Is it possible to see the interface of a loaded module in interactive OCaml? I have (unsuccessfully) tried searching for such a possibility, and online docs/sources are not what I am looking for.
The standard trick for this is to define a synonym for the module, which induces the toplevel to list the interface.
$ ocaml
OCaml version 4.00.1
# #load "str.cma";;
# module S = Str;;
module S :
sig
type regexp = Str.regexp
val regexp : string -> regexp
val regexp_case_fold : string -> regexp
val quote : string -> string
val regexp_string : string -> regexp
val regexp_string_case_fold : string -> regexp
val string_match : regexp -> string -> int -> bool
. . .
val first_chars : string -> int -> string
val last_chars : string -> int -> string
end
Update
(Note that this answer is from 2013. Recent revisions of OCaml provide a toplevel directive to show a module interface:
# #show_module Str;;
module Str :
sig
type regexp
val regexp : string -> regexp
. . .
val first_chars : string -> int -> string
val last_chars : string -> int -> string
end
So the semi-clever workaround is no longer required.
(There are many new directives. Type #help;; at toplevel to get a list.)
Both utop and ocaml interpreters added the #show directive since a moment. It does exactly what you want, as in the following example :
│ Welcome to utop version 1.19.3 (using OCaml version 4.04.0) │
└──────────────────────────────────────────────────────────────┘
Type #utop_help for help about using utop.
─( 15:12:33 )─< command 0 >──────────────────────────────────────{ counter: 0 }─
utop # #show List;;
module List :
sig
val length : 'a list -> int
val cons : 'a -> 'a list -> 'a list
val hd : 'a list -> 'a
...
val fast_sort : ('a -> 'a -> int) -> 'a list -> 'a list
val sort_uniq : ('a -> 'a -> int) -> 'a list -> 'a list
val merge : ('a -> 'a -> int) -> 'a list -> 'a list -> 'a list
end
PS:i'm using 4.04 version but i know that it also works for 4.03.> and maybe before that too.
In OCaml toplevel version 4.04.0, the trick of defining a module's synonym works no more:
# module L = List;;
module L = List
#
but you can use the include directive:
# module L = struct include List end;;
module L :
sig
val length : 'a list -> int
val cons : 'a -> 'a list -> 'a list
val hd : 'a list -> 'a
val tl : 'a list -> 'a list
val nth : 'a list -> int -> 'a
val rev : 'a list -> 'a list
...
val sort_uniq : ('a -> 'a -> int) -> 'a list -> 'a list
val merge : ('a -> 'a -> int) -> 'a list -> 'a list -> 'a list
end
#
Related
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 = ()
I am creating monads in OCaml and need to compose them, so I created transformers. I implemented the regular monad in terms of the transformer with the Identity monad:
module type MONAD = sig
type 'a m
val (>>=) : 'a m -> ('a -> 'b m) -> 'b m
val return : 'a -> 'a m
end
module Identity : MONAD = struct
type 'a m = 'a
let (>>=) m f = f m
let return x = x
end
module OptionT (M : MONAD) : MONAD with type 'a m := ('a option) M.m = struct
type 'a m = ('a option) M.m
let (>>=) m f = M.(>>=) m (fun option ->
match option with
| Some x -> f x
| None -> M.return None)
let return x = M.return ## Some x
end
module Option = OptionT(Identity)
However, I can't do this:
open Option
let _ = (Some 1) >>= (fun x -> Some (x + 1))
The errors are:
(Some 1)
This expression has type 'a option
but an expression was expected of type 'b option Identity.m
Some (x + 1)
This expression has type 'a option
but an expression was expected of type 'b option Identity.m
If I try to fix the error with module Identity : MONAD with type 'a m = 'a I get an error at module Option = OptionT(Identity) that states that
The type `m' is required but not provided
It seems that now, 'a has replaced 'a m in the signature.
Doing
module Option : MONAD with type 'a m := 'a option = struct
type 'a m = 'a option
let (>>=) m f =
match m with
| Some x -> f x
| None -> None
let return x = Some x
end
works just fine.
