Rx js understanding the lift method

Question

I want to create a new operator and I find in the documentation that one of the ways is to do something like this: ``` class MyObservable extends Observable { lift(operator) { const observable = new MyObservable() observable.source = this; observable.operator = operator; return observable; } // put it here .. or .. customOperator() { /* do things and return an Observable */ } } // ... put it here... MyObservable.prototype.mySimpleOperator = mySimpleOperator; ``` I don't understand what is the `lift` method and what is going on here, can someone help, please?

Accepted Answer

`lift` is used all the time internally in RxJS 5. The principle of lift is that you prepare a new Observable that upon subscribe will forward the events in the way the operator defines. There is a good video about it by Paul Taylor (https://youtu.be/QhjALubBQPg?t=19m). Lift is a very fundamental building block.

Instead of creating a new class - extending Observable - you could also just create the Operator itself. Users of the operator can then call it by writing:

```
Observable.of(1, 2, 3)
  .lift(new MyCustomOperator)
  .subscribe()
```

This means no-one has to learn that yet another operator is available in the Observable API, but instead sees that it is something defined elsewhere.

Ideally you could write

```
someObservable::myCustomOperator();
```

but unfortunately the bind-operator might be long away / never gonna happen, so the `.lift(operator)` seems like the most explicit / clean way.

Rx js understanding the lift method

Solution

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