class - 如何在 Object Pascal "class of interface"(或 "interface of interface")类型中创建

Question

看看这个示例:

//----------------------------------------------------------------------------
type

  ISomeInterface = interface
    procedure SomeMethod;
  end;

  // this is wrong, but illustrates that, what i need:
  TSomeClassWhichImplementsSomeInterface = class of ISomeInterface;

var
  gHardCodedPointer: Pointer; // no matter

procedure Dummy(ASomeClassToWorkWith: TSomeClassWhichImplementsSomeInterface);
begin
  // actually, type of ASomeClassToWorkWith is unknown (at least TObject), but it
  // must implement SomeMethod, so i can make something like this:
  ASomeClassToWorkWith(gHardCodedPointer).SomeMethod;
end;

...

type

  TMyClass = class(TInterfacedObject, ISomeInterface)
  end;

...

// TMyClass implements ISomeInterface, so i can pass it into Dummy:
Dummy(TMyClass);
//----------------------------------------------------------------------------

当然，我可以继承 TMyClass 并使用它的子类，但我不需要这个。我想使用具有自己的层次结构的其他类，只需添加 ISomeInterface 的实现(因为在 Object Pascal 中没有可用的多重继承，就像在 C++ 中一样)。我知道这可能看起来很疯狂，不要问我为什么需要这个，只是说 - 它可能会实现或不会。非常感谢!

Answer 1

我认为您正在寻找的是:

procedure Dummy; 
var Intf : ISomeInterface;
begin
  if Assigned(gHardCodedPointer) and Supports(gHardCodedPointer,ISomeInterface,Intf) then
    Intf.SomeMethod
end;

如果不是，我不知道你想在那里实现什么......