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CMU CS 15213 - Lecture

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Floating Point ArithmeticSeptember 28, 2000Topics• IEEE Floating Point Standard• Rounding• Floating Point Operations• Mathematical properties• IA32 floating pointclass10.ppt15-213“The course that gives CMU its Zip!”CS 213 F’00– 2 –class10.pptFloating Point Puzzles• For each of the following C expressions, either:– Argue that is true for all argument values– Explain why not true• x == (int)(float) x• x == (int)(double) x• f == (float)(double) f• d == (float) d• f == -(-f);• 2/3 == 2/3.0• d < 0.0 ⇒ ((d*2) < 0.0)• d > f ⇒ -f < -d• d * d >= 0.0• (d+f)-d == fint x = …;float f = …;double d = …;Assume neitherd nor f is NANCS 213 F’00– 3 –class10.pptIEEE Floating PointIEEE Standard 754• Estabilished in 1985 as uniform standard for floating point arithmetic–Before that, many idiosyncratic formats• Supported by all major CPUsDriven by Numerical Concerns• Nice standards for rounding, overflow, underflow• Hard to make go fast–Numerical analysts predominated over hardware types in definingstandardCS 213 F’00– 4 –class10.pptFractional Binary NumbersRepresentation• Bits to right of “binary point” represent fractional powers of 2• Represents rational number:bibi–1b2b1b0b–1b–2b–3b–j• • •• • • .1242i–12i• • •• • •1/21/41/82–jbkkk ji⋅=−∑2CS 213 F’00– 5 –class10.pptFractional Binary Number ExamplesValue Representation5-3/4 101.1122-7/8 10.111263/64 0.1111112Observation• Divide by 2 by shifting right• Numbers of form 0.111111…2 just below 1.0–Use notation 1.0 – εLimitation• Can only exactly represent numbers of the form x/2k• Other numbers have repeating bit representationsValue Representation1/3 0.0101010101[01]…21/5 0.001100110011[0011]…21/10 0.0001100110011[0011]…2CS 213 F’00– 6 –class10.pptNumerical Form• –1s M 2E–Sign bit s determines whether number is negative or positive–Significand M normally a fractional value in range [1.0,2.0).–Exponent E weights value by power of twoEncoding• MSB is sign bit• exp field encodes E• frac field encodes MSizes• Single precision: 8 exp bits, 23 frac bits–32 bits total• Double precision: 11 exp bits, 52 frac bits–64 bits totalFloating Point Representations exp fracCS 213 F’00– 7 –class10.ppt“Normalized” Numeric ValuesCondition• exp ≠ 000…0 and exp ≠ 111…1Exponent coded as biased value E = Exp – Bias–Exp : unsigned value denoted by exp–Bias : Bias value»Single precision: 127 (Exp: 1…254, E: -126…127)»Double precision: 1023 (Exp: 1…2046, E: -1022…1023»in general: Bias = 2m-1 - 1, where m is the number of exponent bitsSignificand coded with implied leading 1 m = 1.xxx…x2– xxx…x: bits of frac–Minimum when 000…0 (M = 1.0)–Maximum when 111…1 (M = 2.0 – ε)–Get extra leading bit for “free”CS 213 F’00– 8 –class10.pptNormalized Encoding ExampleValueFloat F = 15213.0;• 1521310 = 111011011011012 = 1.11011011011012 X 213SignificandM = 1.11011011011012frac = 110110110110100000000002ExponentE = 13Bias = 127Exp = 140 = 100011002Floating Point Representation (Class 02):Hex: 4 6 6 D B 4 0 0Binary: 