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Bi 03: S chm ng

Phm Tun Sn

ptson@fit.hcmus.edu.vn

mailto:ptson@fit.hcmus.edu.vnmailto:ptson@fit.hcmus.edu.vn

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Vn vi biu din s nguyn

S nguyn N bit biu din c 2N gi tr Biu din khng du (Unsigned Integer)

0 2N 1

(N=32, 2N1 = 4,294,967,295)

Biu din b 2-2(N-1) 2(N-1) 1

(N=32, 2(N-1) = 2,147,483,648)

Biu din s rt ln ? S giy / 1 nghn nm

31,556,926,000 (3.1556926 x 1010) Biu din s rt nh ? S giy / 1 nano giy

0.00000000110 (1.010 x 10-9)

Biu din s thp phn 1.5 ?2

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Biu din phn thp phn

Biu din s 5.375 th no ?Cn bao nhiu bit ?

Gi s dng 8 bit lu tr phnnguyn

5 = 4 + 1 = 00000101 Tng t c th dng 8 bit lu tr

phn thp phn0.375 = 0.25 + 0.125 = 01100000

Vy c th biu din5.375 = 00000101.01100000

Tng qut ta c:

=> Biu din s chm tnh (fixed point) 3

0 1.0 1

1 0.5 1/2

2 0.25 1/4

3 0.125 1/8

4 0.0625 1/165 0.03125 1/32

6 0.015625 ...

7 0.0078125

8 0.00390625

9 0.001953125

10 0.000976562511 0.00048828125

12 0.000244140625

13 0.0001220703125

14 0.00006103515625

15 0.000030517578125

i 2-i

1 1 0 1 2. 2

n

i

n m i

i m

x x x x x x x =

= K K

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S chm ng tng

H thp phn 123000000000 ~ 1.231011 v 0.0000000000123 ~ 1.2310-11

Tng t vi h nh phn, ta cx = 00000101.01100000 = 22 + 20 + 2-2 + 2-3

Ta c th vit lix = 1.01011 22

Thay v dng 16 bit lu tr, chcn dng 7 bit (5 bitphn tr + 2 bit phn m)

x = 1.01011 10

Nh vy, Mun tit kim s bit lu tr, ta di chuyn vtr ca du chm

sang phi 14 v tr

Cn lu: phn tr, phn m v phn du

=>y l tng c bn ca s chm ng (floating point)5

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Biu din s chm ng

Biu din s chm ng

Sign (S): phn du

Exponent (E): phn s m

Significand (S): phn nh tr

Gi tr S2E

6

S Exponent Significand

1 bit m bits n bits

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Biu din khoa hc

Gi tr 1 / 1,000,000,000 c th biu din nh sau: 1.010 10-9 Dng chun (Normalized form)

0.110 10-8, 10.010 10-10 Dng khng chun(Denormalized form)

Dng chun: phn nguyn gm 1 ch s khc 0

7

6.0210 x 1023

radix (base)

decimal point

mantissa exponent1.02 x 2

-1

binary point

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Chun s chm ng IEEE 754

Biu din s chm ng Single Precision (32 bit)

S: du (Sign) 0: dng, 1: m

Exponent: phn s m (lu di dng s biased)

Significand: phn nh tr Ngm nh bt u l 1 + phn tr ~ (1 + 23) bits

Dng chun: +/-1.xxxx22yyyy

2 V d:

Biu din: 0 10000001 01011000000000000000000

C gi tr: +1.010110000210000001 ~ +(1+2-2 + 2-4 + 2-5) 22 =5.375

8

031S Exponent30 23 22

Significand

1 bit 8 bits 23 bits

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Chuyn t biu din nh phnsang thp phn

9

Du: 0 dng

M: 0110 1000 c gi tr (dng biased) l

104 - 127 = -23

Tr:1 + 1x2-1+ 0x2-2 + 1x2-3 + 0x2-4 + 1x2-5 +...

=1+2-1+2-3 +2-5 +2-7 +2-9 +2-14 +2-15 +2-17 +2-22= 1.0 + 0.666115

Kt qu: 1.6661152-23 ~ 1.98610-7

(~ 2/10,000,000)

0 0110 1000 101 0101 0100 0011 0100 0010

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Chuyn t biu din thp phnsang nh phn

-2.340625 x 101

1. Khng chun ha: -23.40625

2. Chuyn phn nguyn:

23 = 16 + 4 + 2 + 1 = 101113. Chuyn phn thp phn:

.40625 = .25 + .125 + .03125 = .01101

4. Kt hp v chun ha:

10111.01101 = 1.011101101 x 24

5. Chuyn phn m: 127 + 4 = 10000011

10

1 1000 0011 011 1011 0100 0000 0000 0000

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Chuyn t biu din thp phnsang nh phn (tt)

1/3= 0.33333

= 0.25 + 0.0625 + 0.015625 + 0.00390625 +

= 1/4 + 1/16 + 1/64 + 1/256 +

= 2-2 + 2-4 + 2-6 + 2-8 + = 0.0101010101 * 20

= 1.0101010101 * 2-2

Du: 0

M

= -2 + 127 = 125 = 01111101 Tr = 0101010101

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0 0111 1101 0101 0101 0101 0101 0101 010

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Cc sc bit

Phn m = 0, phn tr = 0 S zero

Phn m = 0, phn tr 0

S dng khng chun (denormalized) Phn m ton bit 1, phn tr = 0

S v cng (infinity)

Phn m ton bit 1, phn tr 0 S bo li (NaN - Not a Number)

Signaling NaN

Quiet NaN12

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Phn b, phm vi biu din

Phn b

14

Phm vi biu din. Chng minh ?

