`shallowRef` value incompatible with parameter of the same type

[andrewbrennanfr]

Vue version

3.3.12

Link to minimal reproduction

https://stackblitz.com/edit/vue3-vite-starter-nvhwqh?file=src%2Fissue.ts

Steps to reproduce

• Use 3.3.12

• Create a shallowRef with a generic type that extends an interface T extends Base

• The value of shallowRef is incompatible as a parameter to functions that accept T

• Reproduce the same case with 3.3.11

• There is no issue

ℹ️ This is still an issue on 3.4.x also.

What is expected?

Here I would expect the value of my shallowRef T to be compatible with T.

What is actually happening?

Argument of type 'Base | T' is not assignable to parameter of type 'T'.

System Info

System:
  OS: macOS 14.3.1
  CPU: (8) arm64 Apple M1
  Memory: 335.19 MB / 16.00 GB
  Shell: 5.9 - /bin/zsh
Binaries:
  Node: 20.11.0 - ~/.nvm/versions/node/v20.11.0/bin/node
  Yarn: 1.22.21 - ~/.nvm/versions/node/v20.11.0/bin/yarn
  npm: 10.2.4 - ~/.nvm/versions/node/v20.11.0/bin/npm
  pnpm: 8.15.1 - ~/.nvm/versions/node/v20.11.0/bin/pnpm
  bun: 1.0.15 - ~/.bun/bin/bun
  Watchman: 2023.06.12.00 - /opt/homebrew/bin/watchman
Browsers:
  Chrome: 121.0.6167.184
  Chrome Canary: 123.0.6312.0
  Edge: 121.0.2277.128
  Safari: 17.3.1
npmPackages:
  vue: ^3.3.12 => 3.3.12

Any additional comments?

Likely related to this?
#9839

And wondering if it's not related to this comment specifically?

But then how should I correct do what I've described? ☝️

[Alfred-Skyblue]

Due to the fact that generics require types to be explicitly determined at their usage, unexpected types may arise within functions. It is suggested to use the following approach:

interface Base {
  start: number
}

export function useComposable<T extends Base>(callback: (value: T) => void) {
  const myRef: ShallowRef<T | null> = shallowRef(null)
  if (myRef.value) {
    callback(myRef.value)
  }
}

[andrewbrennanfr]

unexpected types may arise within functions

@Alfred-Skyblue do you have an example?
In this case - under what circumstance could myRef.value ever be incompatible with T in this case?

[Alfred-Skyblue]

@andrewbrennanfr #9564

[andrewbrennanfr]

@Alfred-Skyblue this example is for ref. shallowRef does not (at least not explicitly) return any Ref<UnwrapRef< types. Playpground

In my original reproduction case, changing shallowRef -> ref does introduce a warning about UnwapRef, but this isn't what's reported in my issue.

Also the answer you've linked, was written before version 3.3.11 (the version introducing my type issue) was released.

[Alfred-Skyblue]

I'm just giving a similar example. Due to the uncertainty of generics, we can't guarantee that it returns the passed generic type.

[andrewbrennanfr]

Ok seems a bit odd - then why even offer this in the first place, if the recommendation is to cast 😅 But OK understood. I can update my code. Thanks 🙏