Why does this function only deal with ints instead of generic data types? The current method definition is as follows: `val ith : int list -> int -> int -> int = <fun>`

how do I make it match `val ith: 'a list -> int 0> 'a -> 'a = <fun>`

?

```
let rec ith l i d =
match l with
| [] -> d
| h::t ->
let f baseV elem = if baseV = 0 then elem else (baseV - 1) in
let base = i in
List.fold_left f base l;;
```

It inferred this type

`val ith : int list -> int -> int -> int = <fun>`

because of the following reasoning:

```
let rec ith l i d = match l with
| []-> d
| h::t -> List.fold_left (fun a x -> if a = 0 then x else a-1) i l;;
```

- Think of the function type as a bunch of unknowns:

```
l : 'a
i : 'b
d : 'c
res : 'd
```

- Now you match on
`l`

with list constructors, so you know`l`

is a list. Plus, the first case returns`d`

of type`'c`

, so the result type of the whole function is`'c`

```
l : 'a list
i : 'b
d : 'c
res : 'c
```

- Next notice that you fold over the list, with accumulator
`i`

of type`'b`

. So`'b == 'c`

```
l : 'a list
i : 'c
d : 'c
res : 'c
```

- The
`fold`

function can replace the accumulator with an element`x`

of the list (first branch), so the list must contain elements of type`'c`

```
l : 'c list
i : 'c
d : 'c
res : 'c
```

- Finally, note that the other branch of the
`fold`

function decrements the accumulator, therefore the accumulator (variable`i`

, of type`'c`

) is an`int`

. We have`'c == int`

.

[Alternatively notice that you compare the accumulator with `0`

, therefore it must be an `int`

.]

```
l : int list
i : int
d : int
res : int
```

You canâ€™t have any other type if you write this function this way, because of how the type inference algorithm works.