Typescript: Generic type of function parameter signature
I try to declare a type which defines a transformation of a function type generating a type with the same arguments but another return type.
The goal is to write a function createAction which expects a function and returns a new function expecting the same arguments as the given one but returning another type.
Input: function expecting a string and a number, returning type R :
const fn = (s: string, n: number) => ({ s, n, b: false })
const g = createAction(fn)
Expected type of g is: function expecting a string and a number, returning type { payload: R } :
typeof g === (s: string, n: number) => ({ payload: { s: string, n: number, b: boolean } })
This would allow me to write another function createActions expecting a function map and returning the same map but with transformed functions:
const t = createActions({
func1: (s: string) => ({}),
func2: (n: number, b: boolean) => ({ n }),
});
typeof t === {
func1: (s: string) => ({ payload: {} }),
func2: (n: number, b: boolean) => ({ payload: { n: number } }),
}
I could create a type overload with different number of generic arguments to cover a number of different function signatures (see below) but I am looking for a more generic solution which allows inferring the parameter number and types of the given function automatically. Using type overloads makes the function map described above impossible to implement.
A basic solution using generics by hard-coding the number of arguments:
type Transform<D, T1, R> = {
[K in keyof D]: (p1: T1) => ({
payload: R;
});
}
interface CreateActions<D, T1, R> {
(def: D): Transformed<D, T1, R>;
}
Thanks for the help!
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