I still cannot correctly grasp the `where`

syntax in Julia.

I’d like to know if one can use Julia’s `where`

keyword as in Haskell; e.g., similar to the following Haskell snippet:

```
len x y = sqrt (sq x + sq y)
where
sq a = a * a
```

the Julia code for which could be

```
len(x, y) = sqrt( sq(x) + sq(y) ) where
sq(a) = a * a
end
```

Please let me know if this is (or will be) possible.