In OCaml, I obtain an error I do not understand when passing arguments which should work to Printf.printf. It is probably because I do not understand that function completely, but I cannot pin down what does not work.
First, I define a function (used for logging):
utop # let log verbosity level str =
if level <= verbosity then (
Printf.printf "\nLevel %i: " level;
Printf.printf str);;
val log : int -> int -> (unit, out_channel, unit) format -> unit = <fun>
All seems well, but then I obtain this:
utop # log 0 0 "%i" 0;;
Error: This function has type
int -> int -> (unit, out_channel, unit) format -> unit
It is applied to too many arguments; maybe you forgot a `;'.
although the following works:
utop # Printf.printf;;
- : ('a, out_channel, unit) format -> 'a = <fun>
utop # Printf.printf "%i" 0;;
0- : unit = ()
So, how can I define a function which does what log intends to do ?
Edit: Indeed, log 0 0 "%i" 0;; looks like too many arguments (4 instead of 3), but so does Printf.printf "%i" 0;; (2 instead of 1), and it works. With partial application, this gives this:
utop # Printf.printf "%i";;
- : int -> unit = <fun>
utop # log 0 0 "%i";;
Error: This expression has type (unit, unit) CamlinternalFormatBasics.precision
but an expression was expected of type
(unit, int -> 'a) CamlinternalFormatBasics.precision
Type unit is not compatible with type int -> 'a
The printf-like functions are variadic, in the sense that they accept a variable number of arguments. It is not really specific to printf and the family, you can define your own variadic functions in OCaml, this is fully supported by the type system. The only magic of printf is that the compiler translates a string literal, e.g., "foo %d to a value of type format.
Now, let's look at the type of the printf function,
('a, out_channel, unit) format -> 'a
Notice that it returns 'a which is a type variable. Since 'a could be anything it could be also a function. The ('a, out_channel, unit) format is the type of the format string that defines the type of function that is generated by this format string. It is important to understand though, that despite that "foo %d" looks like a string, in fact, it is a special built-in value of type _ format, which has a literal that looks like a string (though not all valid strings are valid literals for the _ format type.
Just to demonstrate that the first argument of printf is not a string, let's try the following,
# Printf.printf ("foo " ^ "%d");;
Line 1, characters 14-29:
1 | Printf.printf ("foo " ^ "%d");;
^^^^^^^^^^^^^^^
Error: This expression has type string but an expression was expected of type
('a, out_channel, unit) format
Now, when we know that printf is not a typical function, let's define a printf-like function ourselves. For that we need to use the kprintf-family of functions, e.g.,
# #show Printf.ksprintf;;
val ksprintf : (string -> 'd) -> ('a, unit, string, 'd) format4 -> 'a
This function takes the function which receives the resulting string which we can log, for example,
# let log fmt = Printf.ksprintf (fun s -> print_endline ("log> "^s)) fmt;;
val log : ('a, unit, string, unit) format4 -> 'a = <fun>
# log "foo";;
log> foo
- : unit = ()
This resulting function looks more like sprintf, i.e., it will play nicely with pretty-printing function that work with string as their output devices (this is a different topic). You may find it more easier to define your logging functions, using either Printf.kfprintf or, much better, using Format.kasprintf or Format.kfprintf. The latter two functions have the following types,
val kasprintf : (string -> 'a) -> ('b, formatter, unit, 'a) format4 -> 'b
val kfprintf : (formatter -> 'a) -> formatter ->
('b, formatter, unit, 'a) format4 -> 'b
But the type of format works with the formatter type (which is an abstraction of the output device) that is the type that the pretty printers (conventionally named pp) are accepting. So the log function defined using the Format module will play better with the existing libraries.
So, using Format.kasprintf we can define your log function as,
# let log verbosity level =
Format.kasprintf (fun msg ->
if level <= verbosity then
Format.printf "Level %d: %s#\n%!" level msg);;
val log : int -> int -> ('a, Format.formatter, unit, unit) format4 -> 'a = <fun>
And here is how it could be used,
# log 0 0 "Hello, %s, %d times" "world" 3;;
Level 0: Hello, world, 3 times
- : unit = ()
A common pattern in functional programming languages with a sufficiently advanced type system to to have a type of "heterogeneous lists". For instance, given a list defined as:
data List a = Nil | Cons a (List a)
(Note: For concreteness, I will use Idris in this question, but this could also be answered in Haskell (with the right extensions), Agda, etc...)
We can define HList:
data HList : List a -> Type where
Nil : HList []
(::) : a -> HList as -> HList (a :: as)
This is a list which holds a different type (specified by the type-level List a) at each "position" of the List data type. This made me wonder: Can we generalize this construction? For instance, given a simple tree-like structure:
data Tree a = Branch a [Tree a]
Does it make sense to define a heterogenous tree?
where HTree : Tree a -> Type where
...
More generally in a dependently-typed language, is it possible to define a general construction:
data Hetero : (f : Type -> Type) -> f a -> Type where
....
that takes a data type of kind Type -> Type and returns the "heterogeneous container" of shape f? Has anyone made use of this construction before if possible?
