Take, for instance, the creation of a type that functions as follows.
class C1 {}
type TGenericType<T> = typeof T;
let example: TGenericType<C1>; //=> typeof C1
I aim to avoid the need to explicitly pass typeof as a generic parameter, as shown below.
let example: TGenericType<typeof C1>
To recreate or encapsulate the TypeScript typeof operator as a generic type is not possible.
The TypeScript typeof operator accepts a value like a variable or property as input, and returns the corresponding TypeScript data type of that value. For example, using typeof C1 interprets C1 as the value named C1, which represents the constructor of the class C1.
However, generic type parameters such as T in the TGenericType<T> defined type alias signify types rather than values. Therefore, the expression typeof T becomes illogical since T itself is a type. Consequently, this results in a compiler error:
type TGenericType<T> = typeof T; // error!
// -------------------------> ~
// 'T' only refers to a type, but is being used as a value here.
If attempting to resolve this issue, one might end up writing
type TGenericType<T> = T;
let test: TGenericType<typeof C1> = C1;
This approach offers no improvement over directly using typeof C1; alternatively, one may consider that having TGenericType<C1> output a constructor type where C1 signifies the instance:
type TGenericType<T> = new () => T;
let test: TGenericType<C1> = C1; // valid
Nonetheless, this does not directly relate to the C1 value, leading to potential loss of static properties if utilized to construct new instances of C1:
class C1 {
static foo = 1;
bar = 2;
}
new C1().bar; // okay
new test().bar; // okay
C1.foo; // okay
test.foo // error
The preferred approach depends on the specific use case or the choice to relinquish the endeavor altogether.
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