程序代写代做代考 compiler assembly C case study Carnegie Mellon

程序代写 CS代考 / case study, compiler, C语言代写

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14
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513
18
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613

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Floating Point
15-213/18-213/14-513/15-513/18-613: Introduction to Computer Systems 4th Lecture, Sept. 10, 2020
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Announcements
 Lab 0 due today 11:59 pm ET
 Lab 1 went out on Tuesday, due 9/17 ▪ Puzzles can be tricky, so start early
 Written Assignment 1 released yesterday, due 9/16 ▪ Available on canvas. Hand-in via canvas
 Bootcamp 3 is Friday 7-9 pm ET ▪ Debugging & gdb
 First Recitations are Monday
▪ Students requesting in-person recitations will get assigned to an in-person
recitation section
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Today: Floating Point
 Background: Fractional binary numbers  IEEE floating point standard: Definition  Example and properties
 Rounding, addition, multiplication
 Floating point in C  Summary
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Fractional binary numbers  What is 1011.1012?
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Fractional Binary Numbers
bi-1
•••
b1
b0
b-1
b-2
b-3
•••
b-j
1/2
1/4
1/8
 Representation
▪ Bits to right of “binary point” represent fractional powers of 2 ▪ Represents rational number:
2-j
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2i
•••
2i-1
4
2
1
bi
b2
•••

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Fractional Binary Numbers: Examples
 Value
5 3/4 = 23/4 2 7/8 = 23/8 1 7/16 = 23/16
Representation
101.112 010.1112 001.01112
= 4 + 1 + 1/2 + 1/4
= 2 + 1/2 + 1/4 + 1/8 = 1 + 1/4 + 1/8 + 1/16
 Observations
▪ Divide by 2 by shifting right (unsigned)
▪ Multiply by 2 by shifting left
▪ Numbers of form 0.111111…2 are just below 1.0
▪ 1/2 + 1/4 + 1/8 + … + 1/2i + … ➙ 1.0 ▪ Use notation 1.0 – ε
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Representable Numbers
 Limitation #1
▪ Can only exactly represent numbers of the form x/2k
▪ Other rational numbers have repeating bit representations
▪ Value ▪ 1/3 ▪ 1/5
▪ 1/10
Representation
0.0101010101[01]…2 0.001100110011[0011]…2 0.0001100110011[0011]…2
 Limitation #2
▪ Just one setting of binary point within the w bits
▪ Limited range of numbers (very small values? very large?)
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Today: Floating Point
 Background: Fractional binary numbers  IEEE floating point standard: Definition  Example and properties
 Rounding, addition, multiplication
 Floating point in C  Summary
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IEEE Floating Point
 IEEE Standard 754
▪ Established in 1985 as uniform standard for floating point arithmetic
▪ Before that, many idiosyncratic formats
▪ Supported by all major CPUs
▪ Some CPUs don’t implement IEEE 754 in full e.g., early GPUs, Cell BE processor
 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
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This is important!
 Ariane 5 explodes on maiden voyage: $500 MILLION dollars lost ▪ 64-bit floating point number assigned to 16-bit integer (1996)
▪ Legacy code from Ariane 4 with a lower top speed
▪ Causes rocket to get incorrect value of horizontal velocity and crash
 Patriot Missile defense system misses scud – 28 people die
▪ System tracks time in tenths of second
▪ Converted from integer to floating point number.
▪ Accumulated rounding error causes drift. 20% drift over 8 hours.
▪ Eventually (on 2/25/1991 system was on for 100 hours) causes range mis-
estimation sufficiently large to miss incoming missiles. Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition
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(Binary) Scientific Notation
 What are the parts of a number in scientific notation?
1.11011011011012 x 213
Significand
Exponent
 What value does the significand always begin with in scientific
notation?
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Floating Point Representation  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
 Encoding
▪ 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)
Example:
1521310 = (-1)0 x 1.11011011011012 x 213
s
exp
frac
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Precision options  Single precision: 32 bits
 7 decimal digits, 10±38 1 8-bits
 Double precision: 64 bits  16 decimal digits, 10±308
1 11-bits
23-bits
s
exp
frac
s
exp
frac
 Other formats: half precision, quad precision Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition
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52-bits

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Three “kinds” of floating point numbers
s
exp
frac
1 e-bits
f-bits
00…00
denormalized
normalized
special
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exp ≠ 0 and exp ≠ 11…11
11…11

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“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” Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition
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v = (–1)s M 2E

