Async loop in sequence

Question

I would like to ask why this async loop works, but when i change it to use function as parameter it just prints out 1 testing and then just jumps at the end like nothing happens. Any Idea ? Is it cause function can be used just once or ?. Or do you have a better way how to do a async loop in sequence ?

let array=[10,9,7,6,5,4];
fnBlockLoop(array)

    function timeoutTest(delay){
      return new Promise((resolve)=>{
        console.log("testing")
        setTimeout(resolve,delay)
      })
    }

    async function fnBlockLoop(input,func){
        for(const item of input){
            let test = await timeoutTest(2000);
            console.log(test)
        }
        console.log("done")
    }

2

    let array=[10,9,7,6,5,4];
    fnBlockLoop(array,timeoutTest(2000))


function timeoutTest(delay){
  return new Promise((resolve)=>{
    console.log("testing")
    setTimeout(resolve,delay)
  })
}

async function fnBlockLoop(input,func){
    for(const item of input){
        let test = await func;
        console.log(test)
    }
    console.log("done")
}

In the second example, you do not pass the function, but its results as a parameter. Hence, there is nothing to wait for in the loop and it will just proceed. — Sirko
– Sirko, Commented Feb 13, 2018 at 14:39
@trixo: You pass a function by passing the function rather than calling it. So instead of fnBlockLoop(array,timeoutTest(2000)), do fnBlockLoop(array,timeoutTest) — Matt Burland
– Matt Burland, Commented Feb 13, 2018 at 14:41
fnBlockLoop(array, async function(){ timeoutTest(2000) } ) or fnBlockLoop(array, timeoutTest, 2000 ). 2nd variant needs some adaptations within the function body of fnBlockLoop() — Sirko
– Sirko, Commented Feb 13, 2018 at 14:42

Sirko · Accepted Answer · 2018-02-13 15:10:23Z

1

What you want, is to pass the function reference timeoutTest along. What you currently do, is to pass the result of a function call timeoutTest( 2000 ) along. The later is a Promise, which can only resolve once.

To pass the actual function reference along you have to options:

Add an anonymous function wrapper. This will let you use other functions as input: fnBlockLoop(array, async function(){ await timeoutTest(2000) } )
Pass the function reference and its parameters. This needs some adjustments in your fnBlockLoop() function: fnBlockLoop(array, timeoutTest, 2000 )

edited Feb 13, 2018 at 15:10

answered Feb 13, 2018 at 14:53

Sirko

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4 Comments

trixo Over a year ago

I feel like first option does not work tho :) but second does

Sirko Over a year ago

@trixo Change let test = await func; to let test = await func(); and it should work also. Here you did the same error the other way around: using the reference instead of a call.

trixo Over a year ago

hm cant make it run Have no idea where s problem :) Could you try to recreate first option in fiddle ? :)

Sirko Over a year ago

@trixo My Bad. Forgot a keyword myself. Adding that await makes it work. See this fiddle.

Dathan · Accepted Answer · 2018-02-13 14:52:28Z

0

You should be passing the function reference like this.

   let array=[10,9,7,6,5,4];
    fnBlockLoop(array,timeoutTest)


function timeoutTest(delay){
  return new Promise((resolve)=>{
    console.log("testing")
    setTimeout(resolve,delay)
  })
}

async function fnBlockLoop(input,func){
    for(const item of input){
        let test = await func(2000);
        console.log(test)
    }
    console.log("done")
}

answered Feb 13, 2018 at 14:52

Dathan

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