I'm trying to write a Caml function and I'm having a few troubles with the typing. The function is :
let new_game size count gens =
let rec continueGame board = function
0 -> ()
|n -> drawBoard board size;
continueGame (nextGeneration board) (n-1)
in
continueGame (seedLife (matrix 0 size) count) (gens) ;;
Here are the types of other functions :
val drawBoard : int list list -> int -> unit = <fun>
val seedLife : int list list -> int -> int -> int list list = <fun>
val nextGeneration : int list list -> int list list = <fun>
val matrix : 'a -> int -> 'a list list = <fun>
When trying to evaluate new_Game I have the following error :
continueGame (seedLife (matrix 0 size) count) (gens);;
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error: This expression has type int -> int list list
but is here used with type int list list
Why is this error occuring and how can I resolve it?
seedLife takes 3 arguments, but it's only passed 2.
Related
I am working on the following exercise:
Define a function libDiv which computes the list of natural divisors of some positive integer.
First define libDivInf, such that libDivInf n i is the list of divisors of n which are lesser than or equal to i
libDivInf : int -> int -> int list
For example:
(liDivInf 20 4) = [4;2;1]
(liDivInf 7 5) = [1]
(liDivInf 4 4) = [4;2;1]
Here's is my attempt:
let liDivInf : int -> int -> int list = function
(n,i) -> if i = 0 then [] (*ERROR LINE*)
else
if (n mod i) = 0 (* if n is dividable by i *)
then
i::liDivInf n(i-1)
else
liDivInf n(i-1);;
let liDiv : int -> int list = function
n -> liDivInf n n;;
I get:
ERROR: this pattern matches values of type 'a * 'b ,but a pattern
was expected which matches values of type int
What does this error mean? How can I fix it?
You've stated that the signature of liDivInf needs to be int -> int -> int list. This is a function which takes two curried arguments and returns a list, but then bound that to a function which accepts a single tuple with two ints. And then you've recursively called it in the curried fashion. This is leading to your type error.
The function keyword can only introduce a function which takes a single argument. It is primarily useful when you need to pattern-match on that single argument. The fun keyboard can have multiple arguments specified, but does not allow for pattern-matching the same way.
It is possible to write a function without using either.
let foo = function x -> x + 1
Can just be:
let foo x = x + 1
Similarly:
let foo = function x -> function y -> x + y
Can be written:
let foo x y = x + y
You've also defined a recursive function, but not included the rec keyword. It seems you're looking for something much more like the following slightly modified version of your attempt.
let rec liDivInf n i =
if i = 0 then
[]
else if (n mod i) = 0 then
i::liDivInf n (i-1)
else
liDivInf n (i-1)
I'm doing a school assignment in OCaml and I had a question regarding the meaning of an expression.
When defining function if I, for example, wrote:
let iter : int * (int -> int) -> (int -> int)
= fun (n,f) ->
What does (int -> int) mean? I understand the function itself receives a pair as an argument, but I don't fully understand what the parentheses mean...
The parentheses are there to disambiguate between a function of type (int -> int) - meaning it takes in a parameter of type int and returns an int - and possibly just two regular ints taken as parameters of that function. Without the first pair of parentheses for example, your iter would expect an(int, int) tuple, and in case no other parameter is present, expect an int -> int -> int as return type.
Note that the second pair of parentheses is not strictly necessary, but it can be a good indicator that you are expecting a function in return. Without that pair of parentheses, the function can be read as expect a tuple of (int, int -> int) plus another int, returning an int for example.
An example of a function with the same signature as your iter could be:
let random_func: int * (int -> int) -> (int -> int) =
fun (n, f) -> f
Find TL;DR below.
In lambda calculus (please bear with me), which is what ML-languages are rooted from, the core idea is all about abstracting an application or mapping of function to an argument. Only one argument.
λx[x + 1]
The λ in the above reads abstracting function x + 1 into an application waiting for a value for x, guard it from changing, and apply (replace the x in the function with the value and calculate).
