Conditionally pushing to an array with spread operator

Question

I am pushing elements to an array based on a condition as explained here http://2ality.com/2017/04/conditional-literal-entries.html

const arr = [
  ...(cond ? ['a'] : []),
  'b',
];

Now, this works fine, but when I try

const arr = [
  ...(cond && ['a']),
  'b',
];

instead, it stops working.

I would like to know why it's not working anymore, and if there is a way to conditionally push using spread operator and && instead of ?.

Thank you

Answer 1

No, it is not possible, because all iterable objects are truthy.

If cond is falsey, you have a value which is not spreadable by Symbol.iterator

The built-in types with a @@iterator method are:

• Array.prototype[@@iterator]()
• TypedArray.prototype[@@iterator]()
• String.prototype[@@iterator]()
• Map.prototype[@@iterator]()
• Set.prototype[@@iterator]()

 var cond = false; const arr = [ ...(cond && ['a']), // throws error, function expected 'b', ]; console.log(arr);

Answer 2

Yes , it's possible. _{But maybe that's an overkill that performs worse and decreases the readability.}

 const arr = []; arr.push(...[false && 'nope'].filter(v => v)); arr.push(...[true && 'yep'].filter(v => v)); arr.push(...[false && 'no', true && 'yes'].filter(v => v)); console.info(arr);

As @Nina Scholz indicated an iterable is required for spread operator to work. By using a second array (which may be empty) we can eventually reach the following state ( arr.push() ).