12/11/2005 CSE 378 Floating-point 1How to represent real numbers• In decimal scientific notation: 5.280 x 103– sign– fraction– base (i.e., 10) to some power• Most of the time, usual representation 1 digit at left of decimal point– Example: - 0.1234 x 106• A number is normalized if the leading digit is not 0– Example: -1.234 x 10512/11/2005 CSE 378 Floating-point 2Real numbers representation inside computer• Use a representation akin to scientific notationsign x mantissa x baseexponent• Many variations in choice of representation for– mantissa (could be 2’s complement, sign and magnitude etc.)– base (could be 2, 8, 16 etc.)– exponent (cf. mantissa)• Arithmetic support for real numbers is called floating-point arithmetic12/11/2005 CSE 378 Floating-point 3Floating-point representation: IEEE Standard• Basic choices– A single precision number must fit into 1 word (4 bytes, 32 bits)– A double precision number must fit into 2 words– The base for the exponent is 2– There should be approximately as many positive and negative exponents• Additional criteria– The mantissa will be represented in sign and magnitude form– Numbers will be normalized12/11/2005 CSE 378 Floating-point 4Example: MIPS representation of IEEE Standard• A number is represented as : (-1)S. F .2E• In single precision the representation is:s exponent mantissa31 2322 08 bits 23 bits12/11/2005 CSE 378 Floating-point 5MIPS representation (continued)• Bit 31 sign bit for mantissa (0 pos, 1 neg)• Exponent 8 bits (“biased” exponent, see next slide)• mantissa 23 bits : always a fraction with an implied binary point at left of bit 22• Number is normalized (see implication next slides)• 0 is represented by all zero’s.• Note that having the most significant bit as sign bit makes it easier to test for positive and negative12/11/2005 CSE 378 Floating-point 6Biased exponent• The “middle” exp. (01111111) will represent exponent 0, i.e. 127• All exps starting with a “1” will be positive exponents .– Example: 10000001 is exponent 2 (10000001 -01111111)• All exps starting with a “0” will be negative exponents– Example 01111110 is exponent -1 (01111110 - 01111111)• The largest positive exponent will be 11111111, about 1038• The smallest negative exponent is about 10-3812/11/2005 CSE 378 Floating-point 7Normalization• Since numbers must be normalized, there is an implicit “one” at the left of the binary point• No need to put it in (improves precision by 1 bit)• But need to reinstate it when performing operations.• In summary, in MIPS a floating-point number has the value:(-1)S .(1 + mantissa) .2 (exponent - 127)12/11/2005 CSE 378 Floating-point 8Double precision• Takes 2 words (64 bits)• Exponent 11 bits (instead of 8)• Mantissa 52 bits (instead of 23)• Still biased exponent and normalized numbers• Still 0 is represented by all zeros• We can still have overflow (the exponent cannot handle super big numbers) and underflow (the exponent cannot handle super small numbers)12/11/2005 CSE 378 Floating-point 9Floating-Point Addition• Quite “complex” (more complex than multiplication)• Need to know which of the addends is larger (compare exponents)• Need to shift “smaller” mantissa• Need to know if mantissas have to be added or subtracted (since it’s a sign/magnitude representation)• Need to normalize the result• Correct round-off procedures are not simple (not covered in detail here)12/11/2005 CSE 378 Floating-point 10One of the 4 round-off modes• Round to nearest even– Example 1: in base 10. Assume 2 digit accuracy.3.1 *100 + 4.6 * 10 -2 = 3.146 * 100 clearly should be rounded to 3.1 * 100 – Example 2: 3.1 *100 + 5.0* 10 -2 = 3.15 * 100By convention, round-off to nearest “even” number 3.2 * 100• Other round-off modes: towards 0, +∞, -∞12/11/2005 CSE 378 Floating-point 11F-P add (details for round-off omitted)1. Compare exponents . If e1 < e2, swap the 2 operands such thatd = e1 - e2 >= 0. Tentatively set exponent of result to e1.2. Insert 1’s at left of mantissas. If the signs of operands differ, replace 2nd mantissa by its 2’s complement. 3. Shift 2nd mantissa d bits to the right (this is an arithmetic shift, i.e., insert either 1’s or 0’s depending on the sign of the second operand)4. Add the (shifted) mantissas. (There is one case where the result could be negative and you have to take the 2’s complement; this can happen only when d = 0 and the signs of the operands are different.)12/11/2005 CSE 378 Floating-point 12F-P Add (continued)5. Normalize (if there was a carry-out in step 4, shift right once; else shift left until the first “1” appears on msb)6. Modify exponent to reflect the number of bits shifted in previous step12/11/2005 CSE 378 Floating-point 13Example• Add decimal: 0.375 + 0.75• 3/23+ 3/22= 0.011 + 0.11 = 1.1 x 2-2+ 1.1 x 2-1• Now add:• Align fractions: 0.11 x 2-1+ 1.1 x 2-1• Add fractions: 10.01 x 2-1• Normalize: 10.01 x 2-1= 1.001 x 20• Round: Not needed• 1.001 x 20= 1 + 1/23= 1 + 1/8 = 1.125 decimal• In IEEE single precision 0.75 i.e. 0.11 is0 0111 1110 100 0000 0000 0000 0000 0000 why?12/11/2005 CSE 378 Floating-point 14Using pipelining• Stage 1– Exponent compare• Stage 2– Shift and Add• Stage 3– Round-off , normalize and fix exponent• Most of the time, done in 2 stages.12/11/2005 CSE 378 Floating-point 15Floating-point multiplication• Conceptually easier1. Add exponents (careful, subtract one “bias”)2. Multiply mantissas (don’t have to worry about signs)3. Normalize and round-off and get the correct sign12/11/2005 CSE 378 Floating-point 16Pipelining• Use tree of “carry-save adders” (cf. CSE 370) Can cut-it off in several stages depending on hardware available• Have a “regular” adder in the last stage.12/11/2005 CSE 378 Floating-point 17Special Values• Allow computation to continue in face of exceptional conditions – For example: divide by 0, overflow, underflow• Special value: NaN (Not a Number; e.g., sqrt(-1))– Operations such as 1 + NaN yield NaN• Special values: +∞ and -∞ (e.g, 1/0 is +∞)• Can also use “denormal” numbers for underflow and overflow allowing a wider range of
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