BUG: Wrong Pointer Value When Nested Classes Have Same NameLast reviewed: July 24, 1997Article ID: Q143082 |
The information in this article applies to:
SYMPTOMSWhen 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.
RESOLUTIONThis happens only when the nested classes have the same name. Change the names, for example "A::NestedA" and "B::NestedB".
STATUSMicrosoft has confirmed this to be a bug in the Microsoft products listed at the beginning of this article. We are researching this problem and will post new information here in the Microsoft Knowledge Base as it becomes available.
MORE INFORMATION
Sample Code to Reproduces Error
// Compile option needed: none // File test.cpp struct A { struct Nested { virtual void A_Func() = 0; }; }; struct B { struct Nested { virtual void B_Func() = 0; }; }; struct 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. |
Keywords : CPPLangIss vcbuglist400 vcbuglist500 kbbuglist kbprg
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