How do I call a super constructor in Dart?

86,302

Solution 1

Yes, it is, the syntax is close to C#, here is an example with both default constructor and named constructor:

class Foo {
  Foo(int a, int b) {
    //Code of constructor
  }

  Foo.named(int c, int d) {
    //Code of named constructor
  }
}

class Bar extends Foo {
  Bar(int a, int b) : super(a, b);
}

class Baz extends Foo {
  Baz(int c, int d) : super.named(c, d);  
}

If you want to initialize instance variables in the subclass, the super() call must be last in an initializer list.

class CBar extends Foo {
  int c;

  CBar(int a, int b, int cParam) :
    c = cParam,
    super(a, b);
}

You can read about the motivation behind this super() call guideline on /r/dartlang.

Solution 2

This is a file I am sharing with you, run it as is. You'll learn how to call super constructor, and how to call super parameterized constructor.

// Objectives
// 1. Inheritance with Default Constructor and Parameterised Constructor
// 2. Inheritance with Named Constructor

void main() {
    var dog1 = Dog("Labrador", "Black");
    var dog2 = Dog("Pug", "Brown");
    var dog3 = Dog.myNamedConstructor("German Shepherd", "Black-Brown");
}

class Animal {
    String color;

    Animal(String color) {
        this.color = color;
        print("Animal class constructor");
    }

    Animal.myAnimalNamedConstrctor(String color) {
        print("Animal class named constructor");
    }
}

class Dog extends Animal {
    String breed;

    Dog(String breed, String color) : super(color) {
        this.breed = breed;
        print("Dog class constructor");
    }

    Dog.myNamedConstructor(String breed, String color) : super.myAnimalNamedConstrctor(color) {
        this.breed = breed;
        print("Dog class Named Constructor");
    }
}

Solution 3

case of a constructor with Optional Parameters

class Foo {
  String a;
  int b;
  Foo({this.a, this.b});
}

class Bar extends Foo {
  Bar({a,b}) : super(a:a, b:b);
}

Solution 4

Can I call a private constructor of the superclass?

Yes, but only if the superclass and the subclass you are creating are in the same library. (Since private identifiers are visible across the whole library they are defined in). Private identifiers are those that start with underscore.

class Foo {    
  Foo._private(int a, int b) {
    //Code of private named constructor
  }
}

class Bar extends Foo {
  Bar(int a, int b) : super._private(a,b);
}

Solution 5

As dart supports implementing a class as interface (Implicit interfaces), you can't call the parent constructor if you implemented it you should use extends. If you use implements change it to extends and use Eduardo Copat's Solution.

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Eduardo Copat
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Eduardo Copat

Software developer.

Updated on April 11, 2022

Comments

  • Eduardo Copat
    Eduardo Copat about 2 years

    How do I call a super constructor in Dart? Is it possible to call named super constructors?

  • Piotr Temp
    Piotr Temp over 3 years
    In your constructor of Bar 'a' and 'b' parameters are 'dynamic' type. You missed advantages of statically typed language.
  • manish kiranagi
    manish kiranagi over 2 years
    The dynamic type can be removed by class Bar extends Foo { Bar({required String a, required int b}) : super(a:a, b:b); }