Promise with recursion

Question

I\'ve taken a look at a few questions on recursion in promises and am confused on how to implement them properly:

• Recursive Promise in javascript
• Angu

Answer 1

it would be better if you made i a parameter of the function instead of relying upon external state

const countToTen = (i = 0) =>
  new Promise ((resolve, _) =>
    i < 10
      ? (console.log (i), resolve (countToTen (i + 1)))
      : resolve (i))
      
      
countToTen () .then (console.log, console.error)
// 0 1 2 3 4 5 6 7 8 9 10

And even better if you made 10 a parameter too

const countTo = (to, from = 0) =>
  new Promise ((resolve, _) =>
    from < to
      ? (console.log (from), resolve (countTo (to, from + 1)))
      : resolve (from))

countTo (7, 2) .then (console.log, console.error)
// 2 3 4 5 6 7

A more generic approach is a reverse fold - or unfold

const unfold = (f, init) =>
  f ( (x, acc) => [ x, ...unfold (f, acc) ]
    , () => []
    , init
    )

const countTo = (to, from = 0) =>
  unfold
    ( (next, done, acc) =>
        acc <= to
          ? next (acc, acc + 1)
          : done ()
    , from
    )

console.log (countTo (10))
// [ 0, 1, 2, 3, 4, 5, 6,  7, 8, 9, 10 ]

console.log (countTo (7, 2))
// [ 2, 3, 4, 5, 6, 7 ]

But you want a asynchronous unfold, asyncUnfold. Now the user-supplied function f can be async and we get a Promise of all collected values

const asyncUnfold = async (f, init) =>
  f ( async (x, acc) => [ x, ...await asyncUnfold (f, acc) ]
    , async () => []
    , init
    )

const delay = (x, ms = 50) =>
  new Promise (r => setTimeout (r, ms, x))

const countTo = (to, from = 0) =>
  asyncUnfold
    ( async (next, done, acc) =>
        acc <= to
          ? next (await delay (acc), await delay (acc + 1))
          : done ()
    , from
    )

countTo (10) .then (console.log, console.error)
// [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]

countTo (7, 2) .then (console.log, console.error)
// [ 2, 3, 4, 5, 6, 7 ]

Here's a more practical example where we have a database of records and we wish to perform a recursive look-up, or something...

• db.getChildren accepts a node id and returns only the node's immediate children

• traverse accepts a node id and it recursively fetches all descendant children (in depth-first order)

const data =
  { 0 : [ 1, 2, 3 ]
  , 1 : [ 11, 12, 13 ]
  , 2 : [ 21, 22, 23 ]
  , 3 : [ 31, 32, 33 ]
  , 11 : [ 111, 112, 113 ]
  , 33 : [ 333 ]
  , 333 : [ 3333 ]
  }

const db =
  { getChildren : (id) =>
      delay (data [id] || [])
  }

const Empty =
  Symbol ()

const traverse = (id) =>
  asyncUnfold
    ( async (next, done, [ id = Empty, ...rest ]) =>
        id === Empty
          ? done ()
          : next (id, [ ...await db.getChildren (id), ...rest ])
    , [ id ]
    )

traverse (0) .then (console.log, console.error)
// [ 0, 1, 11, 111, 112, 113, 12, 13, 2, 21, 22, 23, 3, 31, 32, 33, 333, 3333 ]

Answer 2

Many members already mentioned, need to resolve with the promise returned by the recursion.

I would like to share code into async/await syntax.

const printNumber = (i) => console.log("i is now: " + i);

// recursive function to call number increment order
const recursiveCallNumber = async (i, checkCondition) => {
    // if false return it, other wise continue to next step
    if (!checkCondition(i)) return;
    // then print
    printNumber(i); 
     // then call again for print next number
    recursiveCallNumber(++i, checkCondition);
}

await recursiveCallNumber(1, (i) => i <= 10);

Answer 3

If you look at your code as long as i is less than 10 you are recursing and never resolving the promise. You eventually resolve a promise. but it is not the promise the initial caller gets.

You need to resolve with the promise returned by the recursion. How the system works if you resolve with a promise it will still not resolve until also the value is resolved:

let i = 0;
const countToTen = () => new Promise((resolve, reject) => {
    if (i < 10) {
      i++;
      console.log("i is now: " + i);
      resolve(countToTen());
    } else {
      resolve(i);
    }
  });

countToTen().then(() => console.log("i ended up at: " + i));

There was an error in the last part as well. You didn't provide a function to then so if you would have done something that actually would have waited you would have got the "i ended up at: 0" first.

Answer 4

Try not to use shared mutable state in your functions (especially when they are asynchronous). You are using window.i but anything can change that value, this is not needed because the i value is only used in your function as a counter:

const later = (milliseconds,value) =>
  new Promise(
    resolve=>
      setTimeout(
        ()=>resolve(value),
        milliseconds
      )
  );

const countTo = toWhat => {
  const recur = counter =>
    later(1000,counter)
    .then(
      i=>{
        console.log(`i is now: ${i}`);
        return (i<toWhat)
          ? recur(i+1)
          : i;
      }
    )
  return recur(1);
}

countTo(10)
.then(
  i=>console.log(`i ended up at: ${i}`)
);