You're mixing up runtime with compile time. TS doesn't know in compile time that 'key1' exists. If you explicitly write const key1 = props['key1'] it will infer it as unknown, because you said that props[string] = unknown. - ThreadSky

About ThreadSky

trekinbami.bsky.social • 42 days ago

You're mixing up runtime with compile time. TS doesn't know in compile time that 'key1' exists. If you explicitly write const key1 = props['key1'] it will infer it as unknown, because you said that props[string] = unknown.

Comments

leekeh.com•42 days ago

but if I write key1: key1 as T['key1']` in the return statement, it will resolve it to string. What exactly is the difference?

leekeh.com•42 days ago

And why would it not know key1, but it would know the restprops?

trekinbami.bsky.social•41 days ago

It doesn't know what's in rest props. It just knows its T without key1 and key2.

leekeh.com•41 days ago

But at the same time the type does resolve properly, the result is typed as string. Anyway maybe my mind is just stubborn on this one ¯\_(ツ)_/¯

trekinbami.bsky.social•41 days ago

Yes, because key1 and key2 lost inference to T after destructuring (rest is inferred as Omit which makes sense). After you've captured T you can explictly tell it that key1 = T['key1']. Heres two ways to go about it.

https://tsplay.dev/NVBkBN

leekeh.com•40 days ago

Alright, somehow I cannot wrap around to how it can infer rest but not key1 and key2, but that's probable a me-problem 😅 The examples you pass are helpful, I guess I just wish it would not be needed to do double bookkeeping.
Thanks anyway for guiding me and I hope you are doing well! 🫶

Posting Rules

Be respectful to others
No spam or self-promotion
Stay on topic
Follow Bluesky's terms of service

Comments

Posting Rules

Reply