How do I use Dart extension functions?

Question

Dart 2.6 introduces a new language feature called \"static extension members\".
However, I do not quite understand how to use it.

I would like t

Answer 1

There is an official video by the Flutter team about extension methods now.

Static extension members

Here is an intuitive example of how extension methods work:

extension FancyNum on num {
  num plus(num other) => this + other;

  num times(num other) => this * other;
}

I simply extend num here and add methods to the class. This could be used like this:

print(5.plus(3)); // Equal to "5 + 3".
print(5.times(8)); // Equal to "5 * 8".
print(2.plus(1).times(3)); // Equal to "(2 + 1) * 3".

Note that the name FancyNum is optional and the following is valid too:

extension on num {}

When you use your extension in another file, you must give it a name.

---

The extension above will make use of implicit extension member invocations as you do not have to explicitly declare your num to be a FancyNum.

You can also explicitly declare your extension, but this is not needed in most cases:

print(FancyNum(1).plus(2));

Flex childCount

The desired behavior from the question can be achieved by extending Row or Column, or even better: you can extend Flex, which is the super class of Row and Column:

extension ExtendedFlex on Flex {
  int get childCount => this.children.length;
}

this. can also be omitted if children is not defined in the current lexical scope of childCount, which means that => children.length is also valid.

---

With this static extension of Flex imported, you can call it on any Flex, i.e. also on every Row and Column.
Row(children: const [Text('one'), Text('two')]).childCount will evaluate to 2.

Answer 2

Dart 2.7 introduced new Extension Method concept.

https://dart.dev/guides/language/extension-methods

extension ParseNumbers on String {
    int parseInt() {
        return int.parse(this);
    }
    double parseDouble() {
        return double.parse(this);
    }
}
main() {
    int i = '42'.parseInt();
    print(i);
}

Answer 3

Extensions can have generic type parameters. For The following examples display the implicit extension resolution when multiple applicable extensions are available.

extension SmartIterable<T> on Iterable<T> {
  void doTheSmartThing(void Function(T) smart) {
    for (var e in this) smart(e);
  }
}
extension SmartList<T> on List<T> {
  void doTheSmartThing(void Function(T) smart) {
    for (int i = 0; i < length; i++) smart(this[i]);
  }
}
...
  List<int> x = ....;
  x.doTheSmartThing(print);

Here both the extensions apply, but the SmartList extension is more specific than the SmartIterable extension because List <: Iterable<dynamic>.