对于简单的浮点截断使用DecimalFormat太多了吗？

Question

I suddenly had a need to just chop off the excessive digits off of a float number, so I looked in the toolbox and saw that DecimalFormat was available.

Though creating a new object just to chop off some extra digits off a number seemed rather expensive, so I threw together a small program to test it.

public class Snippet {

    static float unformatted = -542.347543274623876F;
    static int fractionDigits = 2;

    public static void main(String[] args){

        long a = System.nanoTime();
        System.out.println(stringMethod(unformatted));
        long b = System.nanoTime();
        System.out.println(formatMethod(unformatted));
        long c = System.nanoTime();
        System.out.println(stringMethod2(unformatted));
        long d = System.nanoTime();

        System.out.println("OP1:"+(b-a));
        System.out.println("OP2:"+(c-b));
        System.out.println("OP3:"+(d-c));

    }

    private static float stringMethod(float number){
        String unfStr = String.valueOf(number);
        for(int i=0;i<unfStr.length();i++){
            if(unfStr.charAt(i) == '.'){
                return Float.parseFloat(unfStr.substring(0, i+1+fractionDigits));
            }
        }
        return Float.parseFloat(unfStr);
    }

    private static float stringMethod2(float number){
        String unfStr = String.format("%."+(fractionDigits+1)+"f",number);
        return Float.parseFloat(unfStr.substring(0,unfStr.length()-1));
    }

    private static float formatMethod(float number){
        DecimalFormat df = new DecimalFormat();
        df.setMaximumFractionDigits(fractionDigits);
        df.setRoundingMode(RoundingMode.DOWN);
        return Float.parseFloat(df.format(unformatted));
    }

}

OUTPUT:

-542.34
-542.34
-542.34
OP1:1937181
OP2:32609426
OP3:3111908

No matter how many times I run it, the DecimalFormat method just can't keep up.

So I guess the question here is, is there any reason (apart from code readability) to use DecimalFormat instead of creating your own method for simple float truncation?

Answer 1

This is a numerical way to do it:

double d = -542.347543274623876;
System.out.println(d);
int n = 2; // decimal digits

double p = Math.pow(10,n);
d = Math.floor((int)(p*d))/p;
System.out.println(d);

Try it here: http://ideone.com/wIhBpL

What it does, it to multiply it by 10 times the number of decimal digits you want, convert it to an integer (which chops off the remaining digits), then convert it back to a decimal by dividing by 10 times the number of decimal digits. It should work for float , too, if you use that instead.