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Phn b

t f(1,2) = s lng s chm ng trongkhong 1 v 2

t f(2,3) = s lng s chm ng trongkhong 2 v 3

Hi1. f(1,2) < f(2,3)

2. f(1,2) = f(2,3)

3. f(1,2) > f(2,3)

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S dng khng chun

S dng nh nht c th biu dina = 1.0 2 2-126 = 2-126

L do: ngm nh 1 + phn tr

Gii php: Qui c nu s m = 0 (phn tr 0), khng ngm nh

bt u l 1 S dng khng chun (denormalized) S dng nh nht c th biu din

a = 0.0012 2-126 = 2-23 2-126 = 2-149

16

a0 +-

Gaps!

0

+-

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Mt s loi chm ng

Single Precision (32 bit) 1/8/23 (kiu floattrong C), 10-38 1038

Double Precision (64 bit) 1/11/52 (kiu doubletrong C), 10-308 10308

Half Precision (16 bit) 1/5/10

Quad Precision (8 bit)

1/4/3 IEEE 754-2008 binary128 (128 bit)

1/15/112

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en.wikipedia.org/wiki/Floating_point

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Biu din s chm ng 8 bit

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Single Precision (32 bit) Double Precision (64 bit)Du M Tr Gi t r Du M Tr Gi t r

+0 0 0 0 0 0 0 0 0

0 1 0 0 -0 1 0 0 -0

+ 0 255 (ton bit 1) 0 0 2047 (ton bit 1) 0

1 255 (ton bit 1) 0 - 1 2047 (ton bit 1) 0 -

Quiet NaN 0/ 1 255 (ton bit 1) 0 NaN 0/ 1 2047 (ton bit 1) 0 NaN

Signaling NaN 0/ 1 255 (ton bit 1) 0 NaN 0/ 1 2047 (ton bit 1) 0 NaN

Sdng(dng chun)

0 0

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Khi nim Precision v Accuracy

Precision: s bit c s dng trong my tnh biu din 1 gi tr.

Accuracy: chnh xc m mt kiu biu dintrong my tnh c th biu din c mt gi tr.

Thng th precision cao s dn ti accuracycao.

V d: float pi = 3.14;

pi c biu din bi 24 bit phn tr (precise cao),nhng chc th biu din c gn ng pi (khngaccuracy).

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Lm trn (Rounding)

Khi thc hin cc php ton trn s chm ng, kt qunhn c c th vt ra ngoi kh nng biu din caphn nh tr.

Phn cng phc v cc php ton trn s chm ng

thng c thm 2 bit nhh trcho phn nh tr gipthc hin vic lm trn c c kt qu chnh xcnht c th.

V d: thc hin (1.000021) (1.111120)

21

1.00.00 21

- 0.111.1 21 (= 1.11.1120)0.000..01 21

=1.0000 2-22

1.00.00 00 21

- 0.111.1 10 21 (= 1.11.1120)0.000..00 10 21

=1.0000 00 2-23

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Chun IEEE lm trns chm ng

Lm trn ln (Round up / Round towards +)1.01 10 1.10 , -1.01 10 -1.01

Lm trn xung (Round down / Round towards )1.01 10 1.01, -1.01 10 -1.10

Lm trn v 0 (Truncate / Round towards 0) B gi tr 2 bit nh

Lm trn v gi tr gn nht (Round to nearest): 1.01 01 1.01 , -1.01 11 -1.10

Trng hp 2 bit nh l 10 (halfway) ? Lm trn v s chn gn nht (mc nh), ngha l LSB ca phnnh tr lun bng 01.01 10 1.10 , -1.10 11 -1.10

22

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Cc trng hp lm trn khc

Lm trn cng c thc hin khi thc hin chuyn i: Chuyn i t kiu double precision thnh single precision

Chuyn i t s chm ng thnh s nguyn v ngc li

p kiu t s chm ng thnh s nguyn v ngc li

Hy kho st cc trng hp sau:

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1.Chuyn i float -> int -> float.Kt qu nh ban u ?2.Chuyn i int -> float -> int. Kt qu nh ban u ?3.Php cng s chm ng c tnh kt hp ?

(x+y)+z = x+(y+z)

4.i = (int) (3.14159 * f);

5.f = f + (float) i;

6.if (i == (int)((float) i)) { printf(true); }

7.if (i == (int)((double) i)) { printf(true); }

8.if (f == (float)((int) f)) { printf(true); }

9.if (f == (double)((int) f)) { printf(true); }

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Php cng, tr s chm ng

24

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Php nhn s chm ng

25

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Php chia s chm ng

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Tham kho

Chng 3, P&H Chng 9, William Stallings

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