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
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