Typescript error：Type 'number' is not assignable to type 'never'

Question

interface Opts {
  onFrame: () => void;
  onAudioSample: null;
  emulateSound: boolean;
  sampleRate: number;
}

class NES {
    constructor(opts: Opts) {
        this.opts = {
            onFrame() { },
            onAudioSample: null,
            emulateSound: true,
            sampleRate: 44100,
        }

        if (typeof opts !== "undefined") {
            let key: keyof Opts
            for (key in this.opts) {
                if (typeof opts[key] !== "undefined") {
                    // got err here
                    this.opts[key] = opts[key];
                }
            }
        }
    }
    opts: Opts
}

you can cheak the error on TS playground：here

typescript: v3.8.3 err: Type 'number' is not assignable to type never. I don't understand why it is a never type.

Jeff Bowman · Accepted Answer · 2020-05-08 05:32:39Z

4

You are getting that error because TypeScript can't infer that this.opts[key] and opts[key] are the same type. If you were to cover up each half of the =:

opts[key] could be a Function, null, boolean, or number
this.opts[key] needs to receive a Function, null, boolean, or number depending on the value of key, which we don't know
the only supertype in common for Function, null, boolean, and number is never, so that's the type that TypeScript wants for this.opts[key]

Interestingly, if you were to extract this to an anonymous generic function, it works: Within a single assignment Typescript can infer and use the type Opts[K]. jcalz suggests a similar solution in this similar question.

if (typeof opts[key] !== "undefined") {
  // this.opts[key] = opts[key];
  (<K extends keyof Opts>(k: K) => { this.opts[k] = opts[k]; })(key);
}

typescript playground

That said, I would use the spread operator as in Mukesh Soni's answer, ending your assignment of this.opts with ...opts, or use Object.assign. There's a slight risk that the passed opts contains extra keys, but Typescript should ensure otherwise at compile time. (For that matter, if you're expecting opts to be optional and potentially incomplete, it should probably be defined as opts?: Partial<Opts>.)

class NES {
    constructor(opts?: Partial<Opts>) {
        this.opts = Object.assign({
            onFrame() { },
            onAudioSample: null,
            emulateSound: true,
            sampleRate: 44100,
        }, opts);
    }
    opts: Opts
}

See also: Object spread vs. Object.assign, which notes that the solutions are quite similar and both applicable for default options values.

answered May 8, 2020 at 5:32

Jeff Bowman

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2 Comments

Orange Over a year ago

thank you, your solution is good, I use type assert solved it, seems your solution is better.

Orange Over a year ago

my solution: (this.opts[key] as Opts[typeof key]) = opts[key]

Mukesh Soni · Accepted Answer · 2020-05-08 04:31:57Z

1

I don't know why you are getting that error. One other way to merge both the options might be to use spread operators.

this.opts = {
            onFrame() { },
            onAudioSample: null,
            emulateSound: true,
            sampleRate: 44100,
            ...opts
        };

edited May 8, 2020 at 4:31

answered May 8, 2020 at 4:24

Mukesh Soni

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2 Comments

Will Over a year ago

If they do it this way, then casting with 'as Opts' is unnecessary, right?

Mukesh Soni Over a year ago

True in this case since the instance variable has been typed outside the constructor.

Sifat Haque · Accepted Answer · 2020-05-08 04:24:12Z

0

If you put the as any after the property access, TypeScript will check that key is a valid key

if (typeof opts[key] !== "undefined") {
  (this.opts[key] as any) = opts[key];
}

answered May 8, 2020 at 4:24

Sifat Haque

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