1 / 17

Data Representation: Floating Point for Real Numbers

Data Representation: Floating Point for Real Numbers. Computer Organization and Assembly Language: Module 11. Floating Point Representation. The IEEE-754 Floating Point Standard is a widely used floating point representation from among the many alternative formats

levi
Download Presentation

Data Representation: Floating Point for Real Numbers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Data Representation: Floating Point for Real Numbers Computer Organization and Assembly Language: Module 11

  2. Floating Point Representation • The IEEE-754 Floating Point Standard is a widely used floating point representation from among the many alternative formats • The representation of floating point numbers contains: • a mantissa (variant of a scaled, sign magnitude integer) • an exponent (8-bit, biased-127 integer) • In this way floating point representation resembles scientific notation • Any number N can be represented as M*10^e, where • e = floor(log10N) • M = N/(10^e) • 1 < M < 10

  3. Floating Point Representation • A number N represented in floating point is determined by the mantissam, an exponente, and its sign, s N = (-1)s * m * 2e • If the sign is negative, s = 1. If the sign is positive, s = 0. • The mantissa is normalized, i.e., 1 £m <2 • In the IEEE-754 single precision format, the mantissa is represented with 23 bits (only the fractional part is stored • m = (+/-) 1.f22f21…f1f0 • Double precision floating point works the same way, but the bit fields are larger: 1-bit sign, 11-bit exponent, 52 bits for the fractional part of the mantissa

  4. Conversion to base-2 1. Break the decimal number into two parts: an integer and a fraction 2. Convert the integer into binary and place it to the left of the binary point 3. Convert the fraction into binary and place it to the right of the binary point 4. Write it in base-2 scientific notation and normalize

  5. .62510=.1012 Example Convert 22.625 to floating point representation 1. Convert 22 to binary. 2210 = 101102 2. Convert .625 to binary 2*.625 = 1 + .25 2*.25 = 0 + .5 2*.5 = 1 + 0 3. Thus 22.62510 = 10110.1012 4. In base –2 scientific notation: 10110.101*20 Normalized form: 1.0110101*24

  6. S E F IEEE-754 SPFP Representation • Given the floating point representation N = (-1)s * m * 2ewhere m = 1.f22f21…f1f0 • we can convert it to the IEEE-754 SPFP format using the relations: F = (m-1)*223 (hence F is an integer) E = e + 127 S = s

  7. 1 8 23 S E F Single-Precision Floating Point The IEEE-754 single precision format has 32 bits distributed as • 0£E£255, thus the actual exponent e (interpreted as biased-127) is restricted so that -127£ e£ 128 • But e = -127 and e = 128 have special meaning

  8. Special values and the hidden bit • In IEEE-754 , zero is represented by setting E = F = 0 regardless of the sign bit, thus there are two representations for zero: +0 and -0. • +¥ by S=0, E=255, F=0 • -¥ by S=1, E=255, F=0 • NaNorNot-a-NumberbyE=255, F¹0(may result from 0 divided by 0) • The leading 1 in the fraction is not represented. It is the hidden bit.

  9. Converting to IEEE-754 SPFP 1.Convert into a normalized base-2 representation 2.Bias the exponent. The result will be E. 3.Put the values into the correct field. Note that only the fractional part of the mantissa is stored in F.

  10. S E 0 10000011 F Example Convert 22.625 to IEEE-754 SPFP format 1. In scientific notation: 10110.101*20 Normalized form: 1.0110101*24 2. Bias the exponent: 4 + 127 = 131 13110 = 100000112 3. Place into the correct fields. S = 0 E = 10000011 F = 011 0101 0000 0000 0000 0000 01101010000000000000000

  11. 0 10000011 00010010011001100110011 S E F Example Convert 17.15 to IEEE FPS format 17.1510 = 10001.0010 0110 0110 0110 011*20 1. Normalized form: 1. 0001 0010 0110 0110 0110 011 * 24 2. Bias the exponent: 4 + 127 = 131 13110 = 100000112 3. Place into the correct fields. S = 0 E = 10000011 F = 000 1001 0011 0011 0011 0011

  12. -83.710=-1010011.101100110 Example Convert -83.7 to IEEE FPS format (single precision) 2*.7 = 1 + .4 2*.4 = 0 + .8 2*.8 = 1 + .6 2*.6 = 1 + .2 2*.2 = 0 + .4 2*.4 = 0 + .8 2*.8 = 1 + .6 2*.6 = 1 + .2 2*.2 = 0 + .4 . . .

  13. 1 10000101 01001110110011001100110 S E F 1.In binary scientific notation: -1010011.10110011001100110 * 20 Normalized: -1.01001110110011001100110 * 26 2. Bias the exponent: 6 + 127 = 133 13310 = 100001012 3. Place into the correct fields S = 1 E = 10000101 F = 01001110110011001100110

  14. Representing as hexadecimal • It is difficult for people to read binary • one bit pattern looks much like another • Raw data, which is not being interpreted as representing a particular data type, is often displayed using hexadecimal instead of binary • The final step in many IEEE-754 SPFP problems will be to convert the result to hexadecimal • 11000010101001110110011001100110 • C2A76666

  15. Graceful underflow • Given a single precision floating point number with bit fields S, E, and F (interpreted as unsigned integers), the value of the number is normally calculated as N = (-1)S(1 + F/223)2E-127 • This interpretation is not used when • E = 255 (+, -  , or NaN) • E = 0, F = 0 (+0 or –0) • What about E  0, F  0?

  16. Graceful underflow • Given a single precision floating point number with bit fields S, E = 0, and F (interpreted as unsigned integers), the value of the number is calculated as N = (-1)S(0 + F/223)2-126 • This allows representation of numbers as small as 2-149, though each order of magnitude below 2-126 results in loss of one bit of precision.

  17. 0 00000001 00000000000000000000000 0 00000000 10000000000000000000000 0 00000000 00010000000000000000000 Graceful underflow • Normal interpretation: N = 2(1 – 127) = 2-126 • 24 bits of precision (counting the hidden bit) • E = 0 interpretation: N = 2-126 (.12) = 2-126 (.5) = 2-127 • Only 23 bits of precision • E =0 interpretation: N = 2-126 (.00012) = 2-126(.0625) = 2-130 • Only 20 bits of precision

More Related