Why only once next function running in generators recursion?

Question

In my think. I have to run next function in many times,by below code.

code environment:Google Chrome browser. version: 63.0.3239.132.

const a = [1,2,3,[4,5],6,[7,8]];

function *fun(array){
    for(let i = 0;i<array.length;i++){
        if(Array.isArray(array[i])){
            yield *fun(array[i]);
        }
    }
}

console.log(fun(a).next().done);//true

In usually generators function,just one yield.It have to run twice next function then done become true.

For Example:

function *foo() {
    yield 1;
}

const iterator = foo();
console.log(iterator.next().done);//false
console.log(iterator.next().done);//true

Why in the recursion Example,it just run once then iterator had done?

By the way, this doubt from the book which name is You-Dont-Know-JS in chapter 3.

In your code, there is nothing to yield when a value is not an array. When your iterate over your first array and found an array as item, you delegate to a new generator (the same but on the array value). yield *fun means you delegate to a new iterator. What behaviour are you waiting for ? — Okoch
– Okoch, Commented Jan 19, 2018 at 9:00

Ori Drori · Accepted Answer · 2018-01-19 09:05:59Z

2

You are missing the yield statement for the items that are not arrays:

const a = [1,2,3,[4,5],6,[7,8]];

function *fun(array){
    for(let i = 0;i<array.length;i++){
        if(Array.isArray(array[i])){
            yield *fun(array[i]);
        } else { // if the item is not an array
            yield array[i];
        }
    }
}

const f = fun(a);

console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());
console.log(f.next());

edited Jan 19, 2018 at 9:05

answered Jan 19, 2018 at 8:59

Ori Drori

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