Cannot set observable.value

[vandadnp]

Hi and thank you for MobX, it's amazing, I appreciate the hard work.

Given the following code:

void testIt() {
  final greetings = Observable('Hello');
  final dispose = autorun((_) {
    print(greetings.value);
  });
  greetings.value = 'World';
  dispose();
}

I am getting the following error in Flutter:

════════ Exception caught by widgets library ═══════════════════════════════════
The following assertion was thrown building HomePage(dirty):
Side effects like changing state are not allowed at this point. Please wrap the code in an "action". Tried to modify: Observable@1
'package:mobx[/src/core/context.dart]()':
package:mobx/…/core/context.dart:1
Failed assertion: line 220 pos 18: '_state.isWithinBatch'

It seems like we are not allowed to set the value of an observable like that anymore but the examples all show the code above as being valid.

[amondnet]

@vandadnp

Please wrap the code in an "action".

By default, MobX enforces a rule that all mutations to observables should happen inside an action. This is part of the configuration for a ReactiveContext. If you mutate an observable that is being observed, outside of an action, MobX will throw an Exception. This is a safety measure to ensure there are no stray mutations happening in your application.

Do not mutate observables outside an action.

Although not advisable, you can relax this restriction by setting a different config for the ReactiveContext.

docs: https://mobx.netlify.app/api/action/#safety-first

void testIt() {
  final greetings = Observable('Hello');
  final dispose = autorun((_) {
    print(greetings.value);
  });
  runInAction((){
      greetings.value = 'World';
  });
  dispose();
}

[vandadnp]

Thank you @amondnet , much appreciated

My issue is that the code I pasted is mentioned in the documentations, in the Core Concepts section, and shown as being valid code. Please see the attached screenshot:

I believe we need to update the documentations. Thoughts?

[pavanpodila]

Absolutely...would love a PR from you @vandadnp if you can :-)

[vandadnp]

Sure, I'll take a look 💙