Function as parameter (DART)

Question

I'm trying to understand how dart uses a function as parameter. I wrote this code...

typedef Future<String> myFunction();

void main() {
  A a = A();
  a.f1(a._f1);
}

class A {
  f1(myFunction func) async {
    String x = await _f1;
    print(x);
  }

  Future<String> _f1() {
    Future.delayed(Duration(seconds: 3)).then((f) {
      return "test";
    });
  }
} 

I need the function f1 returns "test", but I have this error : A value of type 'Future Function()' can't be assigned to a variable of type 'String'

If I change String x = await _f1 by Future x = await _f1 I have another error.. I tried a lot of combinations, all of them fail.

Can someone fix my code? Thank you.

Answer 1

The problem is this line:

String x = await _f1;

The parameter for the method is func , so you are referencing the _f1 method directly here rather than the parameter. Furthermore, rather than calling the method you are just referencing the method directly. In essence, you are trying to assign a Function to a variable that expects a String , which is what the error message is trying to tell you.

You need to reference the parameter and then you need to call it.

String x = await func();

---

As an aside, your _f1 method is currently returning null . This is because you are returning a value from the method within the then but you aren't returning anything to _f1 itself, which makes Dart default to returning null . You have to either return the future itself:

Future<String> _f1() {
  return Future.delayed(Duration(seconds: 3)).then((f) {
    return "test";
  });
}

or you need to switch to async/await syntax (my personal recommendation):

Future<String> _f1() async {
  await Future.delayed(Duration(seconds: 3));
  return 'test';
}

Answer 2

To execute a function you need to add brackets ( arguments ) after its name. It doesn't matter function is a variable or predefined one (constant).

Please check out this examples:

Example 1.

Future<int> add(int a, int b) async {
  return Future.delayed(Duration(seconds: 2)).then((f) {
    return a + b;
  });
}

Future<int> test(Future<int> Function(int a, int b) func) async {
  return await func(3, 2);
}

void main() {
  test(add)
    .then(print); /// will return result of 3+2 after 2 seconds
}

Run this code

Example 2.

String Function(String, String) mergeFunction = (String a, String b) {
  return a + b;
};

void main() {
  print(mergeFunction('Hello ', 'world'));
}

Run this code