Float.doubleValue() method

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

The simple primitive-type cast (or even an implicit conversion) will do all you need, if you really want to preserve the value of a float:

float f = 4.95f;
double d = f;

Fundamentally 4.95f is already inaccurate - you can't represent that number exactly as a float. The exact value of floatNumber is represented in doubleNumber too - it's just that that value is not 4.95.

If you really care about the exact decimal digits, you should be using BigDecimal instead (or a scaled integer).

• Guillaume: I was about to say that and suggest BigDecimal, but as usual you beat me to it. Do you live on StackOverflow? Are you even human or a SO bot? ;)
• Romeo: I don't have the option of using BigDecimal. I don't want to preserve the actual value of the float. I want to treat is as that would be double after the value was set to float in another system to which I don't have access to modify its type (I can only receive it as float). And lastly what does Float.doubleValue() actually do?
• Jon Skeet: @Romeo: The actual value of the float is 4.94999980926513671875. That's the nearest exact float value to the original 4.95. So it looks like it's doing the right thing. As for what Float.doubleValue() does - it's basically the normal conversion from float to double, which is safe as every float is exactly representable as a double.

Answer 2

Widening float to double conversion doesn't lose precision in java.

• Romeo: Try that piece of code in the post. It doesn't lose precision, but from a human POV that number is not what I would have expected to be.
• Jon Skeet: @Romeo: That's because you're not storing it in a type which preserves a human POV - you're storing it in a binary floating point type.

Answer 3

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