In some of the answers here, I’ve seen some incorrect use of terminology, which has lead people to provide wrong answers. So, before I give my answer, I’m going to provide a few definitions: Strongly typed A language is strongly typed if it enforces the type safety of a program. That means that it guarantees … Read more
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 … Read more
Usually it’s to signal usage of a class. You can implement IMessage to signal that your class is a message. Other code can then use reflection to see if your objects are meant to be used as messages and act accordingly. This is something that was used in Java a lot before they had annotations. … Read more
You can define the parameter to be optional with ?: interface SomeInterface { SomeMethod(arg1: string, arg2: string, arg3?: boolean); }
// How to extend Validator interface adding isArray() method?? You cannot do this in a file that is a module (some guidance here) and your file is a module because you have import expressValidator. Instead create a extendedValidator.d.ts and add the new stuff for TypeScript’s engine: declare module ExpressValidator { export interface Validator { isArray: … Read more
TL;DR: The closest to interface in Rust is a trait. However, do not expect it to be similar in all point to an interface. My answer does not aim to be exhaustive but gives some elements of comparison to those coming from other languages. If you want an abstraction similar to interface, you need to … Read more
You only have to check if a value implements an interface if you don’t know the value’s type. If the type is known, that check is automatically done by the compiler. If you really want to check anyways, you can do it with the second method you gave: var _ Somether = (*MyType)(nil) which would … Read more
It’s worth noting that Partial<T>, as suggested in the accepted answer, makes all fields optional, which is not necessarily what you need. If you want to make some fields required (e.g. id and email), you need to combine it with Pick: type UserWithOptionalPhone = Pick<IUser, ‘id’ | ’email’> & Partial<IUser> Some explanation: What Pick does … Read more
There are examples in the documentation for anonymous classes, but not for interfaces. This is how I created an instance of an interface: fun TileSet.union(another: TileSet) : TileSet = object : TileSet { override fun contains(x: Int, y: Int) : Boolean = this@union.contains(x, y) || another.contains(x, y) } Notice that, unlike in the example from … Read more
No, interfaces are not supported for DI. With TypeScript interfaces are not available at runtime anymore, only statically and therefore can’t be used as DI tokens. Alternatively you can use strings as keys or InjectionToken provide(‘CoursesServiceInterface’, {useClass: CoursesServiceMock}) // old providers: [{provide: ‘CoursesServiceInterface’, useClass: CoursesServiceMock}] and inject it like constructor(@Inject(‘CoursesServiceInterface’) private coursesService:CoursesServiceInterface) {} See also … Read more