How does works the destructuring array with generator functions?

Question

I find this example and I was surprising the way that the eventually values of the generator are get from the generator. This is the code:

function* foo() {
  yield 'a';
  yield 'b';
  yield 'c';
}

const [...values] = foo();

Based on my understand of generators, i think that destructuring a generator execution is equivalent to:

const array = [];
for(const val of foo()) {
    array.push(val)
}

so, my question is: It is this a valid approach to get the values of a generator function and what other ways exists for achieve this?

"is this a valid approach?" -- does it work? Yes? Then it's valid. "what other ways exist to achieve this?" Lots; likely that bit is too broad a question for SO. — Heretic Monkey
– Heretic Monkey, Commented May 16, 2017 at 17:47
It is valid, but IMHO less readable than const values = [...foo()] or even const values = Array.from(foo()). — le_m
– le_m, Commented May 16, 2017 at 18:01

Badacadabra · Accepted Answer · 2017-05-16 19:18:30Z

7

The iteration (iterable & iterator) protocols offer you an ocean of possibilities...

function* foo() {
  yield 'a';
  yield 'b';
  yield 'c';
}

const [...values] = foo();

console.log(...values);

function* foo() {
  yield 'a';
  yield 'b';
  yield 'c';
}

const [a, b, c] = foo();

console.log(a, b, c);

function* foo() {
  yield 'a';
  yield 'b';
  yield 'c';
};

let values = [];

for (const str of foo()) {
  values.push(str);
}

console.log(values[0], values[1], values[2]);

function* foo() {
  yield 'a';
  yield 'b';
  yield 'c';
}

const [a, b, ...values] = Array.from(foo());

console.log(a, b, values.toString());

let foo = (function* () {
  yield 'a';
  yield 'b';
  yield 'c';
})();

console.log(foo.next().value, foo.next().value, foo.next().value);

edited May 16, 2017 at 19:18

answered May 16, 2017 at 18:07

Badacadabra

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