Use default parameter when using the spread syntax in ES6?

Question

I understand that you can use the spread ~~operator~~ syntax with parameters (Rest Parameters) when defining a function in es6 , like so:

function logEach(...things) {
  things.forEach(function(thing) {
    console.log(thing);
  });
}

logEach("a", "b", "c");
// "a" // "b" // "c"

My question :

Can you use the default parameter along with the spread syntax ? This doesn't seem to work:

function logDefault(...things = 'nothing to Log'){
  things.forEach(function(thing) {
    console.log(thing);
  });
}
//Error: Unexpected token = 
// Note: Using Babel

Why would that make any sense? `things` would be an array with the remaining arguments, with a default being empty array. — Alexander O'Mara, Oct 18 '16 at 21:50
Couldn't you just check `things.length` to determine if nothing was passed? — Rob M., Oct 18 '16 at 21:59
`...things` is called a [rest parameter](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Functions/rest_parameters) (some people describe it as "gather"). The syntax is the same as spread, but it does the opposite. — joews, Oct 18 '16 at 22:04
Related: [Spread Syntax vs Rest Parameter in ES2015 / ES6](https://stackoverflow.com/q/33898512). — Henke, Mar 16 '21 at 12:09

joews · Answer 1 · 2016-10-18T22:07:36.363

2

JavaScript doesn't support a default for rest arguments.

You could split the arguments and merge their values in the function body:

function logDefault(head = "nothing", ...tail) {
  [head, ...tail].forEach(function(thing) {
    console.log(thing);
  });
}

logDefault(); // "nothing"
logDefault("a", "b", "c"); // a, b, c

edited Oct 18 '16 at 22:07

answered Oct 18 '16 at 21:59

joews

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score 2 · Accepted Answer · answered Oct 18 '16 at 22:18

No, a rest parameter gets assigned an empty array when there are no arguments left; there is no way to provide a default for it.

You'll want to use

function logEach(...things) {
  for (const thing of (things.length ? things : ['nothing to Log'])) {
    console.log(thing);
  }
}

Use default parameter when using the spread syntax in ES6?

2 Answers2