Question 309714 in Flutter Question posted in
The official Flutter documentation can be found here.

"is" operator expected behavior in dart flutter

I have slightly strange issue (I am probably missing something)
I decided to iterate over element tree in Flutter and had the following question:

if (element.widget.runtimeType == GestureDetector) {
        print('return true1');
}
if (element.widget.runtimeType is GestureDetector) {
        print('return true2');
}

The runtype of my widget seems to be GestureDetector (that what I see in the debugger).
When I compare with == I see the return true1 as expected.
BUT why when I compare it with is operator I don’t see the return true2
That’s kinda strange,
why the is operator doesn’t return true2?
according to docs:
The result of obj is T is true if obj implements the interface specified by T
link

Tags:

2

Answers


  1. 0

    You should test the object itself, not its runtimeType getter. So it should be like this:

    if (element.widget is GestureDetector) {
        print('return true2');
}

    See the usage of Dart type test operators: https://dart.dev/language/operators#type-test-operators

    Login or Signup to reply.
  2. 0

    runtimeType is a property available to every object in dart. It has type of ‘Type’.

    So when you do

    if (element.widget.runtimeType == GestureDetector) {
        print('return true1');
}

    it is checking if runtimeType is GestureDetector where the value of runtimeType is GestureDetector but type is Type.

    When you are doing,

    if (element.widget.runtimeType is GestureDetector) {
        print('return true2');
}

    It is checking of type of runtimeType is GestureDetector, which is not true as it is Type.

    To check if widget is GestureDetector, you should be doing

    if (element.widget is GestureDetector) {
        print('return true2');
}
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search