Single Precision Floating Point Format.ppt
- 1. 1 Floating Point Numbers Patt and Patel Ch. 2
- 2. 2 Floating Point Numbers Floating Point Numbers • Registers for real numbers usually contain 32 or 64 bits, allowing 232 or 264 numbers to be represented. • Which reals to represent? There are an infinite number between 2 adjacent integers. (or two reals!!) • Which bit patterns for reals selected? • Answer: use scientific notation
- 3. 3 A B A x 10B 0 any 0 1 .. 9 0 1 .. 9 1 .. 9 1 10 .. 90 1 .. 9 2 100 .. 900 1 .. 9 -1 0.1 .. 0.9 1 .. 9 -2 0.01 .. 0.09 Consider: A x 10B, where A is one digit How to do scientific notation in binary? Standard: IEEE 754 Floating-Point Floating Point Numbers
- 4. 4 IEEE 754 Single Precision Floating Point Format Representation: S E F •S is one bit representing the sign of the number •E is an 8 bit biased integer representing the exponent •F is an unsigned integer The true value represented is: (-1)S x f x 2e • S = sign bit • e = E – bias • f = F/2n + 1 • for single precision numbers n=23, bias=127
- 5. 5 S, E, F all represent fields within a representation. Each is just a bunch of bits. S is the sign bit • (-1)S (-1)0 = +1 and (-1)1 = -1 • Just a sign bit for signed magnitude E is the exponent field • The E field is a biased-127 representation. • True exponent is (E – bias) • The base (radix) is always 2 (implied). • Some early machines used radix 4 or 16 (IBM) IEEE 754 Single Precision Floating Point Format
- 6. 6 F (or M) is the fractional or mantissa field. • It is in a strange form. • There are 23 bits for F. • A normalized FP number always has a leading 1. • No need to store the one, just assume it. • This MSB is called the HIDDEN BIT. IEEE 754 Single Precision Floating Point Format
- 7. 7 How to convert 64.2 into IEEE SP 1. Get a binary representation for 64.2 • Binary of left of radix pointis: • Binary of right of radix .2 x 2 = 0.4 0 .4 x 2 = 0.8 0 .8 x 2 = 1.6 1 .6 x 2 = 1.2 1 • Binary for .2: • 64.2 is: 2. Normalize binary form • Produces:
- 8. 8 Floating Point •Since floating point numbers are always stored in normal form, how do we represent 0? •0x0000 0000 and 0x8000 0000 represent 0. •What numbers cannot be represented because of this? 3. Turn true exponent into bias-127 4. Put it together: 23-bit F is: S E F is: In hex:
- 9. 9 IEEE Floating Point Format Other special values: • + 5 / 0 = +∞ • +∞ = 0 11111111 00000… (0x7f80 0000) • -7/0 = -∞ • -∞ = 1 11111111 00000… (0xff80 0000) • 0/0 or + ∞ + -∞ = NaN (Not a number) • NaN ? 11111111 ?????… (S is either 0 or 1, E=0xff, and F is anything but all zeroes) • Also de-normalized numbers (beyond scope)
- 10. 10 IEEE Floating Point What is the decimal value for this SP FP number 0x4228 0000?
- 11. 11 IEEE Floating Point What is 47.62510 in SP FP format?
- 12. 12 What do floating-point numbers represent? • Rational numbers with non-repeating expansions in the given base within the specified exponent range. • They do not represent repeating rational or irrational numbers, or any number too small or too large. Floating Point Format
- 13. 13 IEEE Double Precision FP • IEEE Double Precision is similar to SP – 52-bit M • 53 bits of precision with hidden bit – 11-bit E, excess 1023, representing –1023 <- -> 2046 – One sign bit • Always use DP unless memory/file size is important – SP ~ 10-38 … 1038 – DP ~ 10-308 … 10308 • Be very careful of these ranges in numeric computation
- 14. 14 More Conversions • 113.910 = ??SP FP
- 15. 15 More … • -125.510 = ?SP FP
- 16. 16 And More Conversions • 0xC3066666
- 17. 17 And more … • 0xC3805000 =
- 18. 18 Floating Point Arithmetic Floating Point operations include •Addition •Subtraction •Multiplication •Division They are complicated because…
- 19. 19 Floating Point Addition 1. Align decimal points 2. Add 3. Normalize the result • Often already normalized • Otherwise move one digit 1.0001631 x 103 4. Round result 1.000 x 103 9.997 x 102 + 4.631 x 10-1 9.997 x 102 + 0.004631 x 102 10.001631 x 102 Decimal Review How do we do this?
