Java float 123.129456 to 123.12 without rounding

How do you cut down float primitive in java to two decimal places, without using rounding?:

123.99999 to 123.99
-8.022222 to -8.02

There should be no rounding just cut of the decimal places and leave two.

Second point is how do you validate or count how many decimals are after the point?:

123.99 will give true or 开发者_开发百科2
123.999 will give false or 3

UPDATE

The numbers are String input so I think I will go with this as suggested; and I'll just have int try/catch block for any exceptions. Any suggestions how to make this work any smarter way are welcome:

public static float onlyTwoDecimalPlaces(String number) {
    StringBuilder sbFloat = new StringBuilder(number);
    int start = sbFloat.indexOf(".");
    if (start < 0) {
        return new Float(sbFloat.toString());
    }
    int end = start+3;
    if((end)>(sbFloat.length()-1)) end = sbFloat.length();

    String twoPlaces = sbFloat.substring(start, end);
    sbFloat.replace(start, sbFloat.length(), twoPlaces);
    return new Float(sbFloat.toString());
}

When you use DecimalFormat be aware to the fact that many languages uses "," instead of "." for float. So while you will format your float to "0.00" it will become "0,00" in certain locales (such as German and Polish). This will cause a NullPointerException while you will use this new formatted float in android applications. So what I did in order to cut and not round is to cast it to int after multiply it with 100 then recast it back to float and divide it to 100 This is the line:

myFloat = (float)((int)( myFloat *100f))/100f;

You can try it with log:

float myFloat = 12.349;
myFloat = (float)((int)( myFloat *100f ))/100f;
Log.d(TAG, " myFloat = "+ myFloat);       //you will get myFloat = 12.34

This will cut myFloat two places after the decimal point to format of ("0.00") it will not round it like this line (myFloat = Math.round(myFloat *100.0)/100.0;) it will just cut it.

Do keep in mind that float are floating point values. So there might not even be an exact two decimal representation for a certain number.

Having said that, you might try something like:

float f = -8.022222f;
BigDecimal bd = new BigDecimal(f);
BigDecimal res = bd.setScale(2, RoundingMode.HALF_UP);
f = res.floatValue();
System.out.println(f);

You might need to use a different RoundingMode though. Depends on what you want.

First of all, there is no guarantee that just because a float can represent a.bcde exactly, it is guaranteed to be able to represent a.bc exactly.

So if you're after just the printing part, how about doing it with some string-manipulation? Find the decimal point using indexOf and extract the part with two decimals, using substring.

The easy way to use DecimalFormat:

DecimalFormat df = new DecimalFormat("0.00");
float f = -8.0222222f;
f = Float.parseFloat(df.format(f));
System.out.println(f);

Please try this, this one works for you.

double d = 16.66667;
DecimalFormat decimalFormat= new DecimalFormat("#.##");
decimalFormat.setRoundingMode(RoundingMode.FLOOR);
System.out.println("Result :"+decimalFormat.format(d));