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Type for arbitrary number of function parameters of specified types #4338
Description
I'd like to have a type for a function which takes in zero or more parameters, and then use the type for the parameters, e.g.
// Note: the syntax below of course doesn't do what I'm looking for, something like (params: ...args: T) would be closer
let f = (params: T) => (params2: T) => 0;
// these should be legal:
f(5)(3);
f("foo")("bar");
f(5, "foo")(3, "bar");
// these should be illegal
f(5)("bar");
f("foo")(3);
f(5, 3)("foo", "bar");
This is useful in e.g. typing the spread function here:
DefinitelyTyped/DefinitelyTyped@3eb67c9
I need to use the ...args: any[], which is not quite what I'm looking for.
In essence, the type I want to use is [A?, B?, C?, ...], or maybe something like Params<T> if such an interface would exist.
Activity
JsonFreeman commentedon Aug 24, 2015
Looks like you want something that allows a variable number of parameters with arbitrary types, but then captures them so they can be repeated in another signature. I think you may find a relevant discussion at #3622 (comment) and #3870.
benliddicott commentedon Sep 7, 2015
Do you actually need an infinite number of parameters? If in practice you top out at something like ten or twenty then this might be a good pattern:
It's flexible, doesn't require new syntax, and works today.
JsonFreeman commentedon Sep 8, 2015
That is indeed the recommended pattern to use today.
mhegazy commentedon Feb 22, 2016
looks like a duplicate of #1773