How to call the same function sequentially with different arguments while using setTimeout()

Question

I would like to call the function example multiple times with different arguments, while using setTimeout inside logic. I want this to be sequential (after the first call is made and finished, the second call can begin, and so on). In the following code snippet, example seems to be going through both arrays element-wise:

let arr1 = [1, 2, 3];
let arr2 = [4, 5, 6];

function logic(i, arr) {
    setTimeout(() => {
        console.log(arr[i]);
    }, (i + 1) * 300);
}

function example(arr) {

    for (var i = 0; i < arr.length; i++) {

        logic(i, arr);
    }

}

setTimeout(() => {
    example(arr1)
}, 3000);
setTimeout(() => {
    example(arr2)
}, 3000);

I am aware that I could just set the timer of the second call to 6 seconds for example, and it will work, but I wonder:

Is there is another way to find out that the call to example is done and then trigger the next one? Without statically defining some delays and maybe even without setTimeout?

The expected output would be:

setInterval or Promise.then

eroironico
– eroironico

2022-12-05 11:06:58 +00:00
Commented Dec 5, 2022 at 11:06 — eroironico
– eroironico, Commented Dec 5, 2022 at 11:06

Trevor Dixon · Accepted Answer · 2022-12-05 11:08:04Z

2

logic needs to let the caller know when it's done, then example needs to let the caller know when it's done. Promises will help you here.

let arr1 = [1, 2, 3];
let arr2 = [4, 5, 6];

function logic(i, arr) {
  return new Promise(resolve => {
    setTimeout(() => {
        console.log(arr[i]);
        resolve();
    }, (i + 1) * 300);
  });
}

async function example(arr) {

    for (var i = 0; i < arr.length; i++) {

        await logic(i, arr);
    }

}

(async function() {
  await example(arr1);
  await example(arr2);
})();

answered Dec 5, 2022 at 11:08

Trevor Dixon

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