating Floating Point Number Normalize to have leading 1 Round to fit within fraction Postnormalize to deal with effects of rounding Case Study Convert 8-bit unsigned numbers to tiny floating point format Example Numbers 128 10000000 15 00001101 17 00010001 19 00010011 138 10001010 63 00111111 s exp frac 1 4-bits 3-bits Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Requirement Set binary point so that numbers of form 1.xxxxx Adjust all to have leading one Decrement exponent as shift left Value Binary Fraction Exponent 128 10000000 1.0000000 7 15 00001101 1.1010000 3 17 00010001 1.0001000 4 19 00010011 1.0011000 4 138 10001010 1.0001010 7 63 00111111 1.1111100 5 s exp frac 1 4-bits 3-bits Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Round up conditions Round = 1, Sticky = 1 ➙ > 0.5
Guard = 1, Round = 1, Sticky = 0 ➙ Round to even
Value Fraction GRS Incr? Rounded
128 1.0000000 000 N 1.000
15 1.1010000 100 N 1.101
17 1.0001000 010 N 1.000
19 1.0011000 110 Y 1.010
138 1.0001010 011 Y 1.001
63 1.1111100 111 Y 10.000
Guard bit: LSB of result
Round bit: 1st bit removed

Sticky bit: OR of remaining bits

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Postnormalize
Rounding may have caused overflow
Handle by shifting right once & incrementing exponent
Value Rounded Exp Adjusted Result
128 1.000 7 128
15 1.101 3 15
17 1.000 4 16
19 1.010 4 20
138 1.001 7 134
63 10.000 5 1.000/6 64

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Today: Floating Point
Background: Fractional binary numbers
IEEE floating point standard: Definition
Example and properties
Rounding, addition, multiplication
Floating point in C

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Distribution of Values
6-bit IEEE-like format
e = 3 exponent bits
f = 2 fraction bits
Bias is 23-1-1 = 3

Notice how the distribution gets denser toward zero.
s exp frac
1 3-bits 2-bits

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Rounding Modes (illustrate with $ rounding)

$1.40 $1.60 $1.50 $2.50 –$1.50

Towards zero $1 $1 $1 $2 –$1

Round down (−) $1 $1 $1 $2 –$2

Round up (+) $2 $2 $2 $3 –$1

Nearest Even (default) $1 $2 $2 $2 –$2

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Closer Look at Round-To-Even (Nearest Even)
Default Rounding Mode
Other rounding modes may produce statistically biased results
E.g., sum of positive numbers will consistently be over- or under- estimated
Applying to Other Decimal Places / Bit Positions
When exactly halfway between two possible values
Round so that least significant digit is even
E.g., round to nearest hundredth
7.8949999 7.89 (Less than half way)
7.8950001 7.90 (Greater than half way)
7.8950000 7.90 (Half way—round up)
7.8850000 7.88 (Half way—round down)

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Rounding Binary Numbers
Binary Fractional Numbers
“Even” when least significant bit is 0
“Half way” when bits to right of rounding position = 100…2

Round to nearest 1/4 (2 bits right of binary point)
Value Binary Rounded Action Rounded Value
2 3/32 10.000112 10.002 (<1/2—down) 2 2 3/16 10.001102 10.012 (>1/2—up) 2 1/4
2 7/8 10.111002 11.002 ( 1/2—up) 3
2 5/8 10.101002 10.102 ( 1/2—down) 2 1/2

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Floating Point Operations: Basic Idea
x +f y = Round(x + y)

x f y = Round(x  y)

Basic idea
First compute exact result
Make it fit into desired precision
Possibly overflow if exponent too large
Possibly round to fit into frac

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

FP Multiplication
(–1)s1 M1 2E1 x (–1)s2 M2 2E2
Exact Result: (–1)s M 2E
Sign s: s1 ^ s2
Significand M: M1 x  M2
Exponent E: E1 + E2
If M ≥ 2, shift M right, increment E
If E out of range, overflow
Round M to fit frac precision
Implementation
The biggest task is multiplying significands
s1 exp1 frac1

s2 exp2 frac2

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Mathematical Properties of FP Mult
Multiplication Commutative?
Multiplication is Associative?
Possibility of overflow, inexactness of rounding
Ex: (1e20*1e20)*1e-20= inf,
1e20*(1e20*1e-20)= 1e20
1 is multiplicative identity?
Multiplication distributes over addition?
Possibility of overflow, inexactness of rounding
1e20*(1e20-1e20)= 0.0,
1e20*1e20 – 1e20*1e20 = NaN

Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition

Floating Point Addition
(–1)s1 M1 2E1 + (-1)s2 M2 2E2
Assume E1 > E2

Exact Result: (–1)s M 2E
Sign s, significand M:
Result of signed align & add
Exponent E: E1

If M ≥ 2, shift M right, increment E
if M < 1, shift M left k positions, decrement E by k Overflow if E out of range Round M to fit frac precision Get binary points lined up Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Mathematical Properties of FP Add Commutative? Associative? Overflow and inexactness of rounding (3.14+1e10)-1e10 = 0, 3.14+(1e10-1e10) = 3.14 0 is additive identity? Every element has additive inverse? Yes, except for infinities & NaNs Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Float Point Comparison FP Zero Same as Integer Zero All bits = 0 Can (Almost) Use Unsigned Integer Comparison Must first compare sign bits Must consider −0 = 0 NaNs problematic Will be greater than any other values What should comparison yield? Otherwise OK Denorm vs. normalized Normalized vs. infinity s1 exp1 frac1 s2 exp2 frac2 Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Floating Point in C C Guarantees Two Levels float single precision double double precision Conversions/Casting Casting between int, float, and double changes bit representation double/float → int Truncates fractional part Like rounding toward zero Not defined when out of range or NaN: Generally sets to TMin int → double Exact conversion, as long as int has ≤ 53 bit word size int → float Will round according to rounding mode Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Floating Point Questions For each of the following C expressions, either: Argue that it is true for all argument values Explain why not true 2/3 == 2/3.0 (d+f)-d == f f == (float)(double) f d == (double)(float) d int x = …; float f = …; double d = …; Assume neither d nor f is NaN Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Representation of IEEE Floating Point numbers Operations Rounding, addition, multiplication, casting Not the same as real arithmetic Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition -15-10-5051015 DenormalizedNormalizedInfinity 0.005 0.25 15 0.0625 0.3125 -15 0.125 0.375 0.1875 0.4375 -0.005 0.5 -0.0625 0.625 -0.125 0.75 -0.1875 0.875 Denormalized Normalized 0.02 0.25 0.5 0.75 -0.02 -0.25 -0.5 -0.75 0 0 0 0 0 0 0 0 1 1.25 1.5 1.75 2 2.5 3 3.5 -1 -1.25 -1.5 -1.75 -2 -2.5 -3 -3.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Denormalized Normalized 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com