Question: For `(int -> int) -> int = <fun>` the corresponding expression can be `fun x -> (x 1) + 1`.  

Is it possible to have an expression with `int -> (int -> int) = <fun>` type? If not then why?

Answer: 
You can split your url by `/` and take needed element (first remove `http://` part):

```js
var str = 'http://test.com/en/country/city';

str = str.replace('http://', '');

var parts = str.split('/');
console.log(parts);

alert(parts[2]+', '+parts[3]);
```
You could do something like this:

```
var url = "http://test.com/en/country/city"
var urlParts = url.split("/")
urlParts[urlParts.length - 1] (which would equal to "city")
urlParts[urlParts.length - 2] (which would equal to "country")

```

Basically split on each occurence of `"/"` and pick the correct item from the returned array.