重载派生类中的新运算符

Question

I have overloaded new operator in the Base class. However, when I add additional overloaded new to the Derived class gcc compiler does not find new operator in the Base class. Why?

Best, Alex

#include <stdlib.h>
template <class t> class Base {
  public:
    Base() {}
    void * operator new (size_t size, void *loc) { return loc; }
};

template <class t> class Derived : public Base<t> {
  public:
    Derived() {}
    void * operator new (size_t size, int sz, void *loc) { return loc; }

};

void foo() {
  void *loc = malloc(sizeof(Derived<char>));
  Derived<char> *d = new (loc) Derived<char>();
}

gcc output:

   new.cpp: In function ‘void foo()’:
new.cpp:17:45: error: no matching function for call to ‘Derived<char>::operator new(sizetype, void*&)’
  Derived<char> *d = new (loc) Derived<char>();
                                             ^
new.cpp:17:45: note: candidate is:
new.cpp:11:10: note: static void* Derived<t>::operator new(size_t, int, void*) [with t = char; size_t = unsigned int]
   void * operator new (size_t size, int sz, void *loc) { return loc; }
          ^
new.cpp:11:10: note:   candidate expects 3 arguments, 2 provided

Answer 1

When you invoke the operator new via the placement new expression

new (loc) Derived<char>();

the compiler looks for an overload of operator new in the Derived class (and not the Base class). It finds it, but your overload

void * operator new (size_t size, int sz, void *loc) { return loc; }
//                                ^ additional parameter

accepts more parameters, hence the error.

If you ask why the compiler is not smart enough to invoke the Base 's overload of operator new , it is because of name hiding : the operator new overload in the Derived class hides the one of the Base class. If you want to make the Base::operator new overload visible in your Derived class, use

using Base<t>::operator new;