FIX: Wrong Pointer Value When Nested Classes Have Same Name (143082)


The information in this article applies to:

  • Microsoft Visual C++, 32-bit Editions 2.0
  • Microsoft Visual C++, 32-bit Editions 2.1
  • Microsoft Visual C++, 32-bit Editions 2.2
  • Microsoft Visual C++, 32-bit Editions 4.0
  • Microsoft Visual C++, 32-bit Editions 4.1
  • Microsoft Visual C++, 32-bit Enterprise Edition 4.2
  • Microsoft Visual C++, 32-bit Professional Edition 4.2
  • Microsoft Visual C++, 32-bit Enterprise Edition 5.0
  • Microsoft Visual C++, 32-bit Professional Edition 5.0
This article was previously published under Q143082

SYMPTOMS

When a class is derived from multiple classes such that two or more of the base classes are nested classes of the same name, pointers to either of the nested base classes point to the same address. Consider a class D which is derived from both B1::Nested and B2::Nested. Given an object d, which is of type D, then (B1::Nested *)&d and (B2::Nested *)&d will both resolve to the same address. If either of the nested base class' names are changed to be unique, the behavior is normal.

RESOLUTION

This happens only when the nested classes have the same name. Change the names, for example "A::NestedA" and "B::NestedB".

STATUS

This bug was corrected in Microsoft Visual C++, version 6.0.

MORE INFORMATION

Sample Code to Reproduce Error

// Compile option needed: none

// File test.cpp

truct A {

   struct Nested {
       virtual void A_Func() = 0;
   };

};

truct B {

   struct Nested {
       virtual void B_Func() = 0;
   };

};

truct MyClass : public A::Nested, public B::Nested
{

   void A_Func() { cout << "A_Func() called" << endl; }
   void B_Func() { cout << "B_Func() called" << endl; }
};

void main()
{

   MyClass m;
   cout << "(A::Nested*)&m = " << (void*)(A::Nested*)&m << endl;
   cout << "(B::Nested*)&m = " << (void*)(B::Nested*)&m << endl;
   ((A::Nested*)&m)->A_Func();
   ((B::Nested*)&m)->B_Func();

}
In this example, the compiler confuses (A::Nested*) and (B::Nested*), so the same values are printed for both ((A::Nested*)&m) and ((B::Nested*)&m) when they should be different. This also causes the call to B_Func on the last line to call A_Func instead.
Modification Type:MinorLast Reviewed:7/5/2005
Keywords:kbBug kbfix kbVC600fix KB143082
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