does this array really need the spread syntax?

Question

I was working on this example yesterday and I was wondering if that spread syntax, (the only one used) has any utility in this case? Given that the array with objects won't change at all with that; or I am mistaken? if it does, how so?

const quiz = [{
    name: "Superman",
    realName: "Clark Kent"
  },
  {
    name: "Wonderwoman",
    realName: "Dianna Prince"
  },
  {
    name: "Batman",
    realName: "Bruce Wayne"
  },
];

const game = {
  start(quiz) {

    //    ----> this \/
    this.questions = [...quiz];
    this.score = 0;
    // main game loop
    for (const question of this.questions) {
      this.question = question;
      this.ask();
    }
    // end of main game loop
    this.gameOver();
  },
  ask() {
    const question = `What is ${this.question.name}'s real name?`;
    const response = prompt(question);
    this.check(response);
  },
  check(response) {
    const answer = this.question.realName;
    if (response === answer) {
      alert('Correct!');
      this.score++;
    } else {
      alert(`Wrong! The correct answer was ${answer}`);
    }
  },
  gameOver() {
    alert(`Game Over, you scored ${this.score} point${this.score !== 1 ? 's' : ''}`);
  }
}

game.start(quiz);

It takes a shallow copy of the array. In this particular code there's no need, but it's often good practise not to mutate arrays and/or objects that have been passed to a function unless the specific intention of that function is to perform that mutation. — Alnitak
– Alnitak, Commented Jan 16, 2019 at 0:31
Looks like he's just spreading the elements of quiz into a new array — Tom O.
– Tom O., Commented Jan 16, 2019 at 0:31
@GrzegorzMiśkiewicz he is using template strings, replacing ` with ' would break the code — Nikita Malyschkin
– Nikita Malyschkin, Commented Jan 16, 2019 at 0:35
For completeness: You don't need the spread syntax to clone an array. You can also use arr.slice() or Array.from(arr). — Felix Kling
– Felix Kling, Commented Jan 16, 2019 at 6:35

Nikita Malyschkin · Accepted Answer · 2019-01-16 00:34:26Z

3

what you witness here is the copying of an array. Basically JS does this

a = [1,2,3];
b = a;
a.push(4);
// a == b == [1,2,3,4]

but if you want to make a copy so b is not changed when a is you need to do the spread

a = [1,2,3];
b = [...a];
a.push(4);
// a == [1,2,3,4], b == [1,2,3]

answered Jan 16, 2019 at 0:34

Nikita Malyschkin

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