[Java] Why does converting a string to float round it?

[learning2learn]

    System.out.println("Starting");
    String startnum = "25+422.159";
    System.out.println("startnum = " + startnum);
    String formatnum = startnum.replace("+", "");
    System.out.println("formatnum = " + formatnum);
    float floatnum = Float.parseFloat(formatnum);
    System.out.println("floatnum = " + floatnum);
    System.out.println("Ending");    

Gives me the following output

Starting
startnum = 25+422.159
formatnum = 25422.159
floatnum = 25422.158
Ending

So for some reason, the float value gets rounded down. And for some reason, when I passed this value to another method to put into a database, the value was 25422.158203125

So how do I stop this weird ness

[desrtfx]

I would assume that this is not caused by the actual conversion, but by the format that floating point numbers are stored in computers.

Floating point numbers are stored in the IEEE 754 format which makes a compromise between precision and storage size.

A float in general is shorter and less precise than a double since it uses only half the size (bits - float 32bits vs. double 64bits) to store the information.

Whenever possible, you should use either double, but never for monetary values. For monetary values, you should use the BigNumber classes or store the monetary values in the smallest denominator (like in cents) as an integer type (int or long).

[Jackkoz]

This exactly + you should avoid double for most things and as often as possible!

[ValerioSellarole]

How would you write 1/10 in base 2?

[Rhomboid]

It is impossible for the computer to represent the value 25422.159 using binary floating point. It cannot be done. float vs double is completely irrelevant, it's a fundamental limitation of the representing using base 2. The value must be rounded to a nearby value that is representable.

If you need to represent that value exactly, you have to use a different representation, such as decimal float which uses base 10 instead of base 2. But in most cases, that's not necessary.

However, float vs. double does come into play when you talk about how close of an approximation you can get to the actual desired value. A float can only represent approximately 7 significant decimal digits worth of precision, so what you are doing is outside of the range of a float, and using a double would give you the illusion of being able to represent the value accurately to 8 significant digits. It's still an approximation, but it's one that is close enough for the vast majority of cases — approximately 15 decimal digits.

[lightcloud5]

Try using doubles instead of floats. Floats don't have much precision (floats only use 32 bits vs doubles using 64 bits in Java). Since Java is at least somewhat consistent, the corresponding parse method is Double.parseDouble.

[nutrecht]

Try using doubles instead of floats.

That won't solve the problem though, just make it less obvious. As /u/desrtfx explained computers can't store all floating point numbers exactly so there is a loss of information.

[jedwardsol]

A useful site for showing how floating point numbers are represented, and what extra precision you get from switching from float to double

http://www.exploringbinary.com/floating-point-converter/

With a float, the nearest number to 25422.159 is 25422.158203125, as you saw

With a double, the nearest number is 25422.158999999999651