Upcast mandatory when there are different overload - casting

This is not about windows forms at all it's here only for the "background".
I was toying around Windows Forms when I got an error on an AddRange for a MenuStrip.Items requiring to cast ToolStripMenuItem into ToolStripItem
But I already have an AddRange for a Form.Controls before which didn't require casts.
After a little experimentation I managed to find that the error occurs when there are multiple overload for that AddRange so I tried to validate my thought :
type Foo () = class end
type Bar () = inherit Foo ()
type FooCollection () = class end // not really necessary
type Test1 () =
member __.AddRange (col: FooCollection) = () // could be an int or anything instead
member __.AddRange (foos: Foo []) = ()
type Test2 () = member __.AddRange (foos: Foo []) = ()
let lst1, lst2 = Test1 (), Test2 ()
lst1.AddRange [|Bar ()|] // error: have to explicitely cast => [|Bar () :> Foo|]
lst2.AddRange [|Bar ()|] // works
The question is simply why ; from my point of view the call is not ambiguous

After reading the 14.4.3 F# spec (hinted by Gustavo, kudos to him)
The F# compiler determines whether to insert flexibility after explicit instantiation, but before any
arguments are checked.
I understand that flexibility is never inserted for a method which have overloads because it would need argument checking to choose.

Related

type narrowing not work if type is specified?

class Foo
def initialize(#foo : String | Nil)
end
def foo
#foo
end
end
a = Foo.new "213"
if !a.foo.nil?
puts a.foo, typeof(a.foo)
end
get output of
213
(String | Nil)
but shouldn't type of a.foo be narrowed to String? Is this another design limit?
The compiler doesn't know that #foo doesn't change. Say your class Foo has a setter for #foo. If some concurrently running code uses that setter to set #foo to nil, the second call to Foo#foo inside the if condition might return nil now even though the check passed before.
You can remedy by introducing a local variable which the compiler can reason about:
if foo = a.foo
puts foo, typeof(foo)
end

OCaml - confusion from type alias (warning 40)

I don't understand why OCaml is not able to figure out that there is no room for confusion here: anint below can't be other one but A's.
module A = struct
type test = Graphics.status
end
module type ASIG = sig
type test = A.test
val atest : test
end
module Func (H : ASIG) = struct
let _ = let open H in atest.key
end
However, it raises
Warning 40: key was selected from type Graphics.status.
It is not visible in the current scope, and will not
be selected if the type becomes unknown.
How can I tell it "it's fine" without disabling the warning?
I'm aware I can solve it by opening A. However, if H defines its own functions and types similar---but not equal---to A, then it will have unnecessary clashes. I also know I can duplicate the definition, but that defeats the purpose of type aliasing, and involves lots of unnecessary code duplication. Perhaps there is no solution, but I wonder why OCaml is so blind dumb on this one: type alias should mean also constructor and record fields alias, shouldn't it?
You can simply open the the module defining the original type locally when referring to the field key as in the following:
module A = struct
type test = Graphics.status
end
module type ASIG = sig
type test = A.test
val atest : test
end
module Func (H : ASIG) = struct
let _ = let open H in atest.Graphics.key
end
Or if you need to refer to several fields :
let _ = let open H in Graphics.(atest.key, atest.button)
Well, this happens because the module signature ASIG needs to look the definition of type test for the implementation of A. This often causes problems with visibility of the types, and sometimes require duplication of type definitions, where the contract satisfies the implementation instead of referring to it.
How can we fix this warning? In ASIG, instead of defining type test = A.test, we need to explicitly do type test = { anint: int }as we did in the implementation, so:
module ASIG = sig
type test = { anint: int }
val atest : test
end
module A = struct
type test = { anint: int }
end
module Func (H : ASIG) = struct
let _ = let open H in atest.anint
end
The H module would not be able to view anintin its scope, otherwise, because the signature has a type (contract) that links to the implementation. It is also a core concept of OCaml philosophy isolating signatures and implementations and avoiding signatures depending upon implementations.

