const_set_element_not_primitive_equality

(Previously known as const_set_element_type_implements_equals)

An element in a constant set can't override the '==' operator, or 'hashCode', but the type '{0}' does.

The analyzer produces this diagnostic when the class of object used as an element in a constant set literal implements either the == operator, the getter hashCode, or both. The implementation of constant sets uses both the == operator and the hashCode getter, so any implementation other than the ones inherited from Object requires executing arbitrary code at compile time, which isn't supported.

The following code produces this diagnostic because the constant set contains an element whose type is C, and the class C overrides the implementation of ==:

class C {
  const C();
​
  bool operator ==(Object other) => true;
}
​
const set = {C()};

The following code produces this diagnostic because the constant set contains an element whose type is C, and the class C overrides the implementation of hashCode:

class C {
  const C();
​
  int get hashCode => 3;
}
​
const map = {C()};

If you can remove the implementation of == and hashCode from the class, then do so:

class C {
  const C();
}
​
const set = {C()};

If you can't remove the implementation of == and hashCode from the class, then make the set non-constant:

class C {
  const C();
​
  bool operator ==(Object other) => true;
}
​
final set = {C()};