Nested conditional types make extends clause infer types not visible

[weswigham]

TypeScript Version: 2.7.0-dev.201xxxxx

Code

type Cond<TIn> = TIn extends (infer A extends infer TInside ? infer C : infer D) ? [A, C, D] : never;
// more motivating real example:
type InvertCond<TIn> = TIn extends (infer T extends infer U ? infer A : infer B) ? (T extends U ? B : A) : TIn;

Expected behavior:
At least infer types A, C, and D should be visible within the true branch of the outer conditional type. Arguably, so should TInside, depending on how you expect scoping to work.

Actual behavior:
Cannot find name "A"
Cannot find name "C"
Cannot find name "D"

[ahejlsberg]

Several reasons why this won't work. First, infer X declarations are lexically restricted to the extends type of a conditional type. This holds even when a conditional type is nested in the extends clause of an outer conditional type. I think this is a very reasonable rule. Without it it would be very non-intuitive to understand which conditional type an infer X belongs to.

Now, you could work around the lexical restriction:

type Conditional<T, U, A, B> = T extends U ? A : B;
type InvertCond<X> = X extends Conditional<infer T, infer U, infer A, infer B> ? Conditional<T, U, B, A> : X;

But the issue you'll then face is that we defer resolution of InvertCond (because X is generic) and only resolve once we instantiate for a non-generic type argument. But at that point the "conditionalness" of the type has disappeared, so there's nothing to infer from.

In short, fuhgettaboutit.

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.

[weswigham]

I believe this has been fixed by #22764.