I'm trying to assign a function's parameters to a type alias. Eg for this function:
function foo<T>(bar: T, baz: number) {}
I want to get the signature of its parameters. Ordinarily, I'd do this:
type FooParams = Parameters<typeof foo>;
but the problem is that that results in a signature of [{}, number]
which makes sense because I haven't specified what T
is. However, then I'd expect this to work:
type FooParams<T> = Parameters<typeof foo<T>>; // hopefully [T, number]
but it doesn't, it just results in a bunch of syntax errors.
Is it possible to correctly get the type of a generic function's parameters list?
You can use for that infer
keyword:
type Args<T> = T extends (...args: infer I) => any ? I : never;
type R = Args<(args: number, arg: string) => number >;
// R = [number, string]
It means that type I
is treat as treat as generic, but you don't have to declare it like: let fn = <T>(arg:T): T {...
You can achieve generic types but you have to declare type of function:
type Foo<T> = (bar: T, baz: number) => void;
type R2<Placeholder> = Args<Foo<Placeholder>>; // [Placeholder, number]
type R3 = R2<boolean> // [boolean, number]
My solution is only a workaround which might help. This is not exactly what you want
Here you have non generic solution.I mean it works only with foo
type signature
function foo<T>(bar: T, baz: number) { }
type Fn = (...args: any) => any
type Tail<T extends any[]> = T extends [infer _, ...infer Rest] ? Rest : never
type ReplaceFirstParameter<F extends Fn, Fst> =
(fst: Fst, ...args: Tail<Parameters<F>>) => ReturnType<F>
type Foo<T> = ReplaceFirstParameter<typeof foo, T>
type Result = ReplaceFirstParameter<typeof foo, number> // (fst: number, baz: number) => void
This solution is most more generic. It will replace all unknown
params with provided generic argument.
I assume that every function generic is unknown - this is the weak place.
type MapPredicate<T, V> = T extends unknown ? V : T
type ReplaceUnknown<
Arr extends Array<unknown>,
Value,
Result extends Array<unknown> = []
> = Arr extends []
? []
: Arr extends [infer H]
? [...Result, MapPredicate<H, Value>]
: Arr extends [infer Head, ...infer Tail]
? ReplaceUnknown<[...Tail], Value, [...Result, MapPredicate<Head, Value>]>
: Readonly<Result>;
function foo<T>(bar: T, baz: number) { }
type Fn = (...args: any) => any
type ReplaceParameters<F extends Fn, Value> =
(...args: ReplaceUnknown<Parameters<F>, Value>) => ReturnType<F>
type Result = ReplaceParameters<typeof foo, number> // (fst: number, baz: number) => void
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