Applying implicit generic parameters to a list via ClassTag

I would like to apply a function to all objects in a list, where all objects in the list inherit from a common class. In this function, I would like to use an implicit class to ensure that the correct operation is applied based on the object's type.
For example, I want to ensure that all Employee objects in a list are converted using the employeeConverter below. Calling convert with the Employee directly works just fine, but applying convert to a list of Employee objects is a compiler error.
import scala.reflect.ClassTag
object Example {
abstract class Person { def age: Int }
case class Employee(age: Int) extends Person
class Converter[T] { def convert(t: T) = (t,t) }
def convert[T <: Person:ClassTag](p: T)(implicit converter: Converter[T]) =
converter.convert(p)
def main(args: Array[String]): Unit = {
implicit val employeeConverter = new Converter[Employee]()
println(convert(Employee(1)))
//println(List(Employee(2)) map convert) // COMPILER ERROR
}
}
The above code correctly prints the following:
$ scalac Example.scala && scala Example
(Employee(1),Employee(1))
However, if I uncomment the line indicated with COMPILER ERROR, I get this compiler error:
Example.scala:20: error: could not find implicit value for parameter converter: Example.Converter[T]
println(l map convert)
^
Is this a problem that can be resolved using ClassTag? How can I modify this example to apply convert to a list?
The compiler needs a little bit of hand-holding in this case. This works:
println(List(Employee(2)) map { e => convert(e) })

