function unshift(array, ...int) {
  return [...int].concat(array);
}

function unshift(...array, ...int) {
  return [...int, ...array];
}

The first function is fine but not the second one, so there's no such thing as multiple rest argument in function?

function unshift(array, ...int) {
  return [...int].concat(array);
}

function unshift(...array, ...int) {
  return [...int, ...array];
}

The first function is fine but not the second one, so there's no such thing as multiple rest argument in function?

• javascript
• ecmascript-6

• 7 And what sense would it make? Where do you expect array to finish and int to start? – RaphaMex Commented Jun 11, 2017 at 4:52
• 2 To be pedantic, this is not the "rest operator". It's "rest parameter syntax" or "rest parameter notation". – user663031 Commented Jun 11, 2017 at 5:44

3 Answers 3

Correct. Using the Rest operator on any parameter but the last results in a SyntaxError.

And it makes sense...

The Rest operator on the parameter tells the piler to grab all remaining arguments. If the first parameter had a rest operator, it would grab all the arguments, after which any subsequent parameters would be undefined.

For this reason, the Rest parameter can only be on the last parameter in the argument list.

Example in usage

function foo(bar, ...baz) {
  console.log(bar);
  console.log(baz); 
}

foo(1, 2, 3, 4);
// 1
// [2, 3, 4]

If in some bizarre world, the Rest parameter could be on the first of many parameters, then we'd have something like this

// NOT VALID JS
function foo(...bar, baz) {
  console.log(bar);
  console.log(baz); 
}

foo(1, 2, 3, 4);
// [1, 2, 3, 4]
// undefined

And what use would that be...?

Rest parameter must be last formal parameter error:

function f(a, ...b, c) {
 // ...
}

right:

function f(a, b, ...c) {
 // ...
}

rest will be behave like default argument in other language(c++)