Question

I’m searching for the best practice to convert Float to Double without loosing precision. So far I only found that a proper way to do so is to convert the Float to String and the String to Double.

Searching the Float API I stumbled upon this method doubleValue(). I thought this is a static constructor that will return a double from my Float without loosing precision but the following code behaves like a cast:

public class Main {
  public static void main(String[] args) {

  Float floatNumber= 4.95f;
  Double doubleNumber= floatNumber.doubleValue();

  System.out.println(doubleNumber);

  }
}

The output is 4.949999809265137

Searching any other documentation about this from the Float API I didn’t find any documentation to tell me what exactly happens when I call that method. Does anybody have any idea? Or can someone confirm that all the method does is perform a cast my Float to a double and unbox it?

Answer 1

When you use floating point, you get rounding errors which can be potentially “corrected” by rounding the result (assuming you know the precision you want). BigDecimal handles this by always knowing the precision.

float floatNumber= 4.95f;
double doubleNumber = (double) floatNumber;
System.out.println("After cast " + doubleNumber);
System.out.printf("Show two decimal places. %.2f%n", doubleNumber);
doubleNumber = Math.round(doubleNumber * 100) / 100.0;
System.out.println("After rounding to two places " + doubleNumber);

prints

After cast 4.949999809265137
Show two decimal places. 4.95
After rounding to two places 4.95