Scala Doubles, and Precision

How about :

 val value = 1.4142135623730951

//3 decimal places
println((value * 1000).round / 1000.toDouble)

//4 decimal places
println((value * 10000).round / 10000.toDouble)

Here's another solution without BigDecimals

Truncate:

(math floor 1.23456789 * 100) / 100

Round:

(math rint 1.23456789 * 100) / 100

Or for any double n and precision p:

def truncateAt(n: Double, p: Int): Double = { val s = math pow (10, p); (math floor n * s) / s }

Similar can be done for the rounding function, this time using currying:

def roundAt(p: Int)(n: Double): Double = { val s = math pow (10, p); (math round n * s) / s }

which is more reusable, e.g. when rounding money amounts the following could be used:

def roundAt2(n: Double) = roundAt(2)(n)

You can use scala.math.BigDecimal:

BigDecimal(1.23456789).setScale(2, BigDecimal.RoundingMode.HALF_UP).toDouble

There are a number of other rounding modes, which unfortunately aren't very well documented at present (although their Java equivalents are).

Since no-one mentioned the % operator yet, here comes. It only does truncation, and you cannot rely on the return value not to have floating point inaccuracies, but sometimes it's handy:

scala> 1.23456789 - (1.23456789 % 0.01)
res4: Double = 1.23