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Home/ Questions/Q 867041
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T09:54:36+00:00 2026-05-15T09:54:36+00:00

I’ve recently been able to fetch a TRttiType for an interface using TRttiContext.FindType using

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I’ve recently been able to fetch a TRttiType for an interface using TRttiContext.FindType using Robert Loves “GetType”-workaround (“registering” the interface by an explicit call to ctx.GetType, e.g. RType := ctx.GetType(TypeInfo(IMyPrettyLittleInterface));).

One logical next step would be to iterate the methods of said interface. Consider

program rtti_sb_1;
{$APPTYPE CONSOLE}
uses
  SysUtils, Rtti, mynamespace in 'mynamespace.pas';
var
  ctx:      TRttiContext;
  RType:    TRttiType;
  Method:   TRttiMethod;
begin
  ctx := TRttiContext.Create;
  RType := ctx.GetType(TypeInfo(IMyPrettyLittleInterface));
  if RType <> nil then begin
    for Method in RType.GetMethods do
      WriteLn(Method.Name);
  end;
  ReadLn;
end.

This time, my mynamespace.pas looks like this:

IMyPrettyLittleInterface = interface
  ['{6F89487E-5BB7-42FC-A760-38DA2329E0C5}']
  procedure SomeProcedure;
end;

Unfortunately, RType.GetMethods returns a zero-length TArray-instance. Are there anyone able to reproduce my troubles? (Note that in my example I’ve explicitly fetched the TRttiType using TRttiContext.GetType, not the workaround; the introduction is included to warn readers that there might be some unresolved issues regarding rtti and interfaces.) Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T09:54:36+00:00Added an answer on May 15, 2026 at 9:54 am

    I just traced through what’s going on, and in TRttiInterfaceType.Create, line 5774, it says:

    hasRtti := ReadU16(P);
    if hasRtti = $FFFF then
      Exit;
    

    And in both your interface, and IInterface which it inherits from, HasRtti reads as $FFFF. So apparently no RTTI is being generated for the interface’s methods, and this is even true for the base Interface type. I don’t know why. Not sure who would know why, aside from Barry Kelly.

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