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Normalized Encoding Example
 Value: float F = 15213.0; ▪ 1521310 = 111011011011012
= 1.11011011011012 x 213
 Significand
M = 1.11011011011012
frac=
110110110110100000000002
 Exponent
E = 13
Bias = 127
exp = 140 = 100011002
 Result:
0 10001100 11011011011010000000000
s exp frac
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v = (–1)s M 2E E =exp–Bias

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Denormalized Values  Condition: exp = 000…0
 Exponent value: E = 1 – Bias (instead of exp – Bias) (why?)  Significand coded with implied leading 0: M = 0.xxx…x2
▪ xxx…x: bits of frac
 Cases
▪ exp = 000…0, frac = 000…0
▪ Represents zero value
▪ Note distinct values: +0 and –0 (why?) ▪ exp = 000…0, frac ≠ 000…0
▪ Numbers closest to 0.0 ▪ Equispaced
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v = (–1)s M 2E E = 1–Bias

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Special Values
 Condition: exp = 111…1
 Case: exp = 111…1, frac = 000…0
▪ Represents value  (infinity)
▪ Operation that overflows
▪ Both positive and negative
▪ E.g., 1.0/0.0 = −1.0/−0.0 = +, 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., sqrt(–1),  − ,   0
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C float Decoding Example float: 0xC0A00000
binary:
Bias = 2k-1 – 1 = 127
1 8-bits
23-bits
E = 129
S = 1 -> negative number
M = 1.010 0000 0000 0000 0000 0000 M = 1 + 1/4 = 1.25
v = (–1)s M 2E =
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v = (–1)s M 2E E =exp–Bias

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C float Decoding Example #1
float: 0xC0A00000
binary: 1100 0000 1010 0000 0000 0000 0000 0000
1 8-bits 23-bits
E = 129
S = 1 -> negative number
M = 1.010 0000 0000 0000 0000 0000 M = 1 + 1/4 = 1.25
v = (–1)s M 2E =
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v = (–1)s M 2E E =exp–Bias
1
1000 0001
010 0000 0000 0000 0000 0000

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C float Decoding Example #1 float: 0xC0A00000
Bias = 2k-1 – 1 = 127 binary: 1100 0000 1010 0000 0000 0000 0000 0000
1
1000 0001
010 0000 0000 0000 0000 0000
1 8-bits 23-bits E =exp– Bias = 129 – 127 = 2 (decimal)
S = 1 -> negative number
M = 1.010 0000 0000 0000 0000 0000
M = 1 + 1/4 = 1.25
v = (–1)s M 2E = (-1)1 * 1.25 * 22 = -5
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v = (–1)s M 2E E =exp–Bias

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C float Decoding Example #2
float: 0x001C0000
binary: 0000 0000 0001 1100 0000 0000 0000 0000
1 8-bits 23-bits
E = 129
S = 1 -> negative number
M = 0.010 0000 0000 0000 0000 0000 M = 1 + 1/4 = 1.25
v = (–1)s M 2E =
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v = (–1)s M 2E E =1–Bias
0
0000 0000
001 1100 0000 0000 0000 0000

