Dart generics method not recognizing type

Question

I have an abstract base class validator with a method which takes a generic type as parameter. I will be passing generic type parameter to base class from the subclass inheriting the base class.

Base Class:

abstract class BaseValidator {
   bool isValid<T>(T obj);
}

Child Class:

class IPv4Validator extends BaseValidator{

  final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";

  @override
  bool isValid<String>(String obj) {
    bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);
    return hasMatch;
  }

}

Here hasMatch takes in non nullable string. When I directly pass some string hasMatch doesn't throw an error. But when I try to pass the generic value in the method parameter, it shows an error.

The argument type 'String' can't be assigned to the parameter type 'String'.

I couldn't able to understand why generic type is not accepting, even though its compile-time type.

try to give a type to your regex: `final String IPV4_REGEX = "...";` — john, Nov 19 '22 at 06:44
That won't solve this as I don't have a problem with `IPV4_REGEX` string. The problem is with `obj` String which is a generic type. — Gowtham K K, Nov 19 '22 at 06:50
`isValid(String obj)` defines a generic method named `isValid` whose type parameter happens to be named `String`. It is identical to `isValid(T obj)` but with a more misleading name for its type parameter (and that's why you get a confusing error message about `String` not being assignable to `String`). It is *not* defining a method named `isValid` that accepts only `String` arguments. — jamesdlin, Nov 19 '22 at 07:05
Hi @jamesdlin . Thanks for your answer. Your comment gave me quite an idea. But little confusing. Kasymebek answer solved mine. But I was wondering why type parameter works in classes and not for methods. As far as I know, both are type parameter are only compile type. Then why there is difference between type parameters in class and method. I would be helpful if you could able to provide extended answer about difference between my implementation and the accepted answer. — Gowtham K K, Nov 19 '22 at 18:24
@GowthamKK I'm not sure what you mean. You can have both generic classes and methods. As I stated, `isValid(String obj)` doesn't do what you want because it declares a generic method, not a specialization of an existing generic method. This would be no different than if you declared a generic class with `class SomeGeneric`: in that case, `String` is not the `String` class, it is the arbitrary name of the generic type parameter. — jamesdlin, Nov 19 '22 at 19:00

Kasymbek R. Tashbaev · Accepted Answer · 2022-11-20T06:27:16.187

The following code solves this particular problem. But it may be different from what you intended to implement. On the other hand, the code will be cleaner if you create a new concrete class for different data types.

abstract class BaseValidator<T> {
  bool isValid(T obj);
}

class IPv4Validator extends BaseValidator<String>{
  final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";

  @override
  bool isValid(String obj) {
    bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);

    return hasMatch;
  }
}

Explanation.
In the line class IPv4Validator extends BaseValidator<String> we are not declaring a new class BaseValidator, it is already declared as BaseValidator<T>. Here we are inheriting the specialization of the existing generic class BaseValidator. While in the line bool isValid<String>(String obj), we declare a new function, so the compiler understands it as if we were declaring a new generic function with a parameter type named String. So, here bool isValid<String>(String obj) is equivalent to bool isValid<T>(T obj), just instead of name T we used name String, which is not an object String.

Thanks for your answer. This works. But I was thinking of some other thing. Why type parameter works in classes and not for methods? — Gowtham K K, Nov 19 '22 at 18:17
@GowthamKK I edited my answer to add explanation. It is the same thing what jamesdlin wrote in the comments. — Kasymbek R. Tashbaev, Nov 20 '22 at 06:32

score 0 · Answer 2 · answered Nov 19 '22 at 07:00

another fix that you can do is to use the covariant keyword, to implement that, try this:

abstract class BaseValidator<T> {
  bool isValid(T obj);
}

class IPv4Validator extends BaseValidator {
  final IPV4_REGEX = "^((25[0-5]|(2[0-4]|1d|[1-9]|)d).?\b){4}\$";

  @override
  bool isValid(covariant String obj) {
    bool hasMatch = RegExp(IPV4_REGEX).hasMatch(obj);
    return hasMatch;
  }
}

Dart generics method not recognizing type

2 Answers2