Using Class<T> as a Map key in Haxe - inversion-of-control

I'd like to store instances of models in a common provider using their classes or interfaces as a keys and then pop them up by class references. I have written some code:
class Provider {
public function new() { }
public function set<T:Any>(instance:T, ?type:Class<T>) {
if (type == null)
type = Type.getClass(instance);
if (type != null && instance != null)
map.set(type, instance);
}
public function get<T:Any>(type:Class<T>):Null<T> {
return cast map.get(type);
}
var map = new Map<Class<Any>, Any>();
}
...alas, it's even doesn't compile.
Probably I have to use qualified class name as a key rather than class/interface reference? But I'd like to keep neat get function design that takes type as argument and returns object just of type taken, without additional type casting.
Is it possible or should I change my approach to this problem?

The issue of using Class<T> as a Map key come up every so often, here is a related discussion. The naive approach of Map<Class<T>, T> fails to compile with something like this:
Abstract haxe.ds.Map has no #:to function that accepts haxe.IMap<Class<Main.T>, Main.T>`
There's several different approaches to this problem:
One can use Type reflection to obtain the fully qualified name of a class instance, and then use that as a key in a Map<String, T>:
var map = new Map<String, Any>();
var name = Type.getClassName(Main);
map[name] = value;
For convenience, you would probably want to have a wrapper that does this for you, such as this ClassMap implementation.
A simpler solution is to simply "trick" Haxe into compiling it by using an empty structure type ({}) as the key type. This causes ObjectMap to be chosen as the underlying map implementation.
var map = new Map<{}, Any>();
map[Main] = value;
However, that allows you to use things as keys that are not of type Class<T>, such as:
map[{foo: "bar"}] = value;
The type safety issues of the previous approach can be remedied by using this ClassKey abstract:
#:coreType abstract ClassKey from Class<Dynamic> to {} {}
This still uses ObjectMap as the underlying map implementation due to the to {} implicit cast. However, using a structure as a key now fails at compile time:
var map = new Map<ClassKey, Any>();
map[{foo: "bar"}] = value; // No #:arrayAccess function accepts arguments [...]

Related

Why Class.java don't provide methods like `Type getType()` and `Type getGenericType()`?

Why I have to define a subclass to get the Type of superclass' generic param? Is the limit necessary?
I read the code of Fastjson of Alibaba and tried to figure out why use TypeReference must create an anonymous subclass. Then I found that an object cannot get its own generic param Type even its own Type.
public class TypeReference {
static ConcurrentMap<Type, Type> classTypeCache
= new ConcurrentHashMap<Type, Type>(16, 0.75f, 1);
protected final Type type;
protected TypeReference() {
Type superClass = getClass().getGenericSuperclass();
Type type = ((ParameterizedType) superClass).getActualTypeArguments()[0];
Type cachedType = classTypeCache.get(type);
if (cachedType == null) {
classTypeCache.putIfAbsent(type, type);
cachedType = classTypeCache.get(type);
}
this.type = cachedType;
}
// ...
}
Sorry for my poor English. Thanks for your answers.
Because of Type Erasure.
Consider the following example
List<String> stringList = new ArrayList<>();
List<Number> numberList = new ArrayList<>();
System.out.println(stringList.getClass() == numberList.getClass());
This will print true. Regardless of the generic type, both instances of ArrayList have the same class and a single Class object. So how could this single Class object return the right Type for both objects?
We can even get a step further,
List<String> stringList = Collections.emptyList();
List<Number> numberList = Collections.emptyList();
System.out.println(stringList == (Object)numberList);
Objects do not know their generic type. If a collection is immutable and always empty, it can be used to represent arbitrary empty lists. The same applies to stateless functions
Function<String, String> stringFunction = Function.identity();
Function<Number, Number> numberFunction = Function.identity();
System.out.println(stringFunction == (Object)numberFunction);
Prints true (on most systems; this is not a guaranteed behavior).
Generic types are only retained in some specific cases, like the signatures of field and method declarations and generic super types.
That’s why you need to create a subclass to exploit the fact that it will store the declared generic supertype. While it sometimes would be useful to construct a Type instance in a simpler way and a suitable factory method can be regarded a missing feature, getting the actual generic type of an arbitrary object (or its Class) is not possible in general.

A way to read a String as dart code inside flutter?