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C float Decoding Example #2 float: 0x001C0000
Bias = 2k-1 – 1 = 127 binary: 0000 0000 0001 1100 0000 0000 0000 0000
v = (–1)s M 2E E =1–Bias
0
0000 0000
001 1100 0000 0000 0000 0000
1 8-bits 23-bits
E =1– Bias = 1 – 127 = –126 (decimal) S = 0 -> positive number
M = 0.001 1100 0000 0000 0000 0000 M = 1/8 + 1/16 + 1/32 = 7/32 = 7*2–5
v = (–1)s M 2E = (-1)0 * 7*2–5 * 2–126 = 7*2–131
v ≈ 2.571393892 X 10–39
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Visualization: Floating Point Encodings
− NaN
−Normalized
−Denorm +Denorm −0 +0
+Normalized
+
NaN
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Today: Floating Point
 Background: Fractional binary numbers  IEEE floating point standard: Definition  Example and properties
 Rounding, addition, multiplication
 Floating point in C  Summary
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Tiny Floating Point Example
1 4-bits 3-bits
 8-bit Floating Point Representation ▪ the sign bit is in the most significant bit
▪ the next four bits are the exp, with a bias of 7 ▪ the last three bits are the frac
 Same general form as IEEE Format ▪ normalized, denormalized
▪ representation of 0, NaN, infinity
s
exp
frac
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v = (–1)s M 2E norm: E = exp – Bias denorm: E = 1 – Bias
Normalized numbers
0 0001 000
0 0001 001
…
0 0110 110 0 0110 111 0 0111 000 0 0111 001 0 0111 010 …
0 1110 110 0 1110 111
-6 8/8*1/64
-6 9/8*1/64
-1 14/8*1/2
-1 15/8*1/2
0 8/8*1
0 9/8*1
0 10/8*1
7 14/8*128
7 15/8*128
= 8/512
= 9/512
= 14/16 = 15/16 =1
= 9/8
= 10/8
= 224 = 240
smallest norm
(-1)0(1+1/8)*2-6
closest to 1 below closest to 1 above
largest norm
0 1111 000 n/a inf
Dynamic Range (s=0 only)
Denormalized numbers
0 0000 000 0 0000 001 0 0000 010 …
0 0000 110
0 0000 111
-6 0
-6 1/8*1/64 = 1/512 -6 2/8*1/64 = 2/512
-6 6/8*1/64 = 6/512
-6 7/8*1/64 = 7/512
closest to zero
(-1)0(0+1/4)*2-6
largest denorm
sexpfracE Value
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Distribution of Values
 6-bit IEEE-like format ▪ e = 3 exponent bits
▪ f = 2 fraction bits ▪ Bias is 23-1-1 = 3
1 3-bits
2-bits
s
exp
frac
 Notice how the distribution gets denser toward zero. 8 values
-15 -10 -5 0 5 10 15
Denormalized Normalized Infinity
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Distribution of Values (close-up view)
 6-bit IEEE-like format ▪ e = 3 exponent bits
▪ f = 2 fraction bits ▪ Bias is 3
1 3-bits 2-bits
s
exp
frac
-1 -0.5 0 0.5 1
Denormalized Normalized Infinity
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Special Properties of the IEEE Encoding  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? The answer is complicated. ▪ Otherwise OK
▪ Denorm vs. normalized ▪ Normalized vs. infinity
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Quiz Time!
Check out:
https://canvas.cmu.edu/courses/17808
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Today: Floating Point
 Background: Fractional binary numbers  IEEE floating point standard: Definition  Example and properties
 Rounding, addition, multiplication
 Floating point in C  Summary
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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
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Rounding
 Rounding Modes (illustrate with $ rounding)
$1.40 $1.60 $1.50 $2.50 –$1.50
▪ Towards zero $1 ▪ Round down (−) $1 ▪ Round up (+) $2 ▪ Nearest Even* (default) $1
$1$1$2–$1 $1$1$2–$2 $2$2$3–$1 $2$2$2–$2
*Round to nearest, but if half-way in-between then round to nearest even
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Closer Look at Round-To-Even
 Default Rounding Mode
▪ Hard to get any other kind without dropping into assembly
▪ C99 has support for rounding mode management ▪ All others are statistically biased
▪ Sum of set 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 7.8950001 7.8950000 7.8850000
7.89 (Less than half way) 7.90 (Greater than half way) 7.90 (Half way—round up) 7.88 (Half way—round down)
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Rounding Binary Numbers
 Binary Fractional Numbers
▪ “Even” when least significant bit is 0
▪ “Half way” when bits to right of rounding position = 100…2
 Examples
▪ Round to nearest 1/4 (2 bits right of binary point)
Value 2 3/32 2 3/16 2 7/8 2 5/8
Binary 10.000112 10.001102 10.111002 10.101002
Rounded Action
10.002 (<1/2—down) 10.012 (>1/2—up) 11.002 ( 1/2—up) 10.102 ( 1/2—down)
Rounded Value 2
2 1/4
3
2 1/2
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Rounding
1.BBGRXXX
Sticky bit: OR of remaining bits
 Round up conditions
▪ Round = 1, Sticky = 1 ➙ > 0.5
▪Guard = 1, Round = 1, Sticky = 0 ➙ Round to even
Guard bit: LSB of result Round bit: 1st bit removed
Fraction
1.0000000 1.1010000 1.0001000 1.0011000 1.0001010 1.1111100
GRS Incr?
000 N 100 N 010 N 110 Y 011 Y 111 Y
Rounded
1.000
1.101
1.000
1.010
1.001
10.000
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FP Multiplication
 (–1)s1 M1 2E1 x (–1)s2 M2 2E2  Exact Result: (–1)s M 2E
▪Sign s: ▪SignificandM: ▪ Exponent E:
s1 ^ s2 M1x M2 E1 + E2
 Fixing
▪ If M ≥ 2, shift M right, increment E
▪ If E out of range, overflow
▪ Round M to fit frac precision
 Implementation
▪ Biggest chore is multiplying significands
4 bit significand: 1.010*22 x 1.110*23 = 10.0011*25 = 1.00011*26 = 1.001*26
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Floating Point Addition
 (–1)s1 M1 2E1 + (-1)s2 M2 2E2 ▪Assume E1 > E2
Get binary points lined up
 Exact Result: (–1)s M 2E ▪Sign s, significand M:
▪ Result of signed align & add ▪Exponent E: E1
(–1)s1 M1 +
E1–E2 (–1)s2 M2
 Fixing
▪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 (–1)s M 1.010*22 + 1.110*23 = (0.1010 + 1.1100)*23 = 10.0110 * 23 = 1.00110 * 24 = 1.010 * 24 Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition 40 Carnegie Mellon Mathematical Properties of FP Add  Compare to those of Abelian Group ▪ Closed under addition? Yes ▪ But may generate infinity or NaN ▪ Commutative? Yes ▪ Associative? No ▪ 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  Monotonicity ▪ a ≥ b ⇒ a+c ≥ b+c? ▪ Except for infinities & NaNs Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Yes Almost Almost 41 Carnegie Mellon Mathematical Properties of FP Mult  Compare to Commutative Ring ▪ Closed under multiplication? Yes ▪ But may generate infinity or NaN ▪ Multiplication Commutative? Yes ▪ Multiplication is Associative? No ▪ Possibility of overflow, inexactness of rounding ▪ Ex: (1e20*1e20)*1e-20= inf, 1e20*(1e20*1e-20)= 1e20 ▪ 1 is multiplicative identity? Yes ▪ Multiplication distributes over addition? No ▪ Possibility of overflow, inexactness of rounding ▪ 1e20*(1e20-1e20)= 0.0, 1e20*1e20 – 1e20*1e20 = NaN  Monotonicity ▪ a ≥ b & c ≥ 0 ⇒ a * c ≥ b *c? ▪ Except for infinities & NaNs Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition Almost 42 Carnegie Mellon Today: Floating Point  Background: Fractional binary numbers  IEEE floating point standard: Definition  Example and properties  Rounding, addition, multiplication  Floating point in C  Summary Bryant and O’Hallaron, Computer Systems: A Programmer’s Perspective, Third Edition 43 Carnegie Mellon 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 44 Carnegie Mellon Floating Point Puzzles  For each of the following C expressions, either: ▪ Argue that it is true for all argument values ▪ Explain why not true • x == (int)(float) x • x == (int)(double) x • f == (float)(double) f • d == (double)(float) d • f == -(-f); • 2/3 == 2/3.0 • d < 0.0 ⇒ • d > f ⇒
• d * d >= 0.0
• (d+f)-d == f
int x = …;
float f = …;
double d = …;
Assume neither dnorfis NaN
((d*2) < 0.0) -f > -d
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Summary
 IEEE Floating Point has clear mathematical properties
 Represents numbers of form M x 2E
 One can reason about operations independent of implementation
▪ As if computed with perfect precision and then rounded
 Not the same as real arithmetic
▪ Violates associativity/distributivity
▪ Makes life difficult for compilers & serious numerical applications programmers
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Additional Slides
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Creating Floating Point Number
 Steps
▪ 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 33 00010001 35 00010011
138 10001010
63 00111111
3-bits
s
exp
frac
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1 4-bits

