[Solved] Answer to this Haskell expression [closed]
It works like a foreach loop, so foreach (x in [1..3]) { foreach (y in [x .. x * 2]) { yield y; } } First x is 1, so y in [1 .. 2] Then x is 2, so y in [2 .. 4] Then x is 3, so y in [3 .. 6] … Read more