Closed
Description
TypeScript Version: 2.7.0-dev.20180117
Search Terms:
generic, return, infer, inference, type parameter
Code
type Path<T, V> = Array<string>
function path<T, A extends keyof T>(key: A): Path<T, T[A]>
function path<T>(path: string|Array<string>): Path<T, any> {
if (typeof path === 'string') return [path] as Path<T, any>
else return path as Path<T, any>
}
function field<T, V>(path: Path<T, V>) {
return {path}
}
type User = {name: string}
// Errors
field<User, string>(path('name'))
// Works
field<User, string>(path<User, 'name'>('name'))Expected behavior:
path should have all type information it needs by the fully typed field call and therefor build without explicit type arguments.
If I can freely imagine the steps the compiler would take:
- The compiler detects that it's not possible to infer
Path::Tand therefor alsoPath::Afrom the arguments - The compiler realizes that the return value is
Path<T, T[A]>and is assigned toPath<User, string>and can therefor infer thatT = User - As the compiler now knows T it now knows that A is
A extends keyof User - It can now infer that
A = 'name'from the argument - It now knows that
T[A]isUser['name']which isstring - Finally, as the return value now is fully resolved as
Path<User, string>which is assigned to an argument of typePath<User, string>there should be no build errors
Actual behavior:
Error: Argument of type '"name"' is not assignable to parameter of type 'never'..
Presumably Path::T is inferred to be {} and Path::A naturally then becomes never
Related Issues:
There are many issues related to inference or generics but I can't find any related that I imagines involves the same resolution logic.