Record Type checking

[Frank Pepermans]

I was playing around with the new Record feature in Dart 3 alpha, basically trying out some things with Records when used as generic types,

Doing so, some unexpected failing Record Type checks were failing, so I made a little test to reproduce this.

The first test is the one which fails for me, (int, int) and T also have different hash codes, regardless if that is intentional or not, there is then no way to do something like T is (int, int) for example.

So the question is, is this behavior expected? And if so, would there be any way to properly cast a Record Type, like T is (int, int)?

void main() {
  Test1<(int, int)>(); // false??
  print(identical((int, int), (int, int))); // true
  Test2<int>(); // true
  Test3<Test<Name>>(); // true
}

class Test1<T> {
  Test1() {
    print(identical((int, int), T));
  }
}

class Test2<T> {
  Test2() {
    print(identical(int, T));
  }
}

class Test3<T> {
  Test3() {
    final t = Test<Name>;
    print(identical(Test<Name>, T));
  }
}

class Test<T> {
  final T test;

  Test(this.test);
}

class Name {
  final String value;

  const Name(this.value);

  @override
  String toString() => value;
}

[Erik Ernst]

On Thu, Feb 16, 2023 at 12:06 PM Frank Pepermans <fr...@igindo.com> wrote:

I was playing around with the new Record feature in Dart 3 alpha, basically trying out some things with Records when used as generic types,

Doing so, some unexpected failing Record Type checks were failing, so I made a little test to reproduce this.

The first test is the one which fails for me, (int, int) and T also have different hash codes, regardless if that is intentional or not, there is then no way to do something like T is (int, int) for example.

So the question is, is this behavior expected? And if so, would there be any way to properly cast a Record Type, like T is (int, int)?

void main() {
  Test1<(int, int)>(); // false??

The test fails because `(int, int)` occurs as an expression in the Test1 constructor, and that's a record of type `(Type, Type)` with the value of `int` as an expression in both fields, and that's a different thing than the `Type` for the type `(int, int)`. There is no syntax for type literals denoting record types, but you can use

Type typeOf<X>() => X;

If you use `typeOf<(int, int)>()` rather than `(int, int)` as an expression, you get `true` in the first line of output.

The other cases involve types where we _do_ have a type literal, so there's no need to use `typeOf`.

  print(identical((int, int), (int, int))); // true
  Test2<int>(); // true
  Test3<Test<Name>>(); // true
}

class Test1<T> {
  Test1() {
    print(identical((int, int), T));
  }
}

class Test2<T> {
  Test2() {
    print(identical(int, T));
  }
}

class Test3<T> {
  Test3() {
    final t = Test<Name>;
    print(identical(Test<Name>, T));
  }
}

class Test<T> {
  final T test;

  Test(this.test);
}

class Name {
  final String value;

  const Name(this.value);

  @override
  String toString() => value;
}

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Erik Ernst  -  Google Danmark ApS

Skt Petri Passage 5, 2 sal, 1165 København K, Denmark

CVR no. 28866984

[Bob Nystrom]

Erik has it. More specifically, if you write:

var x = (int, int);

Then x does not hold a type literal for a record type with two int fields. It holds a record object whose two fields are the type literal int. Those are different things.

– bob

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