We can talk about the shape of any functor using map and propositional equality. In Idris 2:
Hetero : (f : Type -> Type) -> Functor f => f Type -> Type
Hetero f tys = (x : f (A : Type ** A) ** map fst x = tys)
The type (A : Type ** A) is the type of non-empty types, in other words, values of arbitrary type. We get heterogeneous collections by putting arbitrarily typed values into functors, then constraining the types elementwise to particular types.
Some examples:
ex1 : Hetero List [Bool, Nat, Bool]
ex1 = ([(_ ** True), (_ ** 10), (_ ** False)] ** Refl)
data Tree : Type -> Type where
Leaf : a -> Tree a
Node : Tree a -> Tree a -> Tree a
Functor Tree where
map f (Leaf a) = Leaf (f a)
map f (Node l r) = Node (map f l) (map f r)
ex2 : Hetero Tree (Node (Leaf Bool) (Leaf Nat))
ex2 = (Node (Leaf (_ ** False)) (Leaf (_ ** 10)) ** Refl)
The function applyRule is supposed to extract the implicit argument n that is used in another arguments it gets, of type VVect.
data IVect : Vect n ix -> (ix -> Type) -> Type where -- n is here
Nil : IVect Nil b
(::) : b i -> IVect is b -> IVect (i :: is) b
VVect : Vect n Nat -> Type -> Type -- also here
VVect is a = IVect is (flip Vect a)
-- just for completeness
data Expression = Sigma Nat Expression
applyRule : (signals : VVect is Double) ->
(params : List Double) ->
(sigmas : List Double) ->
(rule : Expression) ->
Double
applyRule {n} signals params sigmas (Sigma k expr1) = cast n
Without referring to {n}, the code type-checks (if cast n is changed to some valid double). Adding it in, however, results in the following error:
When checking left hand side of applyRule:
Type mismatch between
Double (Type of applyRule signals params sigmas rule)
and
_ -> _ (Is applyRule signals
params
sigmas
rule applied to too many arguments?)
This doesn't seem to make sense to me, because I'm not pattern-matching on any parameter that could have a dependency on n, so I thought that simply putting it in curly braces would bring it into scope.
You can only bring n into scope if it is defined somewhere (e.g. as a variable in the arguments). Otherwise it would be hard to figure out where the n comes from – at least for a human.
applyRule : {is : Vect n Nat} ->
(signals : VVect is Double) ->
(params : List Double) ->
(sigmas : List Double) ->
(rule : Expression) ->
Double
applyRule {n} signals params sigmas (Sigma k expr1) = cast n
I'm trying to implement a simple algebraic structures hierarchy using Idris interfaces. The code is as follows:
module AlgebraicStructures
-- definition of some algebraic structures in terms of type classes
%access public export
Associative : {a : Type} -> (a -> a -> a) -> Type
Associative {a} op = (x : a) ->
(y : a) ->
(z : a) ->
(op x (op y z)) = (op (op x y) z)
Identity : {a : Type} -> (a -> a -> a) -> a -> Type
Identity op v = ((x : a) -> (op x v) = x,
(x : a) -> (op v x) = x)
Commutative : {a : Type} -> (a -> a -> a) -> Type
Commutative {a} op = (x : a) ->
(y : a) ->
(op x y) = (op y x)
infixl 4 <**>
interface IsMonoid a where
empty : a
(<**>) : a -> a -> a
assoc : Associative (<**>)
ident : Identity (<**>) empty
interface IsMonoid a => IsCommutativeMonoid a where
comm : Commutative (<**>)
But, Idris is giving this strange error message:
When checking type of constructor of AlgebraicStructures.IsCommutativeMonoid:
Can't find implementation for IsMonoid a
I believe that Idris interfaces works like Haskell's type classes. In Haskell, it should work. Am I doing something silly?
I believe it may be complaining because I don't know that there's anything that constrains the a in the expression Commutative (<**>) - so it doesn't know that you can invoke <**> on that type.
Explicitly specifying the a seems to work for me - Commutative {a} (<**>) - I hope that that means that the a from the interface signature is in scope and available for explicitly passing to other types.
I am getting an unsolved metavariable for foo in the code below:
namespace Funs
data Funs : Type -> Type where
Nil : Funs a
(::) : {b : Type} -> (a -> List b) -> Funs (List a) -> Funs (List a)
data FunPtr : Funs a -> Type -> Type where
here : FunPtr ((::) {b} _ bs) b
there : FunPtr bs b -> FunPtr (_ :: bs) b
total foo : FunPtr [] b -> Void
How do I convince Idris that foo has no valid patterns to match on?
I've tried adding
foo f = ?foo
and then doing a case split in Emacs on f (just to see what might come up), but that just removes the line, leaving foo as an unsolved meta.
It turns out all I need to do is enumerate all possible patterns for foo's argument, and then Idris is able to figure out, one by one, that they are un-unifyable with foo's type:
foo : FunPtr [] b -> Void
foo here impossible
foo (there _) impossible