Placement new With Overloaded Ordinary new Operator

Question

I have an object of type MyType that, for SSE reasons, needs to be 16-byte aligned. So, I wrote an allocator and overloaded the new operators. Methods in MyType:

inline static void* operator new(size_t size) {
    awesome::my_allocator<MyType,16> alloc;
    return alloc.allocate(size);
}
inline static void* operator new[](size_t size) { return operator new(size); }
inline static void operator delete(void* ptr) {
    awesome::my_allocator<MyType,16> alloc;
    alloc.deallocate(reinterpret_cast<MyType*>(ptr),0); //last arg ignored in my impl
}
inline static void operator delete[](void* ptr) { operator delete(ptr); }

Now, for cache-locality reasons, I need to copy-construct an instance into a particular piece of 64-byte-aligned memory:

void MyType::copy_into(uint8_t* ptr) const {
    new (reinterpret_cast<MyType*>(ptr)) MyType(*this);
}

GCC tells me:

error: no matching function for call to ‘MyType::operator new(sizetype, MyType*)’

ICC tells me:

error : function "MyType::operator new" cannot be called with the given argument list
1>              argument types are: (unsigned __int64, MyType *)

---

As I understand it, the placement new operator is provided by the C++ implementation (or possibly by <new>, which I also tried #includeing?) and simply returns its argument (new makes memory available, and placement new is the programmer saying that the given memory is available).

Curiously, the error does not occur when the (ordinary!) new operators defined above are not in the class. Indeed, MyOtherType, which doesn't define them, works just fine.

Question: what's going on? How should I fix it?

Answer 1

Since you have defined operator new in your class, you need to use the global new to use the placement version of it.

#include <new>

...

::new (reinterpret_cast<MyType*>(ptr)) MyType(*this);