Typescript 错误：类型 'number' 不可分配给类型 '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.

Answer 1

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.

Answer 2

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
        };

Answer 3

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];
}