This also fails.
type
TMyRecord = record
public
a: integer;
b: string;
end;
const
TMyRecord_Default: TMyRecord = (a: 100; b: 'foo');
var
r: TMyRecord = TMyRecord_Default;
> On Nov 10, 2018, at 4:55 PM, Sven Barth via fpc-pascal
> <[email protected]> wrote:
>
> TMyRecord is not yet completely parsed. There could be another field located
> behind the "default" constant. Thus using the type of the record aside from
> Pointers and method parameters and result is not allowed.
>
Regards,
Ryan Joseph
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