Using default parameters in a versatile function

There is a function called zip with the following signature:

function zip<T, U, V>(ts: T[], us: U[], zipper: (t: T, u: U) => V): V[]

An attempt is made to assign a default value of (t, u) => [t, u] to the zipper argument:

function zip<T, U, V>(
    ts: T[],
    us: U[],
    zipper: (t: T, u: U) => V = (t, u) => (<[T, U]>[t, u])
)

This results in a compile error stating that (T, U) => [T, U] cannot be assigned to

(T, U) => V</code.</p>

<p>To resolve this issue, a set of overloads are used:</p>

<pre><code>export function zip<T, U>(ts: T[], us: U[]): [T, U][]
export function zip<T, U, V>(
    ts: T[],
    us: U[],
    zipper: (t: T, u: U) => V
): V[]
export function zip<T, U>(
    ts: T[],
    us: U[],
    zipper: (t: T, u: U) => [T, U] = (t, u) => [t, u]
): [T, U][] {
    /* ... */
}

However, there are two main issues with this approach:

  1. The signature zip(T[], U[]): [T, U][] is repeated twice (first overload and the implementation itself).
  2. The implementation's signature may not be the most general one, leading to potential errors in more complex cases.

Is there a better way to achieve the desired outcome? Is the error in the initial attempt potentially a compiler bug?

Answer №1

It's important to remember that in TypeScript, generic type parameters are determined by the caller of a function, not the implementer. The compiler infers these parameters based on the needs of the caller. This means that whoever calls the function has the freedom to choose the types they want for the parameters. TypeScript warns you when the implementation cannot assume compatibility between types. So, it's up to the developer to ensure safety.


One solution to this issue is overloading. Overload signatures help specify different ways the function can be called. When defining the implementation signature, it's best to keep it more general and make sure optional parameters are clearly asserted. Although the compiler may still raise warnings, as long as the overloads restrict potentially unsafe calls, the implementation should be sound.

insert updated code snippet here...

Using overloads in TypeScript requires developers to take responsibility for ensuring safe implementation. The compiler relies on developers to manage potential risks when using overloads.

Good luck with your coding adventures!

Answer №2

You have the option to limit your V.

function merge<A, B, C extends [A, B]>(
    as: A[],
    bs: B[],
    merger: (a: A, b: B) => C = (a, b) => (<C>[a, b])
)

Alternatively, you can convert to any.

function merge<A, B, C>(
    as: A[],
    bs: B[],
    merger: (a: A, b: B) => C = (a, b) => (<C><any>[a, b])
)

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