Universal variable arguments

Question

Universal variable arguments

Is there a way to modify this function that accepts generic rest parameters to also accept an array parameter without using the spread operator? This would make chaining things much clearer:

function fn<T>(...args: T[]): Something<T> {
}

let something = fn(1, 2, 3, 4);  // all good.

When passing an array as a parameter, the return type becomes Something<T[]>. However, I want the return type to be Something<T>.

let something: Something<number> = fn([1, 2, 3, 4]);

I am looking for a way to achieve a type similar to:

(arg T) | (...args T[])

Not sure how to make it happen yet.

typescript

Answer 1

Answer №1

Utilize the concept of function overloading:

function overloadFunction<T>(args: T[]): Result<T>
function overloadFunction<T>(...args: T[]): Result<T>
function overloadFunction<T>(...args: ([T[]]) | T[]): Result<T> {
  // Implement your logic here
}

Answer 2

Utilize the concept of function overloading:

function overloadFunction<T>(args: T[]): Result<T>
function overloadFunction<T>(...args: T[]): Result<T>
function overloadFunction<T>(...args: ([T[]]) | T[]): Result<T> {
  // Implement your logic here
}

Answer 3

Answer №2

One approach to create distinct signatures for your function

interface FunctionSignature<T> {
  (...params: T[]): Result<T>;
  (params: T[]): Result<T>;
};

const fn: FunctionSignature<number> => (...params) => {
  ...
};

Answer 4

One approach to create distinct signatures for your function

interface FunctionSignature<T> {
  (...params: T[]): Result<T>;
  (params: T[]): Result<T>;
};

const fn: FunctionSignature<number> => (...params) => {
  ...
};

Universal variable arguments

Answer №1

Answer №2

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