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Allow tuple labels passed to satisfies to participate in type inferenceΒ #61415

@ahrjarrett

Description

@ahrjarrett

πŸ” Search Terms

tuple, tuples, label, labels, tuple labels, satisfies, as const

βœ… Viability Checklist

⭐ Suggestion

Hey TypeScript team! First of all, super excited to hear about tsgo.

I've got a feature request for y'all -- the body of this issue was ported from a playground that I created:

https://tsplay.dev/NljeQW

Basically, I often find myself using the satisfies operator to define mutable tuples inline:

const base = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] as const satisfies any[]
//    ^?  const base: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Sometimes when working with tuples of a known length, I use labels, because why not? Free semantics!

But to get the behavior I'm after, I have to add a layer of indirection.

const ex = [0, 0] as const
type Example = never | [x: typeof ex[0], y: typeof ex[1]]
const example: Example = [...ex]
//    ^? const example: [x: 0, y: 0]

This makes the code more confusing to read, which makes the tradeoff basically a wash.

Feature Request

Allow any labels passed to satisfies to participate in type inference, like so:

const point = [0, 0] as const satisfies [x: any, y: any]
// 
//     Desired type:
//     ^? const point: [x: 0, y: 0]
//                      ^     ^

Let me know if you'd like a more complex example, or if there's anything else I can provide.

πŸ“ƒ Motivating Example

Playground:

https://tsplay.dev/NljeQW

πŸ’» Use Cases

  1. What do you want to use this for?

Adding free semantics

  1. What shortcomings exist with current approaches?

Requires a layer of indirection + introduces the possibility of drift

  1. What workarounds are you using in the meantime?

Outlined in the example I shared

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