Here are my types...
unit unitTestInterfaces;
interface
type
IFoo = interface
['{169AF568-4568-429A-A8F6-C69F4BBCC6F0}']
function TestFoo1:string;
function TestFoo:string;
end;
IBah = interface
['{C03E4E20-2D13-45E5-BBC6-9FDE12116F95}']
function TestBah:string;
function TestBah1:string;
end;
TFooBah = class(TInterfacedObject, IFoo, IBah)
//IFoo
function TestFoo1:string;
function TestFoo:string;
//IBah
function TestBah1:string;
function TestBah:string;
end;
implementation
{ TFooBah }
function TFooBah.TestBah: string;
begin
Result := 'TestBah';
end;
function TFooBah.TestBah1: string;
begin
Result := 'TestBah1';
end;
function TFooBah.TestFoo: string;
begin
Result := 'TestFoo';
end;
function TFooBah.TestFoo1: string;
begin
Result := 'TestFoo1';
end;
end.
And here is my code to run the example...
var
fb:TFooBah;
f:IFoo;
b:IBah;
begin
try
fb := TFooBah.Create;
/// Notice we've used IBah here instead of IFoo, our writeln() still outputs the
/// result from the TestBah() function, presumably because it's the "first" method
/// in the IBah interface, just as TestFoo1() is the "first" method in the IFoo
/// interface.
(fb as IUnknown).QueryInterface(IBah,f); //So why bother with this way at all??
//f := fb as IBah; //causes compile error
//f := fb; //works as expected
if Assigned(f)then
begin
writeln(f.TestFoo1); //wouldn't expect this to work since "f" holds reference to IBah, which doesn't have TestFoo1()
end;
(fb as IUnknown).QueryInterface(IBah,b);
if Assigned(f) then
begin
writeln(b.TestBah1);
end;
except on E:Exception do
writeln(E.Message);
end;
end.
It seems that in the first call to QueryInterface, even though we are assigning the wrong type of interface to the "f" variable, it will still try to execute the 'first' method of whatever it's pointing to, as opposed to the method with the name "TestFoo1". Using f := fb works as expected, so is there a reason we would ever use QueryInterface instead of the syntax f := fb?
I guess you are breaking the rules here:
QueryInterface will put the interface into f which you requested. You are responsible that f is of the appropriate type. As the second parameter is untyped the compiler cannot warn you about your fault.