Receiving accurate type inference and helpful editor suggestions simultaneously

How can I create a function that takes a configuration object and returns a processed version with the same keys but different values? This function should:

  1. Offer suggestions for the properties of the configuration object when calling the function.
  2. Inform the compiler about the specific properties that are returned by the function.

.

type Config = {
    a?: number
    b?: string
}

function processConfig(x: Config): Config {
    return x
}

function processConfigExtended<T extends Config>(x: T): T {
    return x
}

In this scenario, the processConfig() function meets the first requirement while the processConfigExtended() function fulfills the second:

processConfig({a: 1}).<CTRL+SPACE> // Indicates the resulting type as "{ a?: number, b?: string }", but I desire it to be "{ a: number }".

processConfigExtended({<CTRL+SPACE> // Does not provide suggestions for the Config type.

Is there a method to design a function that fulfills both requirements simultaneously?

typescript

Answer №1

Utilizing an intersection type between T and Config can greatly enhance code completion suggestions without affecting type inference, based on my observations:

type Config = {
    a?: number
    b?: string
}

function g<T extends Config>(x: T & Config): T {
    return x
}
let justA = g({ a: 0 });// returns { a: number }
g ({ /* Suggestions for a and b appear when using Ctrl Space  */})

Access the Playground here

Answer №2

To enhance the function, make adjustments to the object within it and eliminate the return type from the declaration.

In the given scenario, you will need to modify the object by creating a new one with identical values (try using Object.assign) to ensure that the correct type is inferred. If you return the original object, the compiler may maintain the initial type, even if some optional properties are null.

By excluding the return type from the declaration, you enable type inference to deduce a new type. Depending on your TypeScript version and settings, a specific type declaration may be required. Otherwise, you may receive 'any' and must verify the desired properties later on. The outcome can vary based on the TypeScript version and configurations in your environment.

Answer №3

If you're uncertain about your specific situation, consider this approach: modify function argument types and determine the expected return data types based on input types. Using a mapped type can be helpful in this scenario. Check out this playground

Explore Advanced types documentation for further information.

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