Typescript infer variable type class

Question

I have 3 interfaces:

export interface Foo {a:string}
export interface Foo2 extends Foo {b:boolean}
export interface Foo3 extends Foo {c:number}

How do I make a class assignation without error?

class Goo {
  fooVar: Foo | Foo2 | Foo3;

  constructor() {
    this.fooVar.c = 0; <------ ERROR!
  }
}

All help is welcome :)

Possibly fooVar should use intersection rather than union types (Foo & Foo2 & Foo3), but it also depends on what you're doing specifically — Explosion Pills
– Explosion Pills, Commented Nov 28, 2018 at 21:30

Titian Cernicova-Dragomir · Accepted Answer · 2018-11-28 21:47:57Z

1

There are several options (If fooVar is really supposed to be either one of Foo, Foo2 or Foo3).

The simplest solution is a type assertion :

  class Goo {
    fooVar: Foo | Foo2 | Foo3 = {
      a: ""
    };

    constructor() {
      (this.fooVar as Foo3).c = 0
    }
  }

If c is already in fooVar and you only want to assign it if it already exist (ie fooVar is already Foo3) you can use an in type guard:

  class Goo {
    fooVar: Foo | Foo2 | Foo3 = {
      a: "", c: -1
    };

    constructor() {
      if('c' in this.fooVar) this.fooVar.c = 0
    }
  }

If you don't mind changing the object reference you can use spread :

  class Goo {
    fooVar: Foo | Foo2 | Foo3 = {
      a: "", 
    };

    constructor() {
      this.fooVar = {...this.fooVar, c: 0 }
    }
  }

answered Nov 28, 2018 at 21:47

Titian Cernicova-Dragomir

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2 Comments

kind user Over a year ago

Wouldn't be better to use intersection or extending types? type Foo extends...

Titian Cernicova-Dragomir Over a year ago

@kinduser I am not really sure what he is trying to achieve ... the question asked how you can assign a property on a union type (at least in my reading) I tried to offer options for that.

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