How to use/control errors spread operator on undefined

Question

I have defined following function

const fn = (a) => (console.log(...a));

Function works perfect when I call it with parameters:

fn(['asd','fgh']) // prints 'asd fgh'

However, is there any way to call it with undefined?

fn() //Uncaught TypeError: undefined is not iterable(…)

Only solution I can think about is to check it in the start of the function with an

if (a === undefined) a = '';

Is there any way I can make "...a" returns ' ', for example?

Answer 1

You can use default parameters. You can set up what value each parameter will have, if no value was passed in function call.

let fun = (a=3) => console.log(a);

fun(4); //4
fun(); //3

Answer 2

An alternative is(if it has to be a oneliner):

 fn = a => a!==undefined && typeof a[Symbol.iterator] === 'function'? console.info(...a) : console.info(''); fn(); // result '' fn(1); // result '' fn([1,2,3]); // result 1 2 3 

If the iterator part isn't a problem:

 fn = a => a!==undefined ? console.info(...a) : console.info(''); fn(); // result '' try { fn(1); } catch(e){ console.info("told you:" + e); }; // throws an error fn([1,2,3]); // result 1 2 3