I'm very new to Julia but I've got a some background in Scheme/Rust/F#.
Today I wanted to make yesterday's AoC nicer without an explicit number of nested loops.
I arrived at this working solution, but I don't like the last if
. In the languages mentioned above I would call a function (or use a computation expression) that gives me the first result that is not None
. For Julia, I expected something to do that. It does, but unexpectedly in an eager fashion.
So When I tried return something(find(r, n, start + 1, which), find(r, n - 1, start + 1, extended))
, that also evaluated the second argument when the first already had a result—and thus crashed.
Is there a macro/lazy version or something
that I didn't find? How are you supposed to handle a case like that?
I also thought about (short-circuited) or'ing them together, but I guess Julia's strictness in that matter spoils that.
using DataStructures
function find(r::Array{Int}, n, start = 1, which = nil())::Union{Int,Nothing}
if start <= length(r)
extended = cons(start, which)
with_current = sum(i -> r[i], extended)
if with_current == 2020 && n == 1
return prod(i -> r[i], extended)
else
# Unfortunately no :(
#return something(find(r, n, start + 1, which), find(r, n - 1, start + 1, extended))
re = find(r, n, start + 1, which)
if isnothing(re)
return find(r, n - 1, start + 1, extended)
else
re
end
end
end
end