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FLOATING-POINT UNIT
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so that rounding(s) could be equal to B. A new normalization step would be necessary, that is, substitution of s B by s 1 and e by e 1.
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Algorithm 16.1
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Sum of Positive Numbers
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if e1>=e2 then e:=e1; s:=s1+(s2/B*(e1-e2)); else e:=e2; s:=(s1/B*(e2-e1))+s2; end if; if s>=B then e:=e+1; s:=s/B; end if; s:=round(s); if s>=B then e:=e+1; s:=s/B; end if;
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Examples 16.2 Assume that B 10 and ulp 1024, so that the numbers are represented in the form s.10e where 1 s 9.9999. 1. Compute z (3.4375 103) (2.5491 1021): alignment: z (3:4375 0:00025491) 103 3:43775491 103 , 3:43775491 , 10, rounding: s 3:4378, 3:4378 , 10, z 3:4378 103 2. Compute z (9.4375 103) (8.6247 102): alignment: z (9:4375 0:86247) 103 10:29997 103 , normalization: s 1:029997, e 4, rounding: s 1:0300, 1:0300 , 10, z 1:0300 104 : 3. Compute z (9.4375 103) (5.6247 102): alignment: z (9:4375 0:56247) 103 9:99997 103 , 9:99997 , 10, rounding: s 10:0000, normalization: s 1:0000, e 4, z 1:0000 104 :
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Comment 16.1 The addition of two positive numbers could produce an over ow, as the nal value of e could be greater than emax.
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16.2 ARITHMETIC OPERATIONS
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Difference of Positive Numbers
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Given two positive oating-point numbers s1.Be1 and s2.Be2 their difference s.Be is computed as follows: Assume that e1 is greater than or equal to e2; then (alignment) the difference between s1.Be1 and s2.Be2 can be expressed in the form s.Be, where s s1 s2 =(Be1 e2 ) and e e1 : The value of s belongs to the interval (B ulp) s B ulp: (16:13) (16:12)
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If s is negative, then it is substituted by 2s and the sign of the nal result will be modi ed accordingly. If s is equal to 0, an exception equal_zero could be raised. It remains to consider the case where 0,s B ulp:
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The value of s could be smaller than 1. In order to normalize the signi cand, a procedure
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procedure leading_zeroes(s: in xed_point; k: out natural)
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must be executed: it counts the number of initial 00 s of the representation of s. In other words, it looks for the minimum exponent k such that s.Bk ! 1. Then s is substituted by s.Bk and e by e 2 k. Thus, the relation (16.10) holds, that is, 1 s , B:
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It remains to round (up or down) the signi cand and to normalize it if necessary. Algorithm 16.2 Difference of Positive Numbers
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if e1>=e2 then e:=e1; s:=s1-(s2/B**(e1-e2)); else e:=e2; s:=(s1/B**(e2-e1))-s2; end if; if s<0 then s:=-s; sign:=1; end if; leading_zeroes(s, k); s:=s*(B**k); e:=e-k; s:=round(s); if s>=B then e:=e+1; s:=s/B; end if;
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Examples 16.3 Assume again that B 10 and ulp 1024, so that the numbers are represented in the form s.10e where 1 s 9.9999. For computing the difference, the 10 s complement system is used.
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FLOATING-POINT UNIT
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1. Compute z (3.4518 1021) 2 (7.2471 103): alignment: z (0:00034518 7:2471) 103 (00:00034518 92:75289999 1) 103 92:75324518 103 , change the sign: 7:24675482 ! 1, s 07:24675481 1 7:24675482,
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rounding: s 7:2468, 7:2468 , 10 z 7:2468 103 : 2. Compute z (1.0014 103) 2 (9.9491 102): alignment: z (1:0014 0:99491) 103 (01:0014 99:00508 1) 103 00:00649 103 , 00:00649 , 0, leading zeroes: s 6:4900, e 0, rounding: s 6:4900, 6:4900 , 10, z 6:4900 100 : 3. Compute z (1.0714 104) 2 (7.1403 102): alignment: z (1:0714 0:071403) 104 (01:0714 99:928596 1) 104 00:999997 104 , 00:999997 . 0, leading zeroes: s 9:99997, rounding: s 10:0000, normalization:
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e 3, e 4,
s 1:0000,
z 1:0000 10 : Comment 16.2 The difference of two positive numbers could produce an under ow, as the nal value of e could be smaller than emin. 16.2.3 Addition and Subtraction
Given two oating-point numbers (21)sign1.s1.Be1 and (21)sign2.s2.Be2, and a control variable operation, an algorithm is de ned for computing z ( 1)sign :s:Be ( 1)sign1 :s1 :Be1 ( 1)sign2 :s2 :Be2 , if operation 0, z ( 1)sign :s:Be ( 1)sign1 :s1 :Be1 ( 1)sign2 :s2 :Be2 , if operation 1: