TypeScript 断言在属性为记录时不起作用<number,number></number,number>

Question

I learn from some blog that T as K in typescript only work when T is subType of K or K is subType of T ,I don't know whether it's true of not.

And when I using as when the property is Record<T,K> , as throw some error as below.

interface Test {
    x: Record<number, number>;
    y: number;
}


// this is wrong, but {1: 1} should be subType of Record<number,number>
const x1 = {
    x: { 1: 1 },
} as Test;

// this is okay
const x2 = {
    y: 1,
} as Test;

// this is okay too
const x3 = {
    x: { 1: 1 },
    y: 1,
} as Test;

First, I guess It's because { 1: 1 } doesn't match the Type Record<number, number> , I guess the subType judge is shallow. But here is another sample that can work.

interface SubTest {
    x: 1;
    y: 1;
}
interface Test2 {
    x: SubTest;
    y: number;
}

// { x: 1 } is subType of SubTest, and It works!
const x4 = {
    x: { x: 1 },
} as Test2;

You can see {x:1} doesn't match SubTest too, but this assertion can work, so I guess it's a special problem of Record<T,K> ,can anyone tell me why? thanks a lot!

---

I found out this one can work. so I'm really cofunsed right now...

interface Test {
    x: Record<number, number>;
    y: number;
}

const x4 = {
    x: { a: 1 },
    y: 1,
} as Test;

Answer 1

The problem with below is not with the Record . In interface Test you have defined y as mandatory but you have not provided y here.

const x1 = {
    x: { 1: 1 },
} as Test;

Convert y to optional like below.

Change this

interface Test {
    x: Record<number, number>;
    y: number;
}

To

interface Test {
    x: Record<number, number>;
    y?: number;
}

Now y is made optional with ? . This should make your code compile.