How do I tell the compiler that a module implements a signature so that a type declared in the signature is the same as another type, while still keeping the signature's original type declaration?
It seems that now, 'a has replaced 'a m in the signature.
This the effect of destructive substitution, when you write
module Identity : MONAD with type 'a m := 'a
you are asking the compiler to substitute all instance of 'a m by 'a.
Contrarily, standard with constraint adds a type equality to the module type
module Identity : MONAD with type 'a m = 'a
Looking at your various examples, it seems that you have confused the two, and are using destructive substitution when you meant to add a type constraint:
module OptionT(X:Monad) : MONAD with type 'a m = 'a = …
(* or *) module Option : MONAD with type 'a m = 'a option = …
and not
module OptionT(X:Monad) : MONAD with type 'a m := 'a = …
(* nor *) module Option : MONAD with type 'a m := 'a option = …
This expression has type 'a option but an expression was expected of type 'b option Identity.m
And indeed, the compiler does not know anything about Identity except that its signature is MONAD. (: MONAD) is not something that merely helps the compiler, it hides all information about Identity except that its signature is MONAD.
So, you can add a type equality for that
module Identity : MONAD with type 'a m = 'a = ...
and it works.
I'm trying to create a functor that makes a polynomial ring out of a ring. My underlying type, Ring_elt, has the following signature:
module type Ring_elt = sig
type t
val add : t -> t -> t
val mul : t -> t -> t
val zer : t
val one : t
val neg : t -> t
end;;
My polynomial functor looks like:
module Make_Poly2(Underlying:Ring_elt) = struct
type t = Poly of Underlying.t list
let rec create lst =
match List.rev lst with
| Underlying.zer :: tl -> create List.rev tl
| _ -> Poly of lst
end;;
(so the 'create' function should take a list, remove the leading zeros, and then return the polynomial of the result). However, I get a syntax error and utop underlines the "zer" after "Underlying."
By comparison, the following code (for making integer polynomials) works:
module Make_int_poly = struct
type t = Poly of int list
let rec create lst =
match List.rev lst with
| 0 :: tl -> create (List.rev tl)
| _ -> Poly lst
end;;
Any idea what's going on?
An OCaml pattern is built from constants, data constructors, and new names bound by the pattern match. Underlying.zer isn't any of those things. But 0 is one of them.
Seems like you can just use an if to compare against Underlying.zer.
Jeffrey's answer is good but instead of correcting it with an if construction, what you should do is the following : use algebraic data types
Instead of writing
val zer : t
val one : t
You could write
module type Ring_elt = sig
type t = Zer | One | Num of t
val add : t -> t -> t
val mul : t -> t -> t
val neg : t -> t
end
module Make_int_poly = struct
type t = Poly of int list
let rec create lst =
match List.rev lst with
| Underlying.Zer :: tl -> create (List.rev tl)
| _ -> Poly lst
end
It's a much better way of doing it since you can easily pattern match on it and even add some constants to your type t without problems.
(Disclaimer: I am fairly certain that this is not idiomatic in any way. If there is some alternative tree-traversal idiom in OCaml, I'm all ears :) )
I am writing a toy compiler in OCaml, and I would like to have a visitor for my large syntax tree type. I wrote one using classes, but I thought it would be fun to try and implement one using modules/functors. My type hierarchy is massive, so let me illustrate what I'm trying to do.
Consider the following type definitions (making these up on the spot):
type expr =
SNum of int
| SVarRef of string
| SAdd of expr * expr
| SDo of stmt list
and stmt =
SIf of expr * expr * expr
| SAssign of string * expr
Let me briefly illustrate usage. Say, for example, I wanted to collect all of the SVarRefs inside of the program. If I had a mapping visitor (one which visits every node of the tree and does nothing by default), I could do the following (in a perfect world):
module VarCollector : (sig
include AST_VISITOR
val var_refs : (string list) ref
end) = struct
include MapVisitor
let var_refs = ref []
let s_var_ref (s : string) =
var_refs := s::!var_refs
SVarRef(s)
end
(* Where my_prog is a stmt type *)
let refs = begin
let _ = VarCollector.visit_stmt my_prog in
VarCollector.!var_refs
end
I should note that the advantage of having functions for each specific variant is that my actual codebase has a type which both has a large amount of variants and does not make sense to fragment. Variant-specific functions allow the avoiding of repeated iteration implementations for the other variants of a type.
