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addition sum of two positiv double numbers

master
Dennis Sperzel 11 months ago
parent
commit
ae978dd244
  1. 29
      src/addition.c
  2. 2
      src/addition.h

29
src/addition.c

@ -144,4 +144,33 @@ unsigned long long addition_precision_double(unsigned long long p1, unsigned lon
unsigned long long hinten = 0x00000000ffffffff;
unsigned int vornep1 = p1 >> 32, hintenp1 = p1 & hinten, vornep2 = p2 >> 32, hintenp2 = p2 & hinten;
return ((unsigned long long) addition_uint(vornep1, vornep2) << 32) + (unsigned long long) addition_uint(hintenp1, hintenp2);
}
// addition of two double numbers
double addition_double(double number1, double number2) {
if (number2 == 0.0) return number1;
else if (number1 == 0.0) return number2;
else if (number2 > number1) return addition_double(number2, number1);
unsigned long long e = 0x0010000000000000;
struct datad num1, num2, num3;
num1.number.dnum = number1, num2.number.dnum = number2;
num1.sign = sign_double(num1.number.lnum), num2.sign = sign_double(num2.number.lnum);
num1.exponent = exponent_double(num1.number.lnum), num2.exponent = exponent_double(num2.number.lnum);
num1.precision = precision_double(num1.number.lnum) | e, num2.precision = precision_double(num2.number.lnum) | e;
unsigned int k = (num1.exponent - num2.exponent > 52 ? 53 : num1.exponent - num2.exponent);
num3.precision = addition_precision_double(num1.precision, num2.precision >> k);
if (num3.precision > 0x001fffffffffffff) {
num3.exponent = addition_uint((unsigned int) num1.exponent, 1);
num3.precision = (num3.precision >> 1);
}
else {
num3.exponent = num1.exponent;
}
num3.precision ^= e;
num3.number.lnum = output_double(num1.sign, num3.exponent, num3.precision);
return num3.number.dnum;
}

2
src/addition.h

@ -33,4 +33,6 @@ unsigned long long output_double(unsigned long long sign, unsigned long long exp
unsigned long long addition_precision_double(unsigned long long p1, unsigned long long p2);
double addition_double(double number1, double number2);
#endif // ADDITION_H
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