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Unable to specify return type of class extends T implements IFooable #15001

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AllNamesRTaken opened this issue Apr 4, 2017 · 6 comments
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In Discussion Not yet reached consensus Suggestion An idea for TypeScript

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@AllNamesRTaken
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TypeScript Version: 2.2.1 / nightly (2.2.0-dev.201xxxxx)

Code

// A *self-contained* demonstration of the problem follows...
export type Constructor<T> = new (...args: any[]) => T;
export interface IFooable { FooProp: string; }
export function FooableMixin<T extends Constructor<{}>>(Base: T) {
    return class extends Base implements IFooable {
        FooProp: string;
        constructor(...args: any[]) {
            super(...args);
        }
    }
}
export class BaseBar { BarProp: string = "baz"; }
export class FooableBar extends FooableMixin ( BaseBar ) {}
let foobar = new FooableBar();
foobar.FooProp = foobar.BarProp; 
/* This is ok since FooableBar extends an Anonymous class inheriting from BaseBar and implementing IFooable */

If I turn on "declaration: true" in tsconfig.json this constuct breaks since the Anonymous class counts as private and not exported type (error TS4060) which is kind of natural since it is Anonymous.

What we need is a way to describe Classes as return types.

Expected behavior:

A d.ts file with a type declaration for the Mixin function Fooable that defines a Anonymous Class return type in the style of:

declare function Fooable(T extends Constructor<{}>): (class extends T implements IFooable);

Actual behavior:

TS4060 error when compiling with the "declaration" flag.
If casting the Anonymous Class return value to or T, the interface is lost.

@AllNamesRTaken
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Duplicate of #14075, #14597
Solution for now seem to be to:

  1. add return type : T & ICtor<IFooable> to the Mixin function to get rid of TS4060
  2. declare the union type as const before using it to extend.

This makes mixins much more cumbersome to use.

export const _FooableBar = FooableMixin(BaseBar);
export class FooableBar extends _FooableBar {}

@aluanhaddad
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I wonder if this would be resolvable with #6606

@mhegazy
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mhegazy commented Apr 27, 2017

class expressions are not allowed in type positions. the fix would be to allow it, and then emit it in the declaration emitter.

@mhegazy mhegazy added Suggestion An idea for TypeScript In Discussion Not yet reached consensus labels Apr 27, 2017
@trusktr
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trusktr commented Oct 14, 2019

A partial workaround is to declare the return type using types pre-defined outside of the function, but non-class types can not have protected members, and so inheriting protected members from the mixin class isn't possible (without manual casting).

@trusktr
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trusktr commented Oct 14, 2019

Nice!

Looks like a recent release (3.6.4?) has fixes that allow my lib to specify "types": "./src/index.ts" in package.json (pointing directly to a regular TS file, not a .d.ts declaration file!) without any issues.

This allows me to set "declaration": false in tsconfig.json, and rely on the actual code for type checking, which allows the return types to be inferred!

It does mean that I need to ship the source TS files instead of just declaration files.

@nuthinking
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I too have problems with declarations and mixins. Shipping the source is an option but I would prefer otherwise. Any idea on how to do that?

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