I want to build a method to dynamically save attributes on a specific object
given the attribute name and the value to save I call the "save()" function to update the global targetObj
var targetObj = targetClass();
save(String attribute, String value){
targetObj.attribute = value;
print(targetObj.attribute);
}
But I'm getting the following error:
Class 'targetClass' has no instance setter 'attribute='.
Receiver: Instance of 'targetClass'
Tried calling: attribute="Foo"
The only thing that I can think of is that "attribute" due to being type String results in an error.
That lead me to think if there is a way to read a String as code, something like eval for php.
As #Randal mentioned, you cannot create class..method at runtime. Still, you can try something like this.
A certain class
class Foo {
dynamic bar1;
dynamic bar2;
// ...
}
Your save method
save(Foo fooObject, String attribute, dynamic value) {
if ("bar1" == attribute) fooObject.bar1 = value;
else if ("bar2" == attribute) fooObject.bar2 == value;
// ...
}
Dart (and thus flutter) does not have a way to compile and execute code at runtime (other than dart:mirrors, which is deprecated). You can build additional code that derives from other code using the various builder mechanisms, although it can be rather complicated to implement (and use!).

Why can't I create a callback for the List Find method in Moq?

I created an extension method that lets me treat a List as DbSet for testing purposes (actually, I found this idea in another question here on stack overflow, and it's been fairly useful). Coded as follows:
public static DbSet<T> AsDbSet<T>(this List<T> sourceList) where T : class
{
var queryable = sourceList.AsQueryable();
var mockDbSet = new Mock<DbSet<T>>();
mockDbSet.As<IQueryable<T>>().Setup(m => m.Provider).Returns(queryable.Provider);
mockDbSet.As<IQueryable<T>>().Setup(m => m.Expression).Returns(queryable.Expression);
mockDbSet.As<IQueryable<T>>().Setup(m => m.ElementType).Returns(queryable.ElementType);
mockDbSet.As<IQueryable<T>>().Setup(m => m.GetEnumerator()).Returns(queryable.GetEnumerator());
mockDbSet.Setup(d => d.Add(It.IsAny<T>())).Callback<T>(sourceList.Add);
mockDbSet.Setup(d => d.Find(It.IsAny<object[]>())).Callback(sourceList.Find);
return mockDbSet.Object;
}
I had been using Add for awhile, and that works perfectly. However, when I try to add the callback for Find, I get a compiler error saying that it can't convert a method group to an action. Why is sourceList.Add an Action, but sourceList.Find is a method group?
I'll admit I'm not particularly familiar with C# delegates, so it's likely I'm missing something very obvious. Thanks in advance.
The reason Add works is because the List<T>.Add method group contains a single method which takes a single argument of type T and returns void. This method has the same signature as an Action<T> which is one of the overloads of the Callback method (the one with a single generic type parameter, Callback<T>), therefore the List<T>.Add method group can be converted to an Action<T>.
With Find, you are trying to call the Callback method (as opposed to Callback<T>) which expects an Action parameter (as opposed to Action<T>). The difference here is that an Action does not take any parameters, but an Action<T> takes a single parameter of type T. The List<T>.Find method group cannot be converted to an Action because all the Find methods (there is only one anyway) take input parameters.
The following will compile:
public static DbSet<T> AsDbSet<T>(this List<T> sourceList) where T : class
{
var mockDbSet = new Mock<DbSet<T>>();
mockDbSet.Setup(d => d.Find(It.IsAny<object[]>())).Callback<Predicate<T>>(t => sourceList.Find(t));
return mockDbSet.Object;
}
Note that I have called .Callback<Predicate<T>> because the List<T>.Find method expects and argument of type Predicate. Also note I have had to write t => sourceList.Find(t) instead of sourceList.Find because Find returns a value (which means it doesn't match the signature of Action<Predicate<T>>). By writing it as a lambda expression the return value will be thrown away.
Note that although this compiles it will not actually work because the DbSet.Find method actually takes an object[] for it's parameter, not a Predicate<T>, so you will likely have to do something like this:
public static DbSet<T> AsDbSet<T>(this List<T> sourceList) where T : class
{
var mockDbSet = new Mock<DbSet<T>>();
mockDbSet.Setup(d => d.Find(It.IsAny<object[]>())).Callback<object[]>(keyValues => sourceList.Find(keyValues.Contains));
return mockDbSet.Object;
}
This last point has more to do with how to use the Moq library that how to use method groups, delegates and lambdas - there is all sorts of syntactic sugar going on with this line which is hiding what is actually relevant to the compiler and what isn't.

