Shouldn't there be a "mergeTap"?

[ghost]

Imagine a series of operator functions, like map, mergeMap etc. Using tap along the series works fine for synchronous tasks, but what if I want to perform an asynchronous action (before moving on to the next operator function), without modifying the piped value.

If you do

import { of, tap } from 'rxjs';

of(null)
  .pipe(
    tap(
      () =>
        new Promise((resolve) =>
          setTimeout(() => (console.log('inner'), resolve()), 2e3)
        )
    ),
    tap(() => console.log('outer'))
  )
  .subscribe();

console output will be inner after outer. Using await on the inner promise won't make any difference.

Of course, this would work:

    mergeMap(
      (value) =>
        new Promise((resolve) =>
          setTimeout(() => (console.log('inner'), resolve()), 2e3)
        ).then(() => value)
    ),

but I don't like touching value that I don't want to modify anyway.

So what would be quite handy is some kind of mergeTap operator, which would be like mergeMap but return the input value as is. I could then write

    mergeTap(
      () =>
        new Promise((resolve) =>
          setTimeout(() => (console.log('inner'), resolve()), 2e3)
        )
    ),

[demensky]

I think the delayWhen operator might be a good fit for your purposes.

Example:

interval(3000)
  .pipe(
    delayWhen(
      () =>
        new Promise<void>((resolve) =>
          setTimeout(() => (console.log('inner'), resolve()), 2000)
        )
    )
  )
  .subscribe(console.log);

[demensky]

If you want exactly the operator, then I can offer this option:

function mergeTap<T>(
  handler: (value: T) => ObservableInput<unknown> | void
): MonoTypeOperatorFunction<T> {
  return delayWhen((value) => {
    const result = handler(value);

    return typeof result === 'undefined' ? of(null) : result;
  });
}

Example:

interval(3000)
  .pipe(
    mergeTap(() => {
      console.log('sync');
    }),
    mergeTap(
      () =>
        new Promise<void>((resolve) =>
          setTimeout(() => (console.log('setTimeout'), resolve()), 2000)
        )
    )
  )
  .subscribe(console.log);