F L O A T IN G P O IN T N U M B E R S E n g la n d er C h . 5 ITEC 1011Introduction to Information Technologies

(1)

F L O A T IN G P O IN T N U M B E R S E n g la n d er C h . 5

ITEC 1011Introduction to Information Technologies

E x p o n en ti al N o ta ti o n

The representations differ in that the decimal place – the “point”--“floats”to the left or right (with the appropriate adjustment in the exponent).

• T h e fo ll o w in g a re e q u iv al en t re p re se n ta ti o n s o f 1 ,2 3 4

123,400.0x10-2 12,340.0x10-1 1,234.0x100 123.4x101 12.34x102 1.234x103 0.1234x104 ITEC 1011Introduction to Information Technologies

P ar ts o f a F lo at in g P o in t N u m b er - 0 . 9 8 7 6 x 1 0

-3 Sign of mantissaLocation of decimal pointMantissa

Exponent Sign of exponent Base

E x p o n en t E x ce ss 5 0 R ep re se n ta ti o n

•

W it h 2 d ig it s fo r th e ex p o n en t an d 5 f o r th e m a n ti ss a: f ro m . 0 0 0 0 1 x 1 0

-50

to . 9 9 9 9 9 x 1 0

49

(2)

O v er fl o w s / U n d er fl o w s

•

F ro m .0 0 0 0 1 x 1 0

-50

to . 9 9 9 9 9 x 1 0

49

1 x 1 0

-55

to . 9 9 9 9 9 x 1 0

49

T y p ic al F lo at in g P o in t F o rm at

EC 1011Introduction to Information Technologies

IE E E 7 5 4 S ta n d ar d

•Most common standard for representing floating point numbers •Single precision: 32 bits, consisting of... •Sign bit (1 bit) •Exponent (8 bits) •Mantissa (23 bits) •Double precision: 64 bits, consisting of… •Sign bit (1 bit) •Exponent (11 bits) •Mantissa (52 bits) ITEC 1011Introduction to Information Technologies

S in g le P re ci si o n F o rm at

32 bits Mantissa (23 bits) Exponent (8 bits) Sign of mantissa (1 bit)

(3)

D o u b le P re ci si o n F o rm at

64 bits Mantissa (52 bits) Exponent (11 bits) Sign of mantissa (1 bit) ITEC 1011Introduction to Information Technologies

N o rm al iz at io n • T h e m an ti ss a is

normalized

• H as a n i m p li ed d ec im a l p la ce o n l ef t • H as a n i m p li ed “ 1 ” o n l ef t o f th e d ec im al p la ce • E .g .,

•Mantissa: •Representation:

1010000000000000000000 1.1012=1.62510 ITEC 1011Introduction to Information Technologies

E x ce ss N o ta ti o n

•To include both positive and negative exponents, “excess-n”notation is used •Single precision: excess 127 •Double precision: excess 1023 •The value of the exponent stored is nlarger than the actual exponent •E.g., –excess 127, •Exponent: •Representation:

10000111 135–127=8(value) ITEC 1011Introduction to Information Technologies

E x ce ss N o ta ti o n

-Sample - Represent exponent of 1410in excess 127 form: 12710= + 011111112 1410= +000011102 Representation=100011012

(4)

EC 1011Introduction to Information Technologies

E x ce ss N o ta ti o n

-Sample - Represent exponent of -810in excess 127 form: 12710= + 011111112 -810= -000010002 Representation=011101112 ITEC 1011Introduction to Information Technologies

E x am p le • S in g le p re ci si o n

01000001011000000000000000000000 1.112= 1.7510 130 –127 = 3 0 = positive mantissa +1.75×23= 14.0 EC 1011Introduction to Information Technologies

E x er ci se – F lo at in g P o in t C o n v er si o n ( 1 ) • W h at d ec im al v al u e is r ep re se n te d b y t h e fo ll o w in g 3 2 -b it f lo at in g p o in t n u m b er ? • A n sw er :