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s
exp
frac
Normalize
 Requirement
▪ Set binary point so that numbers of form 1.xxxxx ▪ Adjust all to have leading one
▪ Decrement exponent as shift left
Value Binary
128 10000000 15 00001101 17 00010001 19 00010011
138 10001010 63 00111111
Fraction Exponent
1.0000000 7 1.1010000 3 1.0001000 4 1.0011000 4 1.0001010 7 1.1111100 5
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1 4-bits
3-bits

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

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Interesting Numbers
Description exp
 Zero 00…00
 Smallest Pos. Denorm. 00…00 ▪ Single ≈ 1.4 x 10–45
▪ Double ≈ 4.9 x 10–324
 Largest Denormalized 00…00 ▪ Single ≈ 1.18 x 10–38
▪ Double ≈ 2.2 x 10–308
 Smallest Pos. Normalized 00…01 ▪ Just larger than largest denormalized
{single,double} frac Numeric Value
00…00 0.0
00…01 2– {23,52} x 2– {126,1022}
11…11 (1.0 – ε) x 2– {126,1022} 00…00 1.0 x 2– {126,1022}
 One 01…11 00…00
 Largest Normalized 11…10 11…11 ▪ Single ≈ 3.4 x 1038
▪ Double ≈ 1.8 x 10308
1.0
(2.0 – ε) x 2{127,1023}
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