What is the best way to ensure that all function parameters using a shared generic tuple type have a consistent length?

Understanding that [number, number] | [number] is an extension of [number, ...number[]] is logical, but I'm curious if there's a method to enforce the length of tuples based on the initial parameter so that the second tuple must match that same length.

function addVectors<T extends [number, ...number[]]>(v1: T, v2: T) {
  // not yet implemented
}

// T becomes [number].
addVectors([1], [2]);

// T becomes [number, number] | [number].
addVectors([1, 3], [3]);

I am looking to avoid this scenario:

addVectors([1, 3], [3]);

Is it feasible to accomplish this in TypeScript?

typescripttypescript-generics

Answer №1

After some experimentation, I found a solution by introducing a second type parameter that extends the first one:

function combineArrays<T extends [number, ...number[]], U extends T>(arr1: T, arr2: U) {
  // code implementation here
}

// Using arrays with one element works fine.
combineArrays([1], [2]);

// However, this results in a compilation error.
// "Argument of type '[number]' is not assignable to parameter of type '[number, number]'."
combineArrays([1, 3], [3]);

To enhance the functionality, I modified combineArrays to accept two or more arrays of the same dimension:

function combineArrays<T extends [number, ...number[]], U extends T>(...arrays: [T, U, ...U[]]) {
  // additional implementation here
}

combineArrays([1], [2]);

combineArrays([1, 1], [2, 2], [3, 3]);

// This will not compile due to mismatched dimensions.
combineArrays([1, 1], [2, 2], [3, 3], [4, 4, 4]);

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