11000001011110110000000000000000 Skip answerAnswer ITEC 1011Introduction to Information Technologies

E x er ci se – F lo at in g P o in t C o n v er si o n ( 1 ) • W h at d ec im al v al u e is r ep re se n te d b y t h e fo ll o w in g 3 2 -b it f lo at in g p o in t n u m b er ? • A n sw er : -1 5 .6 8 7 5

11000001011110110000000000000000

Answer

(5)

11000001011110110000000000000000 To decimal form 130 -127 = 31.11110110000000000000000000 1 + .5 + .25 + .125 + .0625 + 0 + .015625 + .0078125 1.960937523 *= 15.6875

- 1 5 .6 8 7 5

( negative )

S te p b y S te p S o lu ti o n

11000001011110110000000000000000 To decimal form 130 -127 = 31.11110110000000000000000000 1111.10110000000000000000000

- 1 5 .6 8 7 5

( negative )

Shift “Point”

Step by Step Solution : Alternative Method ITEC 1011Introduction to Information Technologies

E x er ci se – F lo at in g P o in t C o n v er si o n ( 2 ) • E x p re ss 3 .1 4 a s a 3 2 -b it f lo at in g p o in t n u m b er

•Answer: •(Note: only use 10 significant bits for the mantissa) Skip answerAnswer

E x er ci se – F lo at in g P o in t C o n v er si o n ( 2 ) • E x p re ss 3 .1 4 a s a 3 2 -b it f lo at in g p o in t n u m b er

•Answer:

Answer 01000000010010001111010111000010

(6)

3.14 To Binary:11. 0010001111010111000010 Delete implied left-most “1” and normalize10010001111010111000010 Exponent = 127 + 1 position point moved when normalized10000000 Value is positive: Sign bit = 0 0 1000000010010001111010111000010

Poof !

Detail Solution : 3.14 to IEEE Simple Precision

IE E E S in g le -P re ci si o n F lo at in g P o in t F o rm at

•

E x p o n en t b ia s is 1 2 7 f o r n o rm al iz ed # s

e e Value Type 255 nonenoneInfinity or NaN 254 127 (-1)s××××(1.f1f2...)××××2127Normalized ... ... ... ... 2 -125 (-1)s××××(1.f1f2...)××××2-125Normalized 1 -126 (-1)s××××(1.f1f2...)××××2-126Normalized 0 -126 (-1)s××××(0.f1f2...)××××2-126Denormalized

^

sê

f 1 f 2

. . . f23

signexponentfraction 189310

D ec im al F lo at in g -P o in t A d d a n d S u b tr ac t E x am p le s

OperandsAlignmentNormalize & round 6.144××××1020.06144××××1041.003644××××105 +9.975××××104+9.975 ××××104+ .0005 ××××105 10.03644××××1041.004 ××××105 OperandsAlignmentNormalize & round 1.076××××10-71.076 ××××10-77.7300 ××××10-9 -9.987××××10-8-0.9987××××10-7+ .0005 ××××10-9 0.0773××××10-77.730 ××××10-9

F lo at in g P o in t C al cu la ti o n s: A d d it io n

•Numbers must be aligned: have the same exponent (the larger one, to protect precision) •Add mantissas. If overflow, adjust the exponent •Ex. 0 51 99718 (e = 1) and 0 49 67000 (e = -1) •Align numbers:0 51 99718 0 51 00670 •Add them:99718 +00670 100388Overflow •Round the number and adjust exponent: 0 52 10039

(7)

F lo at in g P o in t C al cu la ti o n s: M u lt ip li ca ti o n

•(a * 10^e) * (b * 10^f) = a * b * 10^(e+f) •Rule: multiply mantissas; add exponents But:(n + e) + (n + f) = 2 * (n + e + f) Must subtract excess n from result •Ex. 0 51 99718 (e = 1) and 0 49 67000 (e = -1) Mantissas:.99718 * .67000 = 0.6681106 Exponents:51 + 49 = 100 and100 –50 = 50 Normalize:.6681106 .66811 Final result:.66811 * 100(since 50 meanse = 0)