Typescript: Why destructuring an object doesn't hold type?

Question

In the following, I would expect this to fail as I've stated afternoon should be of type number, but it doesn't fail.

My guess is I'm destructuring incorrectly as hovering over the word afternoon when it returns shows the type as any (in VS Code).

Alternatively, it could all be wrong, this is literally my first play with Typescript, not getting it, yet!

const greetings =  {
    morning: "Good morning.",
    evening: "Good evening.",
    afternoon: "Good afternoon.",
};

interface AboutGreetingTypes {
    morning: string,
    evening: string,
    afternoon: number,
};

const aboutGreeting = ({copy}): AboutGreetingTypes => {
    const d = new Date();
    const timeInHours = d.getHours();

    const { morning, evening, afternoon } = copy;

    if (timeInHours < 12) {
        return morning;
    }
    if (timeInHours > 18) {
        return evening;
    }
    if (timeInHours > 12) {
        return afternoon;
    }
};

console.log(aboutGreeting({copy: greetings}));

It looks like afternoon has the wrong type? afternoon: "Good afternoon.", vs: afternoon: number, in the interface does not match. I don't get how copy got into the mix. — Zargold
– Zargold, Commented May 2, 2022 at 17:21
You'll want to spend some dedicated time reading through the TypeScript handbook. Here's a link to getting started for JS programmers. — jsejcksn
– jsejcksn, Commented May 2, 2022 at 18:58

Pedro Feltrin · Accepted Answer · 2022-05-02 17:22:14Z

2

Your function parameter type should be {copy: AboutGreetingTypes} not just AboutGreetingTypes

answered May 2, 2022 at 17:22

Pedro Feltrin

4482 silver badges8 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

StudioTime Over a year ago

Not sure what you mean by that.

Alex Wayne · Accepted Answer · 2022-05-02 17:30:56Z

1

const aboutGreeting = ({copy}): AboutGreetingTypes => {

This line declares a function that accepts a single untyped object and returns an AboutGreetingTypes. Since the argument has no declared type, it is assumed to be any, which lets you do anything without without a type error.

What I think you want is a function that accepts an object that has a copy property that has value of an AboutGreetingTypes.

That would look like this:

const aboutGreeting = ({ copy }: { copy: AboutGreetingTypes }) => {

See playground which still has some errors, but those resolving those are answers to different questions.

answered May 2, 2022 at 17:30

Alex Wayne

188k52 gold badges328 silver badges363 bronze badges

Collectives™ on Stack Overflow

Typescript: Why destructuring an object doesn't hold type?

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related