Why won't julia ERROR if a variable is not defined when scheduling?

New to Julia
,
21

I think I’ve found an example that explain things fabulously

julia> function main()
           work = function(x)
               sleep(60)
               println(x)
           end
           v = [3, 4, 5]
           for ȷ = 1:2 Threads.@spawn work( $(v[ȷ]) ) end
           work, v = function(x)
               sleep(0.7)
               println(10x)
           end, -v
       end;

julia> 

julia> main();

julia> 40
30

If the $() were removed, then v is volatile as work is.

Edit: we can also write Threads.@spawn $work($v[ȷ])