What's the right way to parseFloat in Java

Question

I notice some issues with the Java float precision

       Float.parseFloat(\"0.0065\") - 0.001  // 0.0055000000134110451
       new Float(\"0.027\") - 0.001          


        

Answer 1

I would convert your float to a string and then use BigDecimal.

This link explains it well

new BigDecimal(String.valueOf(yourDoubleValue));

Dont use the BigDecimal double constructor though as you will still get errors

Answer 2

See What Every Computer Scientist Should Know About Floating-Point Arithmetic. Your results look correct to me.

If you don't like how floating-point numbers work, try something like BigDecimal instead.

Answer 3

The problem is simply that float has finite precision; it cannot represent 0.0065 exactly. (The same is true of double, of course: it has greater precision, but still finite.)

A further problem, which makes the above problem more obvious, is that 0.001 is a double rather than a float, so your float is getting promoted to a double to perform the subtraction, and of course at that point the system has no way to recover the missing precision that a double could have represented to begin with. To address that, you would write:

float f = Float.parseFloat("0.0065") - 0.001f;

using 0.001f instead of 0.001.

Answer 4

From the Java Tutorials page on Primitive Data Types:

A floating-point literal is of type float if it ends with the letter F or f; otherwise its type is double and it can optionally end with the letter D or d.

So I think your literals (0.001) are doubles and you're subtracting doubles from floats.

Try this instead:

System.out.println((0.0065F - 0.001D)); // 0.005500000134110451
System.out.println((0.0065F - 0.001F)); // 0.0055

... and you'll get:

0.005500000134110451
0.0055

So add F suffixes to your literals and you should get better results:

Float.parseFloat("0.0065") - 0.001F
new Float("0.027") - 0.001F
Float.valueOf("0.074") - 0.001F

Answer 5

Floating point cannot accurately represent decimal numbers. If you need an accurate representation of a number in Java, you should use the java.math.BigDecimal class:

BigDecimal d = new BigDecimal("0.0065");

Answer 6

You're getting the right results. There is no such float as 0.027 exactly, nor is there such a double. You will always get these errors if you use float or double.

float and double are stored as binary fractions: something like 1/2 + 1/4 + 1/16... You can't get all decimal values to be stored exactly as finite-precision binary fractions. It's just not mathematically possible.

The only alternative is to use BigDecimal, which you can use to get exact decimal values.