Seems simple enough, but here's the catch: there are different types of visitors. MapVisitor returns the original syntax tree, so it has type
sig
(** Dispatches to variant implementations *)
val visit_stmt : stmt -> stmt
val visit_expr : expr -> expr
(** Variant implementations *)
val s_num : int -> expr
val s_var_ref : string -> expr
val s_add : (expr * expr) -> expr
val s_do : stmt list -> expr
val s_if : (expr * expr * expr) -> stmt
val s_assign : (string * expr) -> stmt
end
At the same time, one might imagine a folding visitor in which the return type is some t for every function. Attempting to abstract this as much as possible, here is my attempt:
module type AST_DISPATCHER = sig
type expr_ret
type stmt_ret
val visit_expr : expr -> expr_ret
val visit_stmt : stmt -> stmt_ret
end
(** Concrete type designation goes in AST_VISITOR_IMPL *)
module type AST_VISITOR_IMPL = sig
type expr_ret
type stmt_ret
val s_num : int -> expr_ret
val s_var_ref : string -> expr_ret
val s_add : (expr * expr) -> expr_ret
val s_do : stmt list -> expr_ret
val s_if : (expr * expr * expr) -> stmt_ret
val s_assign : (string * expr) -> stmt_ret
end
module type AST_VISITOR = sig
include AST_VISITOR_IMPL
include AST_DISPATCHER with type expr_ret := expr_ret
and type stmt_ret := stmt_ret
end
(** Dispatcher Implementation *)
module AstDispatcherF(IM : AST_VISITOR_IMPL) : AST_DISPATCHER = struct
type expr_ret = IM.expr_ret
type stmt_ret = IM.stmt_ret
let visit_expr = function
| SNum(i) -> IM.s_num i
| SVarRef(s) -> IM.s_var_ref s
| SAdd(l,r) -> IM.s_add (l,r)
| SDo(sl) -> IM.s_do sl
let visit_stmt = function
| SIf(c,t,f) -> IM.s_if (c,t,f)
| SAssign(s,e) -> IM.s_assign (s,e)
end
module rec MapVisitor : AST_VISITOR = struct
type expr_ret = expr
type stmt_ret = stmt
module D : (AST_DISPATCHER with type expr_ret := expr_ret
and type stmt_ret := stmt_ret)
= AstDispatcherF(MapVisitor)
let visit_expr = D.visit_expr
let visit_stmt = D.visit_stmt
let s_num i = SNum i
let s_var_ref s = SVarRef s
let s_add (l,r) = SAdd(D.visit_expr l, D.visit_expr r)
let s_do sl = SDo(List.map D.visit_stmt sl)
let s_if (c,t,f) = SIf(D.visit_expr c, D.visit_expr t, D.visit_expr f)
let s_assign (s,e) = SAssign(s, D.visit_expr e)
end
Running this gives me the following error message, however:
Error: Signature Mismatch:
Values do not match:
val visit_expr : expr -> expr_ret
is not included in
val visit_expr : expr -> expr_ret
I know this means that I am not expressing the relationship between the types correctly, but I cannot figure out what the fix is in this case.
Disclaimer: Modules are just records of values accompanied with type definitions. Since there are no types in your modules there is no need to use them at all, just use plain old record types, and you will get one of the idiomatic AST traversal pattern. Soon, you will find out that you need an open recursion, and will switch to class-based approach. Anyway, this was the main reason, why classes were added to OCaml. You may also wrap your classes into a state monad, to fold AST with arbitrary user data.
What concerning your error message, then it is simple, you hide your types with signatures, a common mistake. The easiest solution is to omit an annotation of the return type of a functor at all, or to propagate type equality with a with type expr = expr annotations.
If you need examples of more idiomatic approaches, then for records you can go for ppx mappers, here is an example of different visitors implemented with classes, including those that are wrapped into a state monad.
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