= Solution
The inclusive interval is nonempty exactly when \b[$l\leq u$]. Reject $l>u$ with a failed <precondition>, rather than assigning a nonexistent normalization to an empty interval. Compute its width in `Long` before taking a reciprocal:
``
class Uniform(l: Int, u: Int) extends FiniteDist[Int] {
require(l <= u, "Empty uniform interval")
private val width = u.toLong - l.toLong + 1L
private val mass = 1.0 / width.toDouble
def getProb(x: Int): Double =
if (l <= x && x <= u) mass else 0.0
def getDom: List[Int] = {
var x = l.toLong
var rev = List.empty[Int]
while (x <= u.toLong) {
rev = x.toInt :: rev
x += 1L
}
rev.reverse
}
}
``
The <loop invariant> says that $l\leq x\leq u+1$ and `rev` contains exactly the <integers> $l,\ldots,x-1$ in reverse order. The <loop variant> is $u-x+1$ while the guard is true. `Long` handles both the full interval width $2^{32}$ and the increment past `Int.MaxValue`. At exit, reversing yields each support value once. All support values have mass $1/(u-l+1)$, so their exact sum is one; `Double` approximates that <rational number>. Very large valid domains require correspondingly large lists and sufficient memory; that resource issue does not make the mathematical endpoint values invalid.
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