Assuming 64-bit IEEE double, there is a 52-bit mantissa and 11-bit exponent. Let’s break it to bits: 1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^0 = 1 The smallest representable number greater than 1: 1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52 Therefore: epsilon = (1 + 2^-52) – … Read more
The Wikipedia page on it is a good place to start. To sum up: float is represented in 32 bits, with 1 sign bit, 8 bits of exponent, and 23 bits of the significand (or what follows from a scientific-notation number: 2.33728*1012; 33728 is the significand). double is represented in 64 bits, with 1 sign … Read more
>>> x = “2342.34” >>> float(x) 2342.3400000000001 There you go. Use float (which behaves like and has the same precision as a C,C++, or Java double).
double.Parse(“3.5”, CultureInfo.InvariantCulture)
The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10308 (if your double is an IEEE 754 64-bit double). It’s an integer. It’s represented exactly. What more do you want? Go on, ask me what … Read more
Java prevent E notation in a double: Five different ways to convert a double to a normal number: import java.math.BigDecimal; import java.text.DecimalFormat; public class Runner { public static void main(String[] args) { double myvalue = 0.00000021d; //Option 1 Print bare double. System.out.println(myvalue); //Option2, use decimalFormat. DecimalFormat df = new DecimalFormat(“#”); df.setMaximumFractionDigits(8); System.out.println(df.format(myvalue)); //Option 3, use … Read more
I think you’ve summarised the advantages quite well. You are however missing one point. The decimal type is only more accurate at representing base 10 numbers (e.g. those used in currency/financial calculations). In general, the double type is going to offer at least as great precision (someone correct me if I’m wrong) and definitely greater … Read more
Use the static Double.isNaN(double) method, or your Double‘s .isNaN() method. // 1. static method if (Double.isNaN(doubleValue)) { … } // 2. object’s method if (doubleObject.isNaN()) { … } Simply doing: if (var == Double.NaN) { … } is not sufficient due to how the IEEE standard for NaN and floating point numbers is defined.
This works: inputValue = Math.Round(inputValue, 2);
You can use Double.parseDouble() to convert a String to a double: String text = “12.34”; // example String double value = Double.parseDouble(text); For your case it looks like you want: double total = Double.parseDouble(jlbTotal.getText()); double price = Double.parseDouble(jlbPrice.getText());