Is there a way to extract the default type parameter of a function in order to make this statement compile successfully?
const fails: string = "" as ReturnType<<T = string>() => T>;
Is there a way to extract the default type parameter of a function in order to make this statement compile successfully?
const fails: string = "" as ReturnType<<T = string>() => T>;
ReturnType<<T = string>() => T>ambiguous and not strictly limited to being just a string:
type F = <T = string>() => T;
type X = ReturnType<F>;
// type X = unknown 😢
Based on the insights shared in microsoft/TypeScript#42064,
The default behavior only takes effect during a direct call when no inference candidates are available.
This implies that utilizing a default generic parameter such as string for T holds significance only if you explicitly invoke a function of the form <T = string>() => T, and if normal type inference fails. However, in the case of ReturnType<F>, there's no actual invocation of a function; rather, it merely examines the type of F and deduces its return type. Given that T can represent any data type, the compiler lacks clear information about its true nature. Consequently, it replaces T with its implicit restriction, which is the unknown type.
To the best of my knowledge, extracting the default value from a generic function type without invoking a function of that type seems unfeasible. Attempting conditional type inference yields unsatisfactory results:
type Y = F extends <T = infer U>(...args: any) => any ? U : never;
// type Y = unknown 😢
One may choose to simulate calling a function of that type, albeit in a convoluted manner. For instance:
let r;
if (false as true) {
// This block will not execute in reality, yet the compiler cannot determine this
const f = null! as F;
r = f();
} else {
r = null!
}
type Z = typeof r; // string
Depending on your specific requirements, employing such tactics might be viable, but generally speaking, the presence of a generic default appears nearly imperceptible within the type system.
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