CODEWITHSUNDEEP
typescripttypescript-typingstypescript-generics
Martín Saposnic
Martín Saposnic

Reputation: 3

Infer parameter type depending on other parameters in method

Suppose I have this function declaration:

function foo<T extends {}>(bar: Array<{ key: keyof T, value: any, otherParam: string}>): T[]

And I have this interface:

interface C {
    d: string
    e: number
}

When I call foo like this:

foo<C>([{key: 'd', value: 'myValue', otherParam: 'other'}, {key: 'e', value: 100, otherParam: 'other'}])

I want to infer the type of the second parameter value and get rid of that awful any.

How can I do that?

Thanks!

Upvotes: 0

Views: 106

Answers (1)

Karol Majewski
Karol Majewski

Reputation: 25790

Use mapped types to convert T to your desired shape.

type Bar<T> = {
    [P in keyof T]: { key: P, value: T[P], otherParam: string }
}[keyof T];

Full solution:

type Bar<T> = {
    [P in keyof T]: { key: P, value: T[P], otherParam: string }
}[keyof T];

declare function foo<T extends object>(bar: Bar<T>[]): T[]

interface C {
    d: string
    e: number
}

foo<C>([{key: 'd', value: 'string', otherParam: 'other'}, {key: 'e', value: 100, otherParam: 'other'}])

TypeScript Playground

Upvotes: 1

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