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Get Last 2 Decimal Places with No Rounding

In C#, I'm trying to get the last two decimal places of a double with NO rounding. I've tried everything from Math.Floor to Math.Truncate and nothing is working.

Samples of the results I'd like:

1,424.2488298 -> 1,424.24
53.5821 -> 53.58
10,209.2991 -> 10,209.29

Any ideas?

like image 488
mint Avatar asked Mar 29 '12 19:03

mint


2 Answers

Well, mathematically it's simple:

var f = 1.1234;
f = Math.Truncate(f * 100) / 100;  // f == 1.12

Move the decimal two places to the right, cast to an int to truncate, shift it back to the left two places. There may be ways in the framework to do it too, but I can't look right now. You could generalize it:

double Truncate(double value, int places)
{
    // not sure if you care to handle negative numbers...       
    var f = Math.Pow( 10, places );
    return Math.Truncate( value * f ) / f;
}
like image 60
Ed S. Avatar answered Oct 07 '22 22:10

Ed S.


My advice: stop using double in the first place. If you need decimal rounding then odds are good you should be using decimal. What is your application?

If you do have a double, you can do it like this:

double r = whatever;
decimal d = (decimal)r;
decimal truncated = decimal.Truncate(d * 100m) / 100m;

Note that this technique will fail if the absolute value of the double is larger than 792281625142643375935439504, because the multiplication by 100 will fail. If you need to handle values that large then you'll need to use special techniques. (Of course, by the time a double is that large, you are well beyond its ability to represent values with two digits after the decimal place anyway.)

like image 39
Eric Lippert Avatar answered Oct 07 '22 20:10

Eric Lippert