What use does the unary & operator have in this piece of C++ code?

Question

In an example for a little game framework (that does work), there are the lines:

ID3D10Device* device;
ID3D10Buffer* pBuffer;

followed by the line

 device()->CreateBuffer(&bd, NULL, &pBuffer());

The third argument requires a pointer, so why do you need to put an ampersand before pBuffer? an ampersand gets a memory address right? But pBuffer is already a pointer so what does the ampersand do to it?

Answer 1

As described in the ID3D10Device::CreateBuffer page, this function simply requires a pointer to a pointer which is used to contain the output of the function.

HRESULT CreateBuffer(
  [in]   const D3D10_BUFFER_DESC *pDesc,
  [in]   const D3D10_SUBRESOURCE_DATA *pInitialData,
  [out]  ID3D10Buffer **ppBuffer
);

As always in C/C++, adding an & simply returns the address of the variable. In this case, the address of the pointer.

Answer 2

The & is also used as a reference operator. It returns the address of the variable it's referencing too. If you had a function that took the pointer directly as a parameter, then the function will be working directly with the pointer (useful for pointer arithmetic).

Pointer pointers to resources are quite common in the D3D APIs. You usually don't want to access them directly as it's not of any benefit.