take_while(p, xs)
Return the longest prefix of elements from the sequence xs that satisfy the predicate p.
take_while(is_positive, [1, 2, 3, -4, 5])
// [1, 2, 3]
function ints(n) n:~ints(n+1)
let s = ints(10)
take_while(^(x) x < 15, s)
// [10, 11, 12, 13, 14]