How to type async function in TypeScript when its a void function

Question

Here is my code:

async buildSomething(): any {
    const requestData = await request;
    requestData.forEach(i => this.table.push(i));
}

How I should type a void function, because it does something but It does not return anything.

In my case I used any but tslint shows me this:

Type 'any' is not a valid async function return type in ES5/ES3 because it does not refer to a Promise-compatible constructor value.

How should I achieve this?

is this unclear? why people put -1? dude, I just need some advice, thats all — Amos Isaila Lucian Onofrei
– Amos Isaila Lucian Onofrei, Commented Oct 28, 2020 at 14:26

Nicholas Tower · Accepted Answer · 2020-10-28 14:29:30Z

4

All async functions return something: they return promises. So you don't want void, you want Promise<void>

async buildSomething(): Promise<void> {
    const requestData = await request;
    requestData.forEach(i => this.table.push(i));
}

answered Oct 28, 2020 at 14:29

Nicholas Tower

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2 Comments

Amos Isaila Lucian Onofrei Over a year ago

yup, ur right, same as above, where I can learn about generics? every page has complex explanations, and Its hard for me trying to understand them

Nicholas Tower Over a year ago

The basics are that it's an [thing] of [another thing]. e.g, "Array<string>" is an "array of strings". Generics let you describe how the type works in general (ie, they let you describe what "of" means in "array of strings"), and then when you use it, you can be more specific. In the case of a promise, the thing you get to be more specific about is what the promise will resolve to. Promise<string> is a promise that resolves to a string; Promise<void> is a promise that resolves to void

ulvesked · Accepted Answer · 2020-10-28 14:29:02Z

1

The return type should be Promise<void>

answered Oct 28, 2020 at 14:29

ulvesked

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