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From: "blarsen at redhat dot com" <sourceware-bugzilla@sourceware.org>
To: gdb-prs@sourceware.org
Subject: [Bug c++/29602] New: cant use typedef class name of a baseclass
Date: Thu, 22 Sep 2022 16:06:29 +0000
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X-Bugzilla-Product: gdb
X-Bugzilla-Component: c++
X-Bugzilla-Version: HEAD
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X-Bugzilla-Who: blarsen at redhat dot com
X-Bugzilla-Status: NEW
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https://sourceware.org/bugzilla/show_bug.cgi?id=3D29602

            Bug ID: 29602
           Summary: cant use typedef class name of a baseclass
           Product: gdb
           Version: HEAD
            Status: NEW
          Severity: normal
          Priority: P2
         Component: c++
          Assignee: unassigned at sourceware dot org
          Reporter: blarsen at redhat dot com
  Target Milestone: ---

Consider the following code:
```
class base{
public:
    int x;
};

typedef base tbase;

class derived: public base{
public:
    int y;
};

int main(){
    derived der;
    der.y =3D 0;
    der.tbase::x =3D 1;
    return der.tbase::x;
}
```

g++ is able to compile it with no problems, but in GDB I get:
(gdb)=20
17          return der.tbase::x;
(gdb) print der.base::x
$1 =3D 1
(gdb) print der.tbase::x
A syntax error in expression, near `'.

GDB should be able to resolve the typedef in the user expression and find t=
he
correct member.

--=20
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