Build a function object with properties in TypeScript

I want to create a function object, which also has some properties held on it. For example in JavaScript I would do:
var f = function() { }
f.someValue = 3;
Now in TypeScript I can describe the type of this as:
var f: { (): any; someValue: number; };
However I can't actually build it, without requiring a cast. Such as:
var f: { (): any; someValue: number; } =
<{ (): any; someValue: number; }>(
function() { }
);
f.someValue = 3;
How would you build this without a cast?
Update: This answer was the best solution in earlier versions of TypeScript, but there are better options available in newer versions (see other answers).
The accepted answer works and might be required in some situations, but have the downside of providing no type safety for building up the object. This technique will at least throw a type error if you attempt to add an undefined property.
interface F { (): any; someValue: number; }
var f = <F>function () { }
f.someValue = 3
// type error
f.notDeclard = 3
This is easily achievable now (typescript 2.x) with Object.assign(target, source)
example:
The magic here is that Object.assign<T, U>(t: T, u: U) is typed to return the intersection T & U.
Enforcing that this resolves to a known interface is also straight-forward. For example:
interface Foo {
(a: number, b: string): string[];
foo: string;
}
let method: Foo = Object.assign(
(a: number, b: string) => { return a * a; },
{ foo: 10 }
);
which errors due to incompatible typing:
Error: foo:number not assignable to foo:string
Error: number not assignable to string[] (return type)
caveat: you may need to polyfill Object.assign if targeting older browsers.
TypeScript is designed to handle this case through declaration merging:
you may also be familiar with JavaScript practice of creating a function and then extending the function further by adding properties onto the function. TypeScript uses declaration merging to build up definitions like this in a type-safe way.
Declaration merging lets us say that something is both a function and a namespace (internal module):
function f() { }
namespace f {
export var someValue = 3;
}
This preserves typing and lets us write both f() and f.someValue. When writing a .d.ts file for existing JavaScript code, use declare:
declare function f(): void;
declare namespace f {
export var someValue: number;
}
Adding properties to functions is often a confusing or unexpected pattern in TypeScript, so try to avoid it, but it can be necessary when using or converting older JS code. This is one of the only times it would be appropriate to mix internal modules (namespaces) with external.
So if the requirement is to simply build and assign that function to "f" without a cast, here is a possible solution:
var f: { (): any; someValue: number; };
f = (() => {
var _f : any = function () { };
_f.someValue = 3;
return _f;
})();
Essentially, it uses a self executing function literal to "construct" an object that will match that signature before the assignment is done. The only weirdness is that the inner declaration of the function needs to be of type 'any', otherwise the compiler cries that you're assigning to a property which does not exist on the object yet.
EDIT: Simplified the code a bit.
Old question, but for versions of TypeScript starting with 3.1, you can simply do the property assignment as you would in plain JS, as long as you use a function declaration or the const keyword for your variable:
function f () {}
f.someValue = 3; // fine
const g = function () {};
g.someValue = 3; // also fine
var h = function () {};
h.someValue = 3; // Error: "Property 'someValue' does not exist on type '() => void'"
Reference and online example.
As a shortcut, you can dynamically assign the object value using the ['property'] accessor:
var f = function() { }
f['someValue'] = 3;
This bypasses the type checking. However, it is pretty safe because you have to intentionally access the property the same way:
var val = f.someValue; // This won't work
var val = f['someValue']; // Yeah, I meant to do that
However, if you really want the type checking for the property value, this won't work.
I can't say that it's very straightforward but it's definitely possible:
interface Optional {
<T>(value?: T): OptionalMonad<T>;
empty(): OptionalMonad<any>;
}
const Optional = (<T>(value?: T) => OptionalCreator(value)) as Optional;
Optional.empty = () => OptionalCreator();
if you got curious this is from a gist of mine with the TypeScript/JavaScript version of Optional
An updated answer: since the addition of intersection types via &, it is possible to "merge" two inferred types on the fly.
Here's a general helper that reads the properties of some object from and copies them over an object onto. It returns the same object onto but with a new type that includes both sets of properties, so correctly describing the runtime behaviour:
function merge<T1, T2>(onto: T1, from: T2): T1 & T2 {
Object.keys(from).forEach(key => onto[key] = from[key]);
return onto as T1 & T2;
}
This low-level helper does still perform a type-assertion, but it is type-safe by design. With this helper in place, we have an operator that we can use to solve the OP's problem with full type safety:
interface Foo {
(message: string): void;
bar(count: number): void;
}
const foo: Foo = merge(
(message: string) => console.log(`message is ${message}`), {
bar(count: number) {
console.log(`bar was passed ${count}`)
}
}
);
Click here to try it out in the TypeScript Playground. Note that we have constrained foo to be of type Foo, so the result of merge has to be a complete Foo. So if you rename bar to bad then you get a type error.
NB There is still one type hole here, however. TypeScript doesn't provide a way to constrain a type parameter to be "not a function". So you could get confused and pass your function as the second argument to merge, and that wouldn't work. So until this can be declared, we have to catch it at runtime:
function merge<T1, T2>(onto: T1, from: T2): T1 & T2 {
if (typeof from !== "object" || from instanceof Array) {
throw new Error("merge: 'from' must be an ordinary object");
}
Object.keys(from).forEach(key => onto[key] = from[key]);
return onto as T1 & T2;
}
This departs from strong typing, but you can do
var f: any = function() { }
f.someValue = 3;
if you are trying to get around oppressive strong typing like I was when I found this question. Sadly this is a case TypeScript fails on perfectly valid JavaScript so you have to you tell TypeScript to back off.
"You JavaScript is perfectly valid TypeScript" evaluates to false. (Note: using 0.95)

Ocaml - Forward Declaration (Classes)

I need to have two classes refering to each other. Is there any way in Ocaml to make Forward Declaration of one of them?
(I don't think it's possible as with easier stuff with word and).
Or maybe it is possible, but different way than how i tried?
Ocaml doesn't have anything like forward declarations (i.e. a promise that something will be defined eventually), but it has recursive definitions (i.e. a block of things that are declared and then immediately defined in terms of each other). Recursive definitions are possible between expressions, types, classes, and modules (and more); mutually recursive modules allow mixed sets of objects to be defined recursively.
You can solve your problem using a recursive definition with the keyword and:
class foo(x : bar) = object
method f () = x#h ()
method g () = 0
end
and bar(x : foo) = object
method h () = x#g()
end
Or you could use parameterized classes. Following the previous example you have:
class ['bar] foo (x : 'bar) =
object
method f () = x#h ()
method g () = 0
end
class ['foo] bar (x : 'foo) =
object
method h () = x#g()
end
The inferred interface is:
class ['a] foo : 'a ->
object
constraint 'a = < h : unit -> 'b; .. >
method f : unit -> 'b
method g : unit -> int
end
class ['a] bar : 'a ->
object
constraint 'a = < g : unit -> 'b; .. >
method h : unit -> 'b
end