Inheritance and a Reference to a Base Pointer

A reader posted this question on one of the C++ newsgroups. Assuming B is a base class for D, his program looks like this:

   void g (B* p) ;  void f (B*& p); // modifies p  int main ()  {    D* p = NULL ;    g (p) ;  // fine    f (p) ;  /* error: "cannot convert parameter 1 from 'class D *' to                    'class B *& ' a reference that is not to 'const' cannot                    be bound to a non-lvalue" */  }

Function f() works as expected, but g() causes a compilation error. The reader wondered what was wrong with the invocation of f() in main(). Can you see what is wrong with it? p is a D*, not a B*, and to convert it to a B*, the implementation creates a temporary pointer (recall that a reference must be bound to a valid object; in this case, a pointer). Now because temporaries are rvalues, and you can’t bind an rvalue to a non-const reference, the compiler complains. Declaring f() as follows:

   void f (D* & p) { /* modifies p */ }

solves this problem.

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