Can you have optional destructured arguments in a Typescript function?

Issue

I'd like to write a function that takes an object argument, uses destructuring in the function signature, and have that argument be optional:

myFunction({opt1, opt2}?: {opt1?: boolean, opt2?: boolean})

However, Typescript doesn't let me ("A binding pattern parameter cannot be optional in an implementation signature.").

Of course I could do it if I didn't destructure:

myFunction(options?: {opt1?: boolean, opt2?: boolean}) {
  const opt1 = options.opt1;
  const opt2 = options.opt1;
  ...

It seems like these should be the same thing, yet the top example is not allowed.

I'd like to use a destructured syntax (1) because it exists, and is a nice syntax, and it seems natural that the two functions above should act the same, and (2) because I also want a concise way to specify defaults:

myFunction({opt1, opt2 = true}?: {opt1?: boolean, opt2?: boolean})

Without destructuring, I have to bury these defaults in the implementation of the function, or have an argument that is actually some class with a constructor...

Solution

Use a default parameter instead:

function myFunction({ opt1, opt2 = true }: { opt1?: boolean; opt2?: boolean; } = {}) {
    console.log(opt2);
}

myFunction(); // outputs: true

It's necessary in order to not destructure undefined:

function myFunction({ opt1, opt2 }) {
}
    
// Uncaught TypeError: Cannot destructure property `opt1` of 'undefined' or 'null'.
myFunction();

Answered By - David Sherret