Microsoft/TypeScript

TypeScript is a superset of JavaScript that compiles to clean JavaScript output.

Bruce Pascoe

@fatcerberus

I had that issue a while ago. ...args: never[] is no good since you can still call it with no args

Never found a good solution :(

Bruce Pascoe

@fatcerberus

If it’s generic though, what about F extends Function?

AnyhowStep

@AnyhowStep

No good, can't use the infer R syntax if you're using extends Function

Also, my concrete F will still end up having never in the param and will break ReturnType<> =x

Even if I'm using Function as the top type for functions

Keith Layne

@keithlayne

IIRC all function has is a name member.

raghanag

@raghanag

hi all, if the typescript class constructor has too many lines, are we going to stub all of it when we are writing unit tests

AnyhowStep

@AnyhowStep

What does that question even mean

Bruce Pascoe

@fatcerberus

@keithlayne Function doesn’t guarantee that it’s callable?

Because that’s kind of sucky if it doesn’t

Keith Layne

@keithlayne

I must have seen an augmentation instead of the actual definition, my bad

Gareth Jones

@G-Rath

ok want to see something interesting?

interface Mx {
  v: string;
}

const promiseTuple = async <T1, T2>(t1: T1, t2: Promise<T2>): Promise<[T1, T2]> =>
  [t1, await t2];

const doThings = async (): Promise<Array<[string, Mx[]]>> => Promise.all([
  promiseTuple('hello', Promise.resolve([{ v: 'sunshine' }])),
  promiseTuple('world', await [{ v: 'peace' }])
]);

doThings().then(r => r.forEach(console.log));

pretty sure that's a bug :grimacing:

AnyhowStep

@AnyhowStep

I'm pretty sure it's not

Gareth Jones

@G-Rath

whys that?

they have the same output

AnyhowStep

@AnyhowStep

(await x) does not give you an expression of type Promise<typeof x>, I'm sure

It's just syntactic sugar

Gareth Jones

@G-Rath

Yeah, that's what I'm questioning

why doesn't it give you an expression of that type? b/c as you say, it's syntactic sugar, so it does give that expression in Node (iirc)

AnyhowStep

@AnyhowStep

If it did return a promise, this would work,

function returnPromise () : Promise<number> {
  return (await 1);
}

Gareth Jones

@G-Rath

which it does

once you add the missing async keyword

AnyhowStep

@AnyhowStep

Exactly. The async part is the problem here

Gareth Jones

@G-Rath

oh ffs

no wait what

async is in my code

for a second I thought it was I'd just missed async, but nope: async is there

so your example doesn't really relate to mine, as far as I can tell

AnyhowStep

@AnyhowStep


function takesNumberOnly (x : number) {
  console.log(x+1);
}
async function returnPromise ()  {
  const x = await 1;
  takesNumberOnly(x);
  takesNumberOnly(await 1);
}

If (await 1) is an expression of type Promise<number>

Then the second call to takesNumberOnly would not work

Gareth Jones

@G-Rath

Yeah I see. and you're right; this is turning how I figured it would

it's contextual, but too complex to type like that

AnyhowStep

@AnyhowStep

It's sytactic sugar for,

function returnPromise2 ()  {
  Promise.resolve(1)
    .then((x) => {
      takesNumberOnly(x);
      Promise.resolve(1)
        .then((y) => {
          takesNumberOnly(y);
        });
    });
}

Gareth Jones

@G-Rath

yeah yeah

AnyhowStep

@AnyhowStep

gdi gitter highlighting

Gareth Jones

@G-Rath

you need to add ts

im actually slightly disppointed in myself for not putting that together XD

but cheers for the sanity check

AnyhowStep

@AnyhowStep

Sometimes, adding ts doesn't make it any better =x

No problem <3

Gareth Jones

@G-Rath

yeah I know

but so yeah: it's b/c of the await in the inner fn

also ugh ffs my whole fn is redundant XD which is annoying b/c I thought that originally :joy:

AnyhowStep

@AnyhowStep

I didn't even notice till you said it =x

I need to up my code review skillz

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