0100 0110 0110 1101 1011 0100 0000 0000140: 100 0110 015213: 1110 1101 1011 01CS 213 F’00– 9 –class10.pptDenormalized ValuesCondition• exp = 000…0Value• Exponent value E = –Bias + 1• Significand value m = 0.xxx…x2–xxx…x: bits of fracCases• exp = 000…0, frac = 000…0–Represents value 0–Note that have distinct values +0 and –0• exp = 000…0, frac ≠ 000…0–Numbers very close to 0.0–Lose precision as get smaller–“Gradual underflow”CS 213 F’00– 10 –class10.pptInteresting NumbersDescription exp frac Numeric ValueZero 00…00 00…00 0.0Smallest Pos. Denorm. 00…00 00…01 2– {23,52} X 2– {126,1022}• Single ≈ 1.4 X 10–45• Double ≈ 4.9 X 10–324Largest Denormalized 00…00 11…11 (1.0 – ε) X 2– {126,1022}• Single ≈ 1.18 X 10–38• Double ≈ 2.2 X 10–308Smallest Pos. Normalized 00…01 00…00 1.0 X 2– {126,1022}• Just larger than largest denormalizedOne 01…11 00…00 1.0 Largest Normalized 11…10 11…11 (2.0 – ε) X 2{127,1023}• Single ≈ 3.4 X 1038• Double ≈ 1.8 X 10308CS 213 F’00– 11 –class10.pptSpecial ValuesCondition• exp = 111…1Cases• exp = 111…1, frac = 000…0–Represents value ∞ (infinity)–Operation that overflows–Both positive and negative–E.g., 1.0/0.0 = −1.0/−0.0 = +∞, 1.0/−0.0 = −∞• exp = 111…1, frac ≠ 000…0–Not-a-Number (NaN)–Represents case when no numeric value can be determined–E.g., sqrt(–1), ∞ − ∞CS 213 F’00– 12 –class10.pptSummary of Floating Point Real Number EncodingsNaNNaN+∞-∞-0 +0+Denorm +Normalized-Denorm-NormalizedCS 213 F’00– 13 –class10.pptTiny floating point example8-bit Floating Point Representation• the sign bit is in the most significant bit.• the next four bits are the exponent, with a bias of 7.• the last three bits are the frac• Same General Form as IEEE Format• normalized, denormalized• representation of 0, NaN, infinity)sexp frac02367CS 213 F’00– 14 –class10.pptValues related to the exponentExp exp E 2E0 0000 -6 1/64 (denorms)1 0001 -6 1/642 0010 -5 1/323 0011 -4 1/164 0100 -3 1/85 0101 -2 1/46 0110 -1 1/27 0111 0 18 1000 +1 29 1001 +2 410 1010 +3 811 1011 +4 1612 1100 +5 3213 1101 +6 6414 1110 +7 12815 1111 n/a (inf, Nan)CS 213 F’00– 15 –class10.pptDynamic Ranges exp frac E Value0 0000 000 -6 00 0000 001 -6 1/8*1/64 = 1/5120 0000 010 -6 2/8*1/64 = 2/512…0 0000 110 -6 6/8*1/64 = 6/5120 0000 111 -6 7/8*1/64 = 7/5120 0001 000 -6 8/8*1/64 = 8/5120 0001 001 -6 9/8*9/64 = 9/512…0 0110 110 -1 14/8*1/2 = 14/160 0110 111 -1 15/8*1/2 = 15/160 0111 000 0 8/8*1 = 10 0111 001 0 9/8*1 = 9/80 0111 010 0 10/8*1 = 10/8…0 1110 110 7 14/8*128 = 2240 1110 111 7 15/8*128 = 2400 1111 000 n/a infclosest to zerolargest denormsmallest normclosest to 1 belowclosest to 1 abovelargest normDenormalizednumbersNormalizednumbersCS 213 F’00– 16 –class10.pptSpecial Properties of EncodingFP Zero Same as Integer Zero• All bits = 0Can (Almost) Use Unsigned Integer Comparison• Must first compare sign bits• Must consider -0 = 0• NaNs problematic–Will be greater than any other values–What should comparison yield?• Otherwise OK–Denorm vs.


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CMU CS 15213 - Lecture

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