Array of generic type with different parameters in TypeScript

Question

Let's say we have a generic interface

interface CompareOp<T>{
  a: T;
  b: T;
  cmp: (v1: T, v2: T) => number;
}

I'm looking for a way to create a type for an array of CompareOp of any type. For example:

// valid, each compare function receives the type of its fields
[
  {a: 1, b:1, cmp: (v1: number, v2: number) => v1 - v2},
  {a: 'a', b: 'b', cmp: (v1: string, v2: string) => v1.localCompare(v2)}
]
// invalid, compare function does not match fields
[
  {a: 1, b:1, cmp: (v1: string, v2: string) => v1.localCompare(v2),
  {a: 'a', b: 'b', cmp: (v1: number, v2: number) => v1 - v2}
]

Is there a way to express the type of this array in TypeScript?

Could you provide an example of which result you're waiting for ? Would you like the second array to be considered as an array of CompareOp ? The issue happening in your second array is that the type used in your operation does not satisfy the generic declaration of your interface: If a & b are of type T, so should be the ones used in your cmp as well. If you want them to possibly be different, then your interface should be CompareOp<T, U>{ a: T; b: T; cmp: (v1: U, v2: U) => number; }. Is that what you're searchinng for ? — lisatr2812
– lisatr2812, Commented Feb 9, 2023 at 13:12

Tobias S. · Accepted Answer · 2023-02-09 13:12:21Z

2

You can't create a stand-alone type that performs individual validation of array elements.

What you can do is to utilize the inference of a generic function and use a reverse mapped type to validate each array element against CompareOp.

interface CompareOp<T> {
  a: T;
  b: T;
  cmp: (v1: T, v2: T) => number;
}

function compareOp<T extends any[]>(
  arr: [...{ [K in keyof T]: CompareOp<T[K]> }]
) {}

compareOp([
  { a: 1, b: 1, cmp: (v1: number, v2: number) => v1 - v2 },
  { a: "a", b: "b", cmp: (v1: string, v2: string) => v1.localeCompare(v2) },
]);

compareOp([
  { a: 1, b: 1, cmp: (v1: string, v2: string) => v1.localeCompare(v2) },
//              ~~~ Type 'number' is not assignable to type 'string'
  { a: "a", b: "b", cmp: (v1: number, v2: number) => v1 - v2 },
//                  ~~~ Type 'string' is not assignable to type 'number'.
]);

Playground

answered Feb 9, 2023 at 13:12

Tobias S.

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4 Comments

ertucode Over a year ago

What does the spread opreator do there?

Tobias S. Over a year ago

The compiler will infer T as an array type if we don't use a variadic tuple here. But we need a tuple type for T so that each element in the array gets its own type element in T

Plargato Over a year ago

This seems exactly like what i need!, But cant seem to find anything regarding the term "reverse mapped type" online, could you exaplain how it works or link something?

Tobias S. Over a year ago

@Plargato - I have seen no official documentation which uses the term "reverse mapped type". I have seen it mostly used by TypeScript-contributors and in the source code of the compiler itself.

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