double

I'm trying to only show the decimal portion of a number but it keeps printing the number with the decimal point: for number 223.50: changeAmountLabel.text = "\(balance.truncatingRemainder(dividingBy: 1))" this will print .5, but what needs to be printed is 50, is there a way to do this? You can do this by using a NumberFormatter . Try this: let formatter = NumberFormatter() formatter.locale = Locale(identifier: "en_US_POSIX") formatter.minimumFractionDigits = 2 formatter.maximumFractionDigits = 3 let number = 223.50 if let str = formatter.string(from: NSNumber(value: number)) as? String { if

I would like to create the Double whose value is closest to, but greater than, Float.MAX_VALUE . I've just written a question similar to this but for Double and Long.MAX_VALUE , see here . How can I repeat the conversion for Double and Float.MAX_VALUE using the standard Java 6 API? My attempt is below, but is incorrect it seems: Long longValue = Long.valueOf(Float.floatToIntBits(Float.MAX_VALUE)); Double value = Double.longBitsToDouble(Double.doubleToLongBits(longValue)+1); if (value < -Float.MAX_VALUE || value > Float.MAX_VALUE) { // Code here should execute but does not. } Sincere thanks.

why this two call to toBinary function compute the same output (at least under VS2010) ? #include <iostream> #include <bitset> #include <limits> using namespace std; template<class T> bitset<sizeof(T)*CHAR_BIT> toBinary(const T num) { bitset<sizeof(T)*CHAR_BIT> mybits; const char * const p = reinterpret_cast<const char*>(&num); for (int i = sizeof(T)*CHAR_BIT-1 ; i >= 0 ; --i) mybits.set(i, (*(p)&(1<<i) )); return mybits; } int main() { cout << toBinary(8.9).to_string() << "\n"; cout << toBinary( 8.9 + std::numeric_limits<double>::epsilon() ).to_string() << "\n"; cin.get(); } That epsilon is