operator new overloading and alignment

Question

I\'m overloading operator new, but I recently hit a problem with alignment. Basically, I have a class IBase which provides operator new

Answer 1

mixins are the right approach, however overloading operator new is not. This will accomplish what you need:

__declspec(align(256))  struct cachealign{};
__declspec(align(4096)) struct pagealign{};
struct DefaultAlign{};
struct CacheAlign:private cachealign{};
struct PageAlign: CacheAlign,private pagealign{};

void foo(){
 DefaultAlign d;
 CacheAlign c;
 PageAlign p;
 std::cout<<"Alignment of d "<<__alignof(d)<<std::endl;
 std::cout<<"Alignment of c "<<__alignof(c)<<std::endl;
 std::cout<<"Alignment of p "<<__alignof(p)<<std::endl;
}

Prints

Alignment of d 1
Alignment of c 256
Alignment of p 4096

For gcc, use

struct cachealign{}__attribute__ ((aligned (256)));

Note that there is automatic selection of the largest alignment, and this works for objects placed on the stack, ones that are new'd, and as members of other classes. Nor does it add any virtuals and assuming EBCO, no extra size to the class (outside of the padding needed for the alignment itself).

Answer 2

This is a possible solution. It will always choose the operator with the highest alignment in a given hierarchy:

#include <exception>
#include <iostream>
#include <cstdlib>

// provides operators for any alignment >= 4 bytes
template<int Alignment>
struct DeAllocator;

template<int Alignment>
struct DeAllocator : virtual DeAllocator<Alignment/2> {
  void *operator new(size_t s) throw (std::bad_alloc) {
    std::cerr << "alignment: " << Alignment << "\n";
    return ::operator new(s);
  }

  void operator delete(void *p) {
    ::operator delete(p);
  }
};

template<>
struct DeAllocator<2> { };

// ........... Test .............
// different classes needing different alignments
struct Align8 : virtual DeAllocator<8> { };
struct Align16 : Align8, virtual DeAllocator<16> { };
struct DontCare : Align16, virtual DeAllocator<4> { };

int main() {
  delete new Align8;   // alignment: 8
  delete new Align16;  // alignment: 16
  delete new DontCare; // alignment: 16
}

It's based on the dominance rule: If there is an ambiguity in lookup, and the ambiguity is between names of a derived and a virtual base class, the name of the derived class is taken instead.

---

Questions were risen why DeAllocator<I> inherits DeAllocator<I / 2>. The answer is because in a given hierarchy, there may be different alignment requirements imposed by classes. Imagine that IBase has no alignment requirements, A has 8 byte requirement and B has 16 byte requirement and inherits A:

class IBAse { };
class A : IBase, Alignment<8> { };
class B : A, Alignment<16> { };

Alignment<16> and Alignment<8> both expose an operator new. If you now say new B, the compiler will look for operator new in B and will find two functions:

            // op new
            Alignment<8>      IBase
                 ^            /
                  \         /
                    \     /
 // op new            \ /
 Alignment<16>         A
            \         /
              \     /
                \ /
                 B 

B ->      Alignment<16>  -> operator new
B -> A -> Alignment<8> -> operator new

Thus, this would be ambiguous and we would fail to compile: Neither of these hide the other one. But if you now inherit Alignment<16> virtually from Alignment<8> and make A and B inherit them virtually, the operator new in Alignment<8> will be hidden:

            // op new
            Alignment<8>      IBase
                 ^            /
                / \         /
              /     \     /
 // op new  /         \ /
 Alignment<16>         A
            \         /
              \     /
                \ /
                 B 

This special hiding rule (also called dominance rule) however only works if all Alignment<8> objects are the same. Thus we always inherit virtually: In that case, there is only one Alignment<8> (or 16, ...) object existing in any given class hierarchy.