I have a type called T
that accepts a generic type Str extends string
. If the Str
extends "hello"
, then the type T
should have an additional property called B
. Something like this:
export type T<Str extends string> = {
A: number;
} & (Str extends "hello" ? { B?: number } : {});
Based on that type, it behaves as follows:
type T1 = T<"hello">; // type T1 = { A: number; B?: number | undefined; }
type T2 = T<"world">; // type T2 = { A: number; }
Now, I want to create a function that accepts this type as an argument, and add an extra logic based on the existence of property B
:
function t<Type extends string>(arg: T<Type>) {
if ("B" in arg) {
// ...
}
}
Although, I get an error that arg
inside the if statement is considered as never
.
It works if you explicitly name the variant:
export type Test<Str extends string> = {
A: number;
} & (Str extends "hello" ? { B?: number } : { });
type HelloVariant = Test<"hello">;
function t<T extends (HelloVariant | Test<string>)>(arg: T) {
if ("B" in arg) {
console.log(arg.B);
}
}
Actually what needs to be named is the possibility of that string being a "hello". Here is more simplified version:
export type Test<Str extends string> = {
A: number;
} & (Str extends "hello" ? { B?: number } : { });
function t<T extends (Test<"hello"> | Test<string>)>(arg: T) {
if ("B" in arg) {
console.log(arg.B);
}
}
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