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Possible regression in conditional types using typescript@4.0.2 #40182
Description
TypeScript Version: 4.0.2
Search Terms: conditional, regression, 4.0.2
Code
type Selector<T> = (selector: T) => any
type Value<T extends Selector<any>> = ReturnType<T> extends string[]
? ReturnType<T>[0][]
: ReturnType<T> extends string
? ReturnType<T>[]
: never
function buildQuery<T = any, S extends Selector<T> = Selector<T>>(selector: S, values: Value<S>) {
values.forEach(value => {
// do things with the value
})
}Expected behavior:
Using typescript@3.9.5 it used to be the following type (string[] & ReturnType<S>)[0] which can be seen here
Actual behavior:
Using typescript@4.0.2 the type definition changes to an implicit any which can be seen here
Playground Link: See above
Related Issues: I did not find any similar issues.
If any more information is required let me know!
Activity
weswigham commentedon Oct 22, 2020
This is working as expected now -
ReturnType<T>whereTis a selector will always resolve toany, andanyresolves to both conditional type branches, so you end up with array element types(string[] & ReturnType<S>)[0]andstring & ReturnType<S>(realistically, whenSis instantiated, these will all collapse toany, but it's not, so it isn't). Practically speaking, we also must consider all branches when looking at constraints while reasoning over generics,anyor no. Regardless, these two options mean we have a union of two different array types, meaningvaluesis a union of (array) types, and we're doing.forEachon it. Thevalueparameter type isn't the same across those two overloads, so we type itany(or require a type annotation by issuing an implicitanyerror withnoImplicitAnyon).Even in 3.9.2, you can actually witness the same behavior if you change
Selectorto returnunknown, rather thanany(so it doesn't trigger the buggy behavior withanywe previously had).