- 20. 20 Floating Point Addition First step: get into SP FP if not already .25 = 0 01111101 00000000000000000000000 100 = 0 10000101 10010000000000000000000 Or with hidden bit .25 = 0 01111101 1 00000000000000000000000 100 = 0 10000101 1 10010000000000000000000 Example: 0.25 + 100 in SP FP Hidden Bit
- 21. 21 Second step: Align radix points – Shifting F left by 1 bit, decreasing e by 1 – Shifting F right by 1 bit, increasing e by 1 – Shift F right so least significant bits fall off – Which of the two numbers should we shift? Floating Point Addition
- 22. 22 Floating Point Addition Shift the .25 to increase its exponent so it matches that of 100. 0.25’s e: 01111101 – 1111111 (127) = 100’s e: 10000101 – 1111111 (127) = Shift .25 by 8 then. Easier method: Bias cancels with subtraction, so Second step: Align radix points cont. 10000101 - 01111101 00001000 100’s E 0.25’s E
- 23. 23 Carefully shifting the 0.25’s fraction S E HB F • 0 01111101 1 00000000000000000000000 (original value) • 0 01111110 0 10000000000000000000000 (shifted by 1) • 0 01111111 0 01000000000000000000000 (shifted by 2) • 0 10000000 0 00100000000000000000000 (shifted by 3) • 0 10000001 0 00010000000000000000000 (shifted by 4) • 0 10000010 0 00001000000000000000000 (shifted by 5) • 0 10000011 0 00000100000000000000000 (shifted by 6) • 0 10000100 0 00000010000000000000000 (shifted by 7) • 0 10000101 0 00000001000000000000000 (shifted by 8) Floating Point Addition
- 24. 24 Floating Point Addition Third Step: Add fractions with hidden bit 0 10000101 1 10010000000000000000000 (100) + 0 10000101 0 00000001000000000000000 (.25) 0 10000101 1 10010001000000000000000 Fourth Step: Normalize the result • Get a ‘1’ back in hidden bit • Already normalized most of the time • Remove hidden bit and finished
- 25. 25 Normalization example S E HB F 0 011 1 1100 + 0 011 1 1011 0 011 11 0111 Need to shift so that only a 1 in HB spot 0 100 1 1011 1 -> discarded Floating Point Addition
- 26. 26 Floating Point Subtraction •Mantissa’s are sign-magnitude •Watch out when the numbers are close 1.23455 x 102 - 1.23456 x 102 •A many-digit normalization is possible This is why FP addition is in many ways more difficult than FP multiplication
- 27. 27 Floating Point Subtraction 1. Align radix points 2. Perform sign-magnitude operand swap if needed • Compare magnitudes (with hidden bit) • Change sign bit if order of operands is changed. 3. Subtract 4. Normalize 5. Round Steps to do subtraction
- 28. 28 S E HB F 0 011 1 1011 smaller - 0 011 1 1101 bigger switch order and make result negative 0 011 1 1101 bigger - 0 011 1 1011 smaller 1 011 0 0010 1 000 1 0000 switched sign Floating Point Subtraction Simple Example:
- 29. 29 Floating Point Multiplication 1. Multiply mantissas 3.0 x 5.0 15.00 2. Add exponents 1 + 2 = 3 3. Combine 15.00 x 103 4. Normalize if needed 1.50 x 104 Decimal example: 3.0 x 101 x 5.0 x 102 How do we do this?
- 30. 30 Floating Point Multiplication Multiplication in binary (4-bit F) 0 10000100 0100 x 1 00111100 1100 Step 1: Multiply mantissas (put hidden bit back first!!) 1.0100 x 1.1100 00000 00000 10100 10100 + 10100 1000110000 10.00110000
- 31. 31 Floating Point Multiplication Second step: Add exponents, subtract extra bias. 10000100 + 00111100 Third step: Renormalize, correcting exponent 1 01000001 10.00110000 Becomes 1 01000010 1.000110000 Fourth step: Drop the hidden bit 1 01000010 000110000 11000000 11000000 - 01111111 (127) 01000001
- 32. 32 Multiply these SP FP numbers together 0x49FC0000 x 0x4BE00000 Floating Point Multiplication
- 33. 33 • True division – Unsigned, full-precision division on mantissas – This is much more costly (e.g. 4x) than mult. – Subtract exponents • Faster division – Newton’s method to find reciprocal – Multiply dividend by reciprocal of divisor – May not yield exact result without some work – Similar speed as multiplication Floating Point Division
- 34. 34 Floating Point Summary • Has 3 portions, S, E, F/M • Do conversion in parts • Arithmetic is signed magnitude • Subtraction could require many shifts for renormalization • Multiplication is easier since do not have to match exponents
- 35. 35 Questions?
- 36. 36