Certain conditional types allow unsound assignments

[mkantor]

🔎 Search Terms

"conditional type" generic assignment

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about conditional types

⏯ Playground Link

https://www.typescriptlang.org/play?#code/C4TwDgpgBAQg9gDwDwBUB8UC8UDeUBuAhgDYCuEAXFClAL4BQ9okUAqgHYBGiAku+xABOqDNhoQEwCOwAmAZ1iIkAS3YAzIVABKGAPxsuvfkKQ6oVPKUMIIMqjQbNoHbsnRYDrvgOHoA2gDkVq62AQC6jDIQAMbEhILQalbRwMpw7FDBSugAFAj2AJRULtlojEnsKWkZEoQAtmDEECJ5hbj0UFBNwFCEHlkIeQUdvR4BwAAWygpyE3CkxDJQnNCENYKCcIIB9LRAA

💻 Code

type Box<T> = { value: T }

type UnboxInner<T> = T extends Box<infer R> ? UnboxInner<R> : { unboxed: T }
type Unbox<T> = UnboxInner<T>['unboxed']

declare function unbox<T>(x: T): Unbox<T>

function example<T>(x: T) {
  let a = unbox(x)
  a = 'this should be an error'
}

🙁 Actual behavior

In example we can assign arbitrary values to a, even though it's typed as Unbox<T>.

🙂 Expected behavior

Attempting to assign invalid values to a (just about anything, since its type depends on a type parameter) should cause a type error.

Additional information about the issue

These issues may possibly be related:

• Conditional type on type argument reports different value when assigned to vs assigned from #52021
• Conditional types incorrect for complex types #35533

cc @webstrand

[Andarist]

This is a fun one.

The constraint of the object type of this type:

type Unbox<T> = UnboxInner<T>['unboxed']

is computed as

type BaseConstraintOfObjectType = { unboxed: unknown; } | { unboxed: T; }

This makes sense because the constraint is computed based on both branches of this conditional type and the infer R is also "replaced" with its own constraint (and that is unknown).

So... later we unpack the 'unboxed' property from this union, so we end up with unknown | T and that is reduced to just unknown (it's not that the reduction thing matters here though). And string can be related to unknown.

[GheorgheP]

But why is allowed the assignment to a variable of type unknown?
Isn't this the real issue?

[Andarist]

You can always assign things to their supertypes and unknown is like an ultimate supertype. Everything is assignable to unknown (except never). If you think about types in terms of sets then unknown is like a set that contains everything.

[GheorgheP]

Hmmm, how then the any type is described and at what point any is different to unknown?
Sure I don't know how to describe the unknown in terms of sets, but this form looks more like any.
Shouldn't unknown be something like bottom type?

[fatcerberus]

any means "this value won't be typechecked". unknown means "any possible value" but doesn't disable typechecking (it's not assignable to anything except itself).

Shouldn't unknown be something like bottom type?

No, the bottom type is never.