How do to do recursive async/await in Javascript?

Question

I have the following code, stripped out of any specifics:

const installation = async() => {
  try {
    const call = await callToServer();
    
    if('finished' in call) {
      return call;
    } else {
      await callToServer();
    }
  } catch(e) {
  //
  }
}

And as you can see, I'd like to recursively call callToServer until I see finished inside its response. I understand why my code isn't working and why it's making only 2 calls (or 1, if it sees finished on the first run), but I don't know how to make it recursive.

I tried with a for loop and, well, it worked as-is, but in my case, I don't know the number of calls I need to make upfront. So I went to a do...while loop, since it seemed to fit:

do {
    const call = await callToServer();
} while(!('finished' in call));

But this only runs once. It sees that the while(!... doesn't actually know about call. How can I make it work?

Have a global call variable. So you would create const call outside the do-while loop. — P.B.UDAY
– P.B.UDAY, Commented Dec 29, 2020 at 9:33
"And as you can see, I'd like to recursively call callToServer" - For recursion something has to call itself again. Nothing in your example does this. So something in installation would have to call installation — Andreas
– Andreas, Commented Dec 29, 2020 at 9:33
Note: for is a generic syntax for loops. You don't need to know how many times you need to lop to use for. For example you can do for ( ; condition ; ) ... which will behave exactly the same as while (condition) — slebetman
– slebetman, Commented Dec 29, 2020 at 9:38
@slebetman Huh, interesting. Let me see if I can find an "empty for" example or some documentation on that. I had always assumed that a for loop needs to know the number of steps it needs to run for initially. However, if for ( ; condition ; ) is valid syntax, then it's just personal preference over do...while, right? It's just weird to have for( i=0; !finished; i++) for this use-case for me. — Vincent Miller
– Vincent Miller, Commented Dec 29, 2020 at 9:42
C programmers have been using for (;;) .. as infinite loop for a long time and it's supported in javascript as well. I've always thought for (;;) .. is a stupid way to do infinite loop because #define ALWAYS 1; while (ALWAYS) .. feels more readable to me. But yeah, for is very flexible — slebetman
– slebetman, Commented Dec 29, 2020 at 10:04

Buh Buh · Accepted Answer · 2020-12-29 09:33:00Z

6

} else {
  return await installation();
}

A recursive function would call the entire function again.

answered Dec 29, 2020 at 9:33

Buh Buh

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2 Comments

Mulan Over a year ago

you're the only person here to actually address the missing return

George Inggs Over a year ago

Can confirm what @Mulan said - my recursive function kept diving out early until I added the return to my recursive function call.

Jamiec · Accepted Answer · 2020-12-29 09:38:15Z

0

You don't necessarily need recursion, a simple loop will work to recall your callToServer.

You were on the right track with a do...while loop. You might consider putting a limit in on the number of calls before bombing out. The below will attempt 10 times before doing so - it uses a random chance on whether it works or not so you might see it bombing out.

const installation = async() => {
  try {
    var call = null;
    var i = 0
    do {
       console.log("attempt",i++)
       call = await callToServer();
    } while(! ('finished' in call) && i <10 );
    return call;
  } catch(e) {
  //
  }
}

const callToServer = async() => {
  return new Promise(resolve => {
    if(Math.random()>0.95) {
      resolve({finished:true})
    }
    else {
      resolve({notfinished:true})
    }
  })
}

const test = async() => {
  var res = await installation()
  console.log(res)
}

test()

answered Dec 29, 2020 at 9:38

Jamiec

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1 Comment

Vincent Miller Over a year ago

You are right! Someone else gave me this idea of not being stuck to just what's happening inside do and I ended up with a finished flag.

P.B.UDAY · Accepted Answer · 2020-12-29 09:34:57Z

0

  const installation = async() => {
  try {
    const call = await callToServer();
    
    if('finished' in call) {
      return call;
    } else {
      return await installation();
    }
  } catch(e) {
  //
  }
}

answered Dec 29, 2020 at 9:34

P.B.UDAY

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1 Comment

Vincent Miller Over a year ago

This can work and if I could accept a sister answer, I'd accept it, but with this approach, I'm always calling callToServer with the same arguments. Assume I wanted to callToServer with arguments that come from call, I can't do that here and making it work this way would make it more complicated than it needs to be.

Louay Al-osh · Accepted Answer · 2020-12-29 09:45:51Z

0

Notice that using async/await is new syntax in ES that is similar to how generators function* and yield work, so we can do for, while, recursive call for async function in a sync syntax.

Here is an example that do recursive call while the target value is not met yet


let counter = 0;
let res;


function delay(ms){
        return new Promise((res,rej) => setTimeout(res, ms));
}

async function callToServer(){
        await delay(100);
        return ++counter;
}







async function recursive(){

        res = await callToServer();
        console.log(`called server with return value = ${res}`);

        if(res < 10){
                return await recursive();
        }

        return 'RESULT';

}


(async function main(){
                
        let finalRes = await recursive();
        console.log(finalRes);

})()

this code will output:

called server with return value = 1
called server with return value = 2
called server with return value = 3
called server with return value = 4
called server with return value = 5
called server with return value = 6
called server with return value = 7
called server with return value = 8
called server with return value = 9
called server with return value = 10
RESULT

answered Dec 29, 2020 at 9:45

Louay Al-osh

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1 Comment

Vincent Miller Over a year ago

Yup, this works, but : stackoverflow.com/questions/65490027/… / so, while the answer is GOOD for this question specifically, stackoverflow.com/a/65490138/14906201 is more flexible while not trading too much readability away.

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