Iterating through generator in ES6

Question

this code

let func = function *(){
    for (let i = 1; i < 5; i++) {
        yield i;
    }
}
let s = func().next();
console.log(s);
let s2 = func().next();
console.log(s2);

returns

Object {value: 1, done: false}
Object {value: 1, done: false}

so basically func yields first value all the time.

But when I change to

let f = func();
let s = f.next();
console.log(s);
let s2 = f.next();
console.log(s2);

it works as expected. Why assigning func to variable makes such a difference?

If it would behave differently, then every generator function could only be used once... — Felix Kling
– Felix Kling, Commented Apr 27, 2016 at 16:25

nils · Accepted Answer · 2016-04-27 16:05:55Z

3

In your first bit of code, you are always creating a new generator instance. I've assigned those instances to separate variables to illustrate this more clearly:

// first generator instance
let g = func();
// first call to next on the first instance
let s = g.next();
console.log(s);
// second generator instance
let g2 = func();
// first call to next on the second instance
let s2 = g2.next();
console.log(s2);

Whereas in your second snippet, you keep iterating the same generator (assigned to the variable f):

let f = func();
// first call to next
let s = f.next();
console.log(s);
// second call to next
let s2 = f.next();
console.log(s2);

answered Apr 27, 2016 at 16:05

nils

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