The above in Ocaml would be equivalent to:
fun x -> x + 1
which has the type int -> int, or input type int and output type int. Now, lambda only deals with one argument at a time. How does that work with functions with multiple arguments like x*x -2*x + c (a polynomial function x2 − 2·x + c)? It evaluates the argument one at a time just like before.
λc[λx[x*x - 2*x + c]]
Thus, the output of the previous application becomes the input of the next one, and so on. The Ocaml equivalent would be
fun c x -> (x * x) - (2 * x) + c
The function has type int -> int -> int or (int -> int) -> int (chain of input -> output) If you apply the function partially to an argument x = 3, you get a reduced function like so:
fun c 3 -> (3 * 3) - (2 * 3) + c
fun c -> 9 - 6 + c
fun c -> 3 + c
at which the resulting function would have the type of int -> int. This is the basis of currying. It might look confusing at first, but it proves to be very useful and under-appreciated in imperative languages. For instance, you could do something like this:
let waiting_for_c_and_x = fun c x -> 2*x + c
let waiting_for_c = waiting_for_c_and_x 10 in
let result = waiting_for_c 2 (* result = 22 *)
TL;DR
However, using parentheses to group these chains of inputs/outputs are tricky but necessary in Ocaml because in reality the compiler cannot guess from e.g. int * int -> int if you mean an application that accepts an int * int pair as an input and return an int as an output (which we could parenthesize as (int * int) -> int) or one that accepts a pair of int and a function of type int -> int as argument (which could be written as int * (int -> int)).
Applied from Stanford Encyclopedia of Philosophy (very good read)
I have an maybe unusual question, but how does one match a function in F# using pattern matching?
Imagine the following:
I have multiple function signatures, which will be used multiple times, like:
binary function: int -> int -> int
unary function: int -> int
boolean function: int -> int -> bool
...
Now imagine the function evaluate, which itself takes a function f. The signature of f must be one of the listed above.
How do I match such a case?
I have tried the following things:
Test No.1 : Using delegates and Unions:
type UnaryFunction = delegate of int -> int
type BinaryFunction = delegate of (int -> int) -> int
type BooleanFunction = delegate of (int -> int) -> bool
type Functions =
| Unary of UnaryFunction
| Binary of BinaryFunction
| Boolean of BooleanFunction
// ...
let evaluate f = // signature: Functions -> string
match f with
| Unary u ->
let test_result = u.Invoke 3
sprintf "the result of the unary function is %d" test_result
| Binary b ->
let test_result = b.Invoke 315 42
sprintf "the result of the binary function is %d" test_result
| Boolean o ->
let test_result = o.Invoke 315 42
if test_result then "yeah" else "nope"
Test No.2 : Using type pattern matching and delegates:
type UnaryFunction = delegate of int -> int
type BinaryFunction = delegate of (int -> int) -> int
type BooleanFunction = delegate of (int -> int) -> bool
let evaluate f =
match f with
| ?: UnaryFunction as u ->
let test_result = u.Invoke 3
sprintf "the result of the unary function is %d" test_result
| ?: BinaryFunction as b ->
let test_result = b.Invoke 315 42
sprintf "the result of the binary function is %d" test_result
| ?: BooleanFunction as o ->
let test_result = o.Invoke 315 42
if test_result then "yeah" else "nope"
| _ -> "invalid function type"
The problem with these examples is, that delegates of ... will be matched instead of actual functions.
I would like to see somethink like this:
let evaluate f =
match f with
| ?: (int -> int) as u ->
let test_result = u 3
sprintf "the result of the unary function is %d" test_result
| ?: ((int -> int) -> int) as b ->
let test_result = b 315 42
sprintf "the result of the binary function is %d" test_result
| ?: ((int -> int) -> bool) as o ->
let test_result = o 315 42
if test_result then "yeah" else "nope"
| _ -> "invalid function type"
Does F# has a special syntax for function pattern matching?
