You have found a bug. It should be reporting this correctly but it isn’t. I
verified it fails on MacOSX.
Greg Clayton
> On Feb 5, 2017, at 1:19 PM, Roman Popov via lldb-dev
> <lldb-dev@lists.llvm.org> wrote:
>
> Hello,
> I'm observing very strange LLDB behavior: it does not always shows a correct
> dynamic type when I ask for.
>
> Originally I was working with LLDB 3.9, but it looks like trunk version
> behaves the same strange way.
>
> I was able to capture this behavior in a small code snippet:
>
> #include <iostream>
> #include <typeinfo>
>
> using namespace std;
>
> struct base_type { virtual ~base_type(){} };
>
> template <class T1, class T2, unsigned SIZE>
> struct derived0 : base_type {};
>
> template <class T1, class T2>
> struct derived1 : base_type {};
>
> int main(int argc, char ** argv) {
>
> base_type * bptr0 = new derived0<int, int, 1024>();
> base_type * bptr1 = new derived1<int, int >();
>
> cout << typeid(*bptr0).name() << endl;
> cout << typeid(*bptr1).name() << endl;
>
> return 0;
> }
>
>
> lldb --version
> lldb version 5.0.0 (http://llvm.org/svn/llvm-project/lldb/trunk
> <http://llvm.org/svn/llvm-project/lldb/trunk> revision 293398)
> clang revision 293398
> llvm revision 293398
>
>
> Testing in LLDB:
> (lldb) break set --file main.cpp --line 22
>
> (lldb) expression -d no-run -- bptr1
> (derived1<int, int> *) $2 = 0x0000000000614c40
>
> (lldb) expression -d no-run -- bptr0
> (base_type *) $3 = 0x0000000000614c20
>
>
> Can someone explain me why for bptr0 I dont get a derived0<int, int, 1024> *
> as I expected?
>
>
> Thanks,
> Roman
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