Typescript types don't compute

sebastian.galczynski

I ran into a typing problem after upgrading a library. This is the distilled form of the bug:

class Bar {
  id: string = '';
}

declare type ExcludeFunctions<T, K extends keyof T> = T[K] extends Function ? never : (K extends symbol ? never : K);


declare type Baz<T> = {
  [K in keyof T as ExcludeFunctions<T, K>]?: T[K];
};


class Base<T extends Bar> {

  public async foo(input: Baz<T>) {
   //Property 'id' does not exist on type 'Baz<T>'.
    input.id = 'x';
  }
}

Why doesn't Baz<Bar> have property 'id'? It looks like "string extends Function" evaluates to true, because when I remove this condition the error disappears. But string clearly doesn't extend Function. What is going on here?

Gribnit

@sebastian-galczynski I don't see what establishes Baz<T> as being an instance of Base<T>

sebastian.galczynski

@Gribnit said in Typescript types don't compute:

@sebastian-galczynski I don't see what establishes Baz<T> as being an instance of Base<T>

@Gribnit Nothing. I don't expect it to be. The problem is that type Baz<T>, where T extends Bar, doesn't have property id, while by my reasoning it should.

sebastian.galczynski

By comparison, this works:

class Bar {
  id: string = '';
}

declare type ExcludeFunctions<T, K extends keyof T> = K extends symbol ? never : K;


declare type Baz<T> = {
  [K in keyof T as ExcludeFunctions<T, K>]?: T[K];
};


class Base<T extends Bar> {

  public async foo(input: Baz<T>) {
    input.id = 'x';
  }
}

Which to me seems to imply that string extends Function, which is false.

Gribnit

@sebastian-galczynski I don't see how T for Baz would be the same as T for Base, so T in Baz doesn't seem to imply Bar. Am I missing a type definition fixing T?, or does Typescript make T consistent automagically? Looks like distinct generic types from here.

sebastian.galczynski

@Gribnit said in Typescript types don't compute:

Am I missing a type definition fixing T?

Yes. T in the method definition is the same T as in the class definition. The second example wouldn't work otherwise.

Gribnit

@sebastian-galczynski OK, I fold. I only understand Java generics.

djls45

@sebastian-galczynski said in Typescript types don't compute:

I ran into a typing problem after upgrading a library.

Well, then stop exploring the Carpal Tunnels of Alexandria.

This is the distilled form of the bug:

djls45

@sebastian-galczynski said in Typescript types don't compute:

~~[code snipped]~~

Why doesn't Baz<Bar> have property 'id'? It looks like "string extends Function" evaluates to true, because when I remove this condition the error disappears. But string clearly doesn't extend Function. What is going on here?

It looks to me like Baz<T extends Bar>.id should exist, I think, if I'm following the generics instantiations correctly.

djls45

@sebastian-galczynski said in Typescript types don't compute:

@Gribnit said in Typescript types don't compute:

Am I missing a type definition fixing T?

Yes. T in the method definition is the same T as in the class definition. The second example wouldn't work otherwise.

I think he means all the Ts through the whole file.

@Gribnit, to answer your question, I think it could be re-written like so:

class Bar {
  id: string = '';
}

declare type ExcludeFunctions<efT, efK extends keyof efT> = efT[efK] extends Function ? never : (efK extends symbol ? never : efK);

declare type Baz<bT> = {
  [bK in keyof bT as ExcludeFunctions<bT, bK>]?: bT[bK];
};

class Base<baseT extends Bar> {

  public async foo(input: Baz<baseT>) {
    //Property 'id' does not exist on type 'Baz<baseT>'.
    input.id = 'x';
  }
}

@sebastian-galczynski, does the fact that the property in Baz is optional make a difference?