Convert float to double without losing precision

Question

I have a primitive float and I need as a primitive double. Simply casting the float to double gives me weird extra precision. For example:

float temp = 14009.35F; System.out.println(Float.toString(temp)); // Prints 14009.35 System.out.println(Double.toString((double)temp)); // Prints 14009.349609375 

However, if instead of casting, I output the float as a string, and parse the string as a double, I get what I want:

System.out.println(Double.toString(Double.parseDouble(Float.toString(temp)))); // Prints 14009.35 

Is there a better way than to go to String and back?

Answer 1

It's not that you're actually getting extra precision - it's that the float didn't accurately represent the number you were aiming for originally. The double is representing the original float accurately; toString is showing the "extra" data which was already present.

For example (and these numbers aren't right, I'm just making things up) suppose you had:

float f = 0.1F; double d = f; 

Then the value of f might be exactly 0.100000234523. d will have exactly the same value, but when you convert it to a string it will "trust" that it's accurate to a higher precision, so won't round off as early, and you'll see the "extra digits" which were already there, but hidden from you.

When you convert to a string and back, you're ending up with a double value which is closer to the string value than the original float was - but that's only good if you really believe that the string value is what you really wanted.

Are you sure that float/double are the appropriate types to use here instead of BigDecimal? If you're trying to use numbers which have precise decimal values (e.g. money), then BigDecimal is a more appropriate type IMO.

Answer 2

I find converting to the binary representation easier to grasp this problem.

float f = 0.27f; double d2 = (double) f; double d3 = 0.27d;  System.out.println(Integer.toBinaryString(Float.floatToRawIntBits(f))); System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(d2))); System.out.println(Long.toBinaryString(Double.doubleToRawLongBits(d3))); 

You can see the float is expanded to the double by adding 0s to the end, but that the double representation of 0.27 is 'more accurate', hence the problem.

   111110100010100011110101110001 11111111010001010001111010111000100000000000000000000000000000 11111111010001010001111010111000010100011110101110000101001000