ok super dumb question - why is the type system angry? ``` val sides = arrayOf<Number>(...

# getting-started

[David Lee]

ok super dumb question - why is the type system angry?

val sides = arrayOf<Number>(1,2,3)
        val sum = sides[0] + sides[1]
// Unresolved reference. None of the following candidates is applicable because of receiver type mismatch: [...much detail]

[ephemient]

there are no arithmetic operations on generic Number, only on concrete subtypes

[Blake MacDade]

You could also just use arrayOf(1,2,3) and let the type be automatically determined. In this case it will be an Array<Int>

[ephemient]

intArrayOf(1, 2, 3) will result in a primitive IntArray

[David Lee]

roger, thanks, makes sense.

[David Lee]

tricky bit for me now is I’ve inherited this constructor type signature from the Exercism problem boilerplate

Triangle<out T : Number>(val a: T, val b: T, val c: T)

and I need to find a way to coax T into something concrete I can add up …

helicopter dogging:

val sides = arrayOf(a,b,c) as Array<Double>

[David Lee]

or

val da: Double = a as Double
        val db: Double = b as Double
        val dc: Double = c as Double

        val sides = arrayOf<Double>(da,db,dc)

seems to compile but isn’t very succinct …

[ephemient]

arrayOf(a, b, c).sumOf { it as Double }

will sum them as doubles, but if you know it's double, you should at least have a Triangle<Double>

[David Lee]

val sides = arrayOf<Double>(a as Double,b as Double,c as Double)

slight improvement 🤷

[Blake MacDade]

val sides = arrayOf(a, b, c).map { it as Double }

[Joffrey]

What type of triangle are you getting as input? As @ephemient said, if you get a Triangle<Double> as input, you shouldn't need to cast. If you don't know (e.g. you get Triangle<*>), you shouldn't assume that you can cast to Double. In that case, you should use .toDouble() instead of as Double so you guarantee you're actually converting to a double in all cases

[ephemient]

none of the conversions are necessarily safe if you don't know what your inputs are. as Double will throw if it's any other type of `Number`; .toDouble() won't throw, but may lose information (9007199254740993L.toDouble() == 9007199254740992.0, 1.toBigDecimal().shiftLeft(1024).toDouble() == Double.POSITIVE_INFINITY, etc.)

[ephemient]

not to mention that Number is an open hierarchy so other types can be implemented, such as https://javadoc.io/doc/org.apache.commons/commons-numbers-complex/latest/index.html