Polymorphism in Object construction

I want to create specific Object according to the type argument.
Pseudo code looks like this.
sub new {
my $type = shift;
if($type eq "S1") {$interface = X->new(); }
if($type eq "S2") {$interface = Y->new(); }
etc...
return $interface;
}
Options might be:
Substitute "package" name with $type argument. Requires package name coordination with $type.
Use Hash{S1 => X} in the Master constructor to select Value according to $type passed. Requires Hash maintenance when adding new
Object types.
I don't like any of above. Looking trully polimorphic way to accomplish that.
Thank You,
k
Your best option would likely be to use a factory pattern. A factory method takes the parameters for creating an instance of your class, then decides which object to instantiate and return from that. This can also make dependency injection easier for testing.
You'd probably be looking at something like this (in Java-esque code), with an employee object:
public class EmployeeFactory
{
public static create(String type)
{
switch (type) {
case type1:
return new EmployeeTypeOne();
case type2:
return new EmployeeTypeTwo();
default:
throw new Exception("Unrecognized type");
}
}
}
Your employees would inherit from a common interface or abstract class. You can use the factory to handle constructor parameters as well if you prefer, just try to keep things fairly reasonable (don't pass a million parameters - the factory should internally handle complex objects)
See http://refactoring.com/catalog/replaceConstructorWithFactoryMethod.html for more information.
You might like Module::PluginFinder for that. Create all your specific types in a specific namespace and give them each some identifying (constant? sub?) that the main dispatcher will then use to identify which class handles a given type.

Implementing TypeScript interface with bare function signature plus other fields

How do I write a class that implements this TypeScript interface (and keeps the TypeScript compiler happy):
interface MyInterface {
(): string;
text2(content: string);
}
I saw this related answer:
How to make a class implement a call signature in Typescript?
But that only works if the interface only has the bare function signature. It doesn't work if you have additional members (such as function text2) to be implemented.
A class cannot implement everything that is available in a typescript interface. Two prime examples are callable signatures and index operations e.g. : Implement an indexible interface
The reason is that an interface is primarily designed to describe anything that JavaScript objects can do. Therefore it needs to be really robust. A TypeScript class however is designed to represent specifically the prototype inheritance in a more OO conventional / easy to understand / easy to type way.
You can still create an object that follows that interface:
interface MyInterface {
(): string;
text2(content: string);
}
var MyType = ((): MyInterface=>{
var x:any = function():string { // Notice the any
return "Some string"; // Dummy implementation
}
x.text2 = function(content:string){
console.log(content); // Dummy implementation
}
return x;
}
);
There's an easy and type-safe way to do this with ES6's Object.assign:
const foo: MyInterface = Object.assign(
// Callable signature implementation
() => 'hi',
{
// Additional properties
text2(content) { /* ... */ }
}
)
Intersection types, which I don't think were available in TypeScript when this question was originally asked and answered, are the secret sauce to getting the typing right.
Here's an elaboration on the accepted answer.
As far as I know, the only way to implement a call-signature is to use a function/method. To implement the remaining members, just define them on this function. This might seem strange to developers coming from C# or Java, but I think it's normal in JavaScript.
In JavaScript, this would be simple because you can just define the function and then add the members. However, TypeScript's type system doesn't allow this because, in this example, Function doesn't define a text2 member.
So to achieve the result you want, you need to bypass the type system while you define the members on the function, and then you can cast the result to the interface type:
//A closure is used here to encapsulate the temporary untyped variable, "result".
var implementation = (() => {
//"any" type specified to bypass type system for next statement.
//Defines the implementation of the call signature.
var result: any = () => "Hello";
//Defines the implementation of the other member.
result.text2 = (content: string) => { };
//Converts the temporary variable to the interface type.
return <MyInterface>result;
})(); //Invokes the closure to produce the implementation
Note that you don't need to use a closure. You could just declare your temporary variable in the same scope as the resulting interface implementation. Another option is to name the closure function to improve readability.
Here's what I think is a more realistic example:
interface TextRetriever {
(): string;
Replace(text: string);
}
function makeInMemoryTextRetriever(initialText: string) {
var currentText = initialText;
var instance: any = () => currentText;
instance.Replace = (newText: string) => currentText = newText;
return <TextRetriever>instance;
}
var inMemoryTextRetriever = makeInMemoryTextRetriever("Hello");