# Test that LLDB displays indirect virtual bases correctly

# REQUIRES: lld, x86, system-windows

# RUN: split-file %s %t

# RUN: %clang_cl_host -Z7 -c /GS- /Fo%t.cv.obj -- %t/main.cpp
# RUN: %clang_cl_host -Z7 -gdwarf -c /GS- /Fo%t.dwarf.obj -- %t/main.cpp

# RUN: lld-link -debug -nodefaultlib -entry:main %t.cv.obj -out:%t.cv.exe
# RUN: lld-link -debug -nodefaultlib -entry:main %t.dwarf.obj -out:%t.dwarf.exe

# RUN: %lldb -f %t.cv.exe -s %t/commands.input 2>&1 | FileCheck %s
# RUN: %lldb -f %t.dwarf.exe -s %t/commands.input 2>&1 | FileCheck %s

#--- main.cpp

struct VBase1 {
  short member = 1;
};
struct VBase2 {
  short member = 2;
};
struct Base1 {
  short member = 3;
};
struct Base2 {
  short member = 4;
};

struct User : public virtual VBase1, public virtual VBase2 {
  short member = 5;
};

struct UserUser : public Base1, public User, public Base2 {
  short member = 6;
};

int main() {
  UserUser useruser;

  return 0; // break here
}

#--- commands.input

br set -p "break here"
r
v useruser
exit

# CHECK:      (lldb) v useruser
# CHECK-NEXT: (UserUser) useruser = {
# CHECK-NEXT:   Base1 = (member = 3)
# CHECK-NEXT:   User = {
# CHECK-NEXT:     VBase1 = (member = 1)
# CHECK-NEXT:     VBase2 = (member = 2)
# CHECK-NEXT:     member = 5
# CHECK-NEXT:   }
# CHECK-NEXT:   Base2 = (member = 4)
# CHECK-NEXT:   member = 6
# CHECK-NEXT: }