ARITHMETIC OPERATIONS in VS .NET

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16.2 ARITHMETIC OPERATIONS
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TABLE 16.1 operation 0 0 0 0 1 1 1 1 sign1 0 0 1 1 0 0 1 1 sign2 0 1 0 1 0 1 0 1 actual operation s1 s2 s1 2 s2 2(s1 2 s2) 2(s1 s2) s1 2 s2 s1 s2 2(s1 s2) 2(s1 2 s2)
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Once the signi cands have been aligned, the actual operation (addition or subtraction of the signi cands) depends on the values of operation, sign1, and sign2 (Table 16.1). The following algorithm, based on Algorithms 16.1 and 16.2 as well as Table 16.1, computes z. Algorithm 16.3 Addition and Subtraction
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if e1>=e2 then e:=e1; s2:=s2/B**(e1-e2); else e:=e2; s1:=s1/B**(e2-e1); end if; sign:=sign1; if operation xor sign1 xor sign2=0 then s:=s1+s2; 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; else s:=s1-s2; if s<0 then s:=-s; sign:=1-sign; 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; end if;
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As regards the hardware implementation, the following equivalent algorithm is better. Algorithm 16.4 Addition and Subtraction, Second Version
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if operation=1 then sign2:=1-sign2; end if; if e1<e2 then swap(sign1, sign2); swap(s1, s2); swap (e1, e2); end if; e:=e1; s2:=s2/B**(e1-e2); sign:=sign1; if sign xor sign2=0 then
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FLOATING-POINT UNIT
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s:=s1+s2; if s>=B then e:=e+1; s:=s/B; end if; else if (e1=e2) and (s1<s2) then swap(s1, s2); sign:=1-sign; end if; s:=s1-s2; leading_zeroes(s, k); s:=s*(B**k); e:=e-k; end if; s:=round(s); if s>=B then e:=e+1; s:=s/B; end if;
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Multiplication
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Given two oating-point numbers (21)sign1.s1.Be1 and (21)sign2.s2.Be2, their product (21)sign.s.Be is computed as follows: sign sign1 xor sign2 , The value of s belongs to the interval 1 s (B ulp)2 , (16:15) s s1 :s2 , e e1 e2 : (16:14)
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and could be greater than or equal to B. If it is the case, that is, if B s (B ulp)2 , (16:16)
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then (normalization) substitute s by s/B, and e by e 1. The new value of s satis es 1 s (B ulp)2 =B B 2:ulp (ulp)2 =B , B ulp (16:17)
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(ulp , B so that 2 2 ulp/B . 1). It remains to round the signi cand and to normalize if necessary.
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Algorithm 16.5
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Multiplication
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sign:=sign1 xor sign2; s:=s1*s2; e:=e1+e2; 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 16.4 Assume that B 10 and ulp 1024, so that the numbers are represented in the form s.10e, where 1 s 9.9999.
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16.2 ARITHMETIC OPERATIONS
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1. Compute z (3.4382 103) (2.5471 1021): z 8:75743922 102 , 8:75743922 , 10, rounding: s 8:7574, 8:7574 , 10, z 8:7574 10 2 : 2. Compute z (9.4300 103) (8.6200 102): z 81:2866 105 , normalization: s 8:12866, rounding: s 8:1287, 8:1287 , 10, z 8:1287 106 : 3. Compute z (4.7619 102) (2.1000 103): z 9:99999 105 , 9:99999 , 10, rounding: s 10:00, normalization: s 1, e 6, z 1:0000 106 : Comment 16.3 The product of two real numbers could produce an over ow as the nal value of e could be greater than emax. 16.2.5 Division Given two oating-point numbers (21)sign1.s1.Be1 and (21)sign2.s2.Be2 their quotient (21)sign.s.Be is computed as follows: sign sign1 xor sign2 , The value of s belongs to the interval 1=B , s B ulp, (16:19) s s1 =s2 , e e1 e2 (16:18) e 6,
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and could be smaller than 1. If that is the case, that is if s s1/s2 , 1, then s1 , s2 , and 1=B , s , 1 ulp=B: (16:20) s1 s2 ulp, s1 =s2 1 ulp=s2 , 1 ulp=B,
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FLOATING-POINT UNIT
Then (normalization) substitute s by s.B, and e by e 2 1. The new value of s satis es 1 , s , B ulp It remains to round the signi cand. Algorithm 16.6 Division (16:21)
sign:=sign1 xor sign2; s:=s1/s2; e:=e1 2 e2; if s<1 then e:=e21; s:=s*B; end if; s:=round(s);
Examples 16.5 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): z 1:3485152 102 , 1:3485152 ! 1, rounding: s 1:3485, z 1:3485 102 : 2. Compute z (2.5491 1021)/(3.4375 103): z 0:74155564 10 4 , normalization: s 7:4155564, e 5, rounding: s 7:4156, z 7:4156 10 5 : Comment 16.4 The quotient of two real numbers could produce an under ow, as the nal value of e could be smaller than emin. 16.2.6 Square Root
Given a positive oating-point number s1.Be1, its square root s.Be is computed as follows: if e1 is even, if e1 is odd, In the rst case (16.22), 1 s (B ulp)1=2 , B ulp: (16:24) s (s1 )1=2 , e e1 =2; s (s1 =B)1=2 , e (e1 1)=2: (16:22) (16:23)