And if not, why so? Am I missing something, or isn't it also important to be able to match functions just as anything else, as this is a functional language?
Instead of using delegates, just define the work using functions directly:
type UnaryFunction = int -> int
type BinaryFunction = int -> int -> int
type BooleanFunction = int -> int -> bool
type Functions =
| Unary of UnaryFunction
| Binary of BinaryFunction
| Boolean of BooleanFunction
// ...
let evaluate f = // signature: Functions -> string
match f with
| Unary u ->
let test_result = u 3
sprintf "the result of the unary function is %d" test_result
| Binary b ->
let test_result = b 315 42
sprintf "the result of the binary function is %d" test_result
| Boolean o ->
let test_result = o 315 42
if test_result then "yeah" else "nope"
Once you've done this, you can call them as needed (as below, showing FSI output):
> evaluate (Unary (fun x -> x + 3));;
val it : string = "the result of the unary function is 6"
> let someBinaryFunction x y = x * y;;
val someBinaryFunction : x:int -> y:int -> int
> Binary someBinaryFunction |> evaluate;;
val it : string = "the result of the binary function is 13230"
I'm completely lost on this. It was explained that functions are right justified so that let add x y = x + y;; has a function type of int -> int -> int or int -> (int -> int).
I'm not sure how I'd define a function of type (int -> int) -> int. I was thinking I'd have the first argument be a function that passes in an int and returns an int. I've tried:
let add = fun x y -> x + y --- int -> int -> int
let add = fun f x = (f x) + 3 --- ('a -> int) -> 'a -> int
What about
let eval (f: int -> int) :int = f 0
?
fun x -> (x 1) + 1;;
- : (int -> int) -> int = <fun>
or
let foo f = (f 1) + 1;;
val foo : (int -> int) -> int = <fun>
it works like
foo (fun x -> x + 1);;
- : int = 3
Your questions is highly associated with the notion of Currying.
But before that, let me say that if you want to write a function that needs a parameter to be a function, you could declare a normal function, and just use its parameter like a function. No need to complicate it. See the ex:
let f x = x(10) + 10
Now comes the currying part. In OCaml, the parameters are semantically evaluated just one at a time, and after evaluating an argument, an anonymous function is returned. This is important because it lets you supply part of the arguments of a function, creating effectively a new function (which is called Partial Application).
In the example bellow, I use + as a function (parenthesis around an operator turn it to a normal function), to create an increment function. And apply it to the previous f function.
let incr = (+) 1
f incr
The code evaluates to f incr = incr(10) + 10 = 21
This link has more information on the topic applied to OCaml.
I need to create a dictionary in sml, but I am having extreme difficulty with an insert function.
type dict = string -> int option
As an example, here is the empty dictionary:
val empty : dict = fn key => NONE
Here is my implementation of an insert function:
fun insert (key,value) d = fn d => fn key => value
But this is of the wrong type, what I need is insert : (string*int) -> dict -> dict.
I've searched everything from lazy functions to implementing dictionaries.
Any help or direction would be greatly appreciateds!
If you are still confused on what I am trying to implement, I drafted up what I should expect to get when calling a simple lookup function
fun lookup k d = d k
- val d = insert ("foo",2) (insert ("bar",3) empty);
val d = fn : string -> int option
- lookup2 "foo" d;
val it = SOME 2 : int option
- lookup2 "bar" d;
val it = SOME 3 : int option
- lookup2 "baz" d;
val it = NONE : int option
You can reason on the signature of the function:
val insert = fn: (string * int) -> dict -> dict
When you supply key, value and a dictionary d, you would like to get back a new dictionary d'. Since dict is string -> int option, d' is a function takes a string and returns an int option.
Suppose you supply a string s to that function. There are two cases which could happen: when s is the same as key you return the associated value, otherwise you return a value by looking up d with key s.
Here is a literal translation:
fun insert (key, value) d = fn s => if s = key then SOME value
else d s