/* This function prints a number in decimal format */
void print_number(int n)
{
int digit, numDigits, divisor, remainder;
numDigits = countNumDigits(n);
divisor = powerOfTen(numDigits - 1);
remainder = n;
for (;numDigits > 0; numDigits--) {
digit = remainder / divisor;
print_char('0' + digit);
remainder -= (digit * divisor);
divisor = divisor / 10;
}
}
std::pair<std::string,std::pair<unsigned,std::string> >
decomposeDirectScaledUnit(const std::string& s) {
if (!isDirectScaledUnit(s)) {
LOG_FREE_AND_THROW("openstudio.QuantityRegex","Cannot decompose " << s
<< " into a numerator and scaled denominator because it is not an direct scaled unit.");
}
std::pair<std::string,std::pair<unsigned,std::string> > result;
boost::match_results<std::string::const_iterator> match;
boost::regex_search(s,match,regexDirectScaledUnit());
result.first = std::string(match[1].first,match[1].second);
std::string powerOfTen(match[2].first,match[2].second);
result.second.first = powerOfTen.size() - 1;
result.second.second = std::string(match[3].first,match[3].second);
return result;
}
double sqrt(double a)
{
/*
find more detail of this method on wiki methods_of_computing_square_roots
*** Babylonian method cannot get exact zero but approximately value of the square_root
*/
double z = a;
double rst = 0.0;
int max = 8; // to define maximum digit
int i;
double j = 1.0;
for(i = max ; i > 0 ; i--){
// value must be bigger then 0
if(z - (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)) >= 0)
{
while( z - (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)) >= 0)
{
j++;
if(j >= 10) break;
}
j--; //correct the extra value by minus one to j
z -= (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)); //find value of z
rst += j * powerOfTen(i); // find sum of a
j = 1.0;
}
}
for(i = 0 ; i >= 0 - max ; i--){
if(z - (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)) >= 0)
{
while( z - (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)) >= 0)
{
j++;
}
j--;
z -= (( 2 * rst ) + ( j * powerOfTen(i)))*( j * powerOfTen(i)); //find value of z
rst += j * powerOfTen(i); // find sum of a
j = 1.0;
}
}
// find the number on each digit
return rst;
}
