Whenever a generic class is specialized, this results in a new, distinct type. These types are assignment compatible if the same template types are used.
Take the following generic definition:
{$mode objfpc}
unit ua;
interface
type
Generic TMyClass<T> = Class(TObject)
Procedure DoSomething(A : T; B : INteger);
end;
Implementation
Procedure TMyClass.DoSomething(A : T; B : Integer);
begin
// Some code.
end;
end.
And the following specializations:
{$mode objfpc}
unit ub;
interface
uses ua;
Type
TB = Specialize TMyClass<string>;
implementation
end.
the following specializations is identical, but appears in a different unit:
{$mode objfpc}
unit uc;
interface
uses ua;
Type
TB = Specialize TMyClass<string>;
implementation
end.
The following will then compile:
{$mode objfpc}
unit ud;
interface
uses ua,ub,uc;
Var
B : ub.TB;
C : uc.TB;
implementation
begin
B:=C;
end.
The types ub.TB and uc.TB are assignment compatible. It does not matter that the types are defined in different units. They could be defined in the same unit as well:
{$mode objfpc}
unit ue;
interface
uses ua;
Type
TB = Specialize TMyClass<string>;
TC = Specialize TMyClass<string>;
Var
B : TB;
C : TC;
implementation
begin
B:=C;
end.
Each specialization of a generic class with the same types as parameters is a new, distinct type, but these types are assignment compatible if the template types used to specialize them are equal.
If the specialization is with a different template type, the types are still distinct, but no longer assignment compatible. i. e. the following will not compile:
{$mode objfpc}
unit uf;
interface
uses ua;
Type
TB = Specialize TMyClass<string>;
TC = Specialize TMyClass<integer>;
Var
B : TB;
C : TC;
implementation
begin
B:=C;
end.
When compiling, an error will result:
Error: Incompatible types: got "TMyClass<System.LongInt>" expected "TMyClass<System.ShortString>"