for...of an iterator without closing it

Question

Is it possible to loop over part of an iterator with for...of without closing the iterator when I break the loop?

Example:

function* numbers(i=0){
  while(true) yield i++;
}

let nums=numbers();

// this loop prints numbers from 0 to 3
for(const n of nums){
  if(n>3) break;
  console.log(n);
}

// this loop doesn't print anything because `nums` has already been closed
for(const n of nums){
  if(n>10) break;
  console.log(n);
}

I know that I can go through the iterator on my own calling iterator.next(). But I wonder if it's possible to do this with the for...of syntax.

Bonneau

Answer 1

No. However you can provide a wrapper that doesn't forward return() to the generator:

function unclosed(iterable) {
  return {
    [Symbol.iterator]() {
      const iterator = iterable[Symbol.iterator]();
      return {
        next: iterator.next.bind(iterator),
        return: null,
        throw: null,
      };
    },
  };
}

function* numbers(i=0) {
  try {
    while (true) yield i++;
  } finally {
    console.log('done');
  }
}

const nums = numbers();

// this loop prints numbers from 0 to 3
for (const n of unclosed(nums)) {
  console.log(n);
  if (n == 3) break;
}
// and this one the numbers from 4 to 10, as well as 'done'
for (const n of nums) {
  console.log(n);
  if (n == 10) break;
}

Lynn

Perfect. Do you know why they chose to return generators when a loop is done? – Bonneau

@BlueNebula Because for (const n of numbers()) … is expected to run the finally – Lynn