How to make a type for arguments accepted by the method in Typescript

Question

I have a method whose structure is given below.

const a = function (...args: any[]) {
  console.log(args);
}

as you can see in the structure of the function, the type of args is any[], I want to make a specific type for argument array accepted by the method. How can I make a specific type for arguments accepted by the method in Typescript? NOTE: The function a can accept any type of argument like a(1, 2, 4, 5, 'asgasg', 124.52, { name: 'sparsh' });

I want to make a specific type for arguments accepted by the method in Typescript.

Answer 1

You can create a custom type:

type x = string | number

and in a case to fit your example:

type User = {
  name: string;
}

type x = string | number | User;

function a(...args: x[]){}

Answer 2

Instead of any[] which would allow any property access including potentially invalid ones, one can use unknown[].
Using unknown[] would allow passing any type of arguments, but will not disable type checking for property access etc. For example:

const a = function (...args: unknown[]) {
  console.log(args);
// console.log(args[0].toString); //disallowed by type checking 
  if(typeof args[0] === 'number') console.log(args[0].toFixed()) //allowed
}

a(2,3,4,5,6,7,"asds"); //ok