Different return types of a function in Typescript when the parameter is either undefined or its type is explicitly asserted vary

Addressing Complexity

This code may seem simple, but the use case it serves is quite complex:

type CustomFunction = <B extends ['A']>(
  param?: B
) => B extends undefined ? 'TYPE_1' : 'TYPE_2'

// Usage examples:
const example1 = customFn() // Returns TYPE_1
const example2 = customFn(['A']) // Returns TYPE_2
  • I want the type to be 'TYPE_1' if param is undefined, and 'TYPE_2' otherwise.
  • param should have dynamic typing, meaning that when values are written within the array, the IDE should provide autocomplete and validation support.
  • The type of param inside the function should be at least (string[] | undefined).

Exploring Different Approaches

I've attempted various configurations, but none of them were successful. You can experiment with these failed attempts in the TypeScript Playground.

The closest I got was with this code:

type MyFn = <
  T,
  B extends ['A']
>(
  param?: T | B
) => T extends string[] ? 'TYPE_2' : 'TYPE_1'

var a: MyFn3 = (param?) => (param ? 'TYPE_1' : 'TYPE_2') as any

const test1 = a() // Returns TYPE_1
const test2  = a(['A']) // Returns TYPE_2

The Dilemma: The inferred type inside the function is any because T has no constraint. However, when we add constraints to the T type, the condition in the return type always evaluates to true.

Please Rescue Me!: Is there any solution that can help overcome this problem?


A Practical Application

This custom type serves the purpose of defining an array of strings as props for addressing an object and returning the corresponding value.
For example, given ['level1', 'level2'], the function should return myObject.level1.level2.

Here's a simplified version of this real-world use case.

Answer №1

Your B type is always guaranteed to have a value and it will never be undefined, since the array ['A'] is never undefined. Consequently, your expression can be simplified by the compiler as follows:

type MyFn = <B extends ['A']>(
  param?: B
) => B extends undefined ? 'TYPE_1' : 'TYPE_2'

This can further be reduced to:

type MyFn = <B extends ['A']>(
  param?: B
) => false ? 'TYPE_1' : 'TYPE_2'

And finally, to:

type MyFn = <B extends ['A']>(
  param?: B
) => 'TYPE_2'

The optional parameter param does not affect the definition or value of B itself.

To address this, you should broaden the definition of B as follows:

type MyFn = <B extends ['A'] | undefined>(

This produces the correct result for the following case:

const test1 = myFn(undefined) // TYPE_1

Additionally, you can set the default value to make it work even when param is omitted:

type MyFn = <B extends ['A'] | undefined = undefined>(

This produces the correct result for the following case:

const test1 = myFn() // TYPE_1

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