I disassemble CreateDevice function from my DirectX code and commented each paratemer:

.text:0043F485                 mov     eax, [ebp+lpParam]
.text:0043F488                 add     eax, 730h
.text:0043F48D                 push    eax  ;LPDIRECT3DDEVICE9
.text:0043F48E                 mov     ecx, [ebp+lpParam]
.text:0043F491                 add     ecx, 0B88h
.text:0043F497                 push    ecx  ;D3DPRESENT_PARAMETERS
.text:0043F498                 mov     edx, [ebp+var_170]
.text:0043F49E                 push    edx  ;vertex processing type
.text:0043F49F                 mov     eax, [ebp+lpParam]
.text:0043F4A2                 mov     ecx, [eax+728h]
.text:0043F4A8                 push    ecx  ;hWnd
.text:0043F4A9                 push    1    ;D3DDEVTYPE_HAL
.text:0043F4AB                 push    0    ;D3DADAPTER_DEFAULT
.text:0043F4AD                 mov     edx, [ebp+lpParam]
.text:0043F4B0                 mov     eax, [edx+734h]
.text:0043F4B6                 mov     ecx, [ebp+lpParam]
.text:0043F4B9                 mov     edx, [ecx+734h]
.text:0043F4BF                 mov     eax, [eax]
.text:0043F4C1                 push    edx  ;??????????????
.text:0043F4C2                 mov     ecx, [eax+40h]   ;this pointer
.text:0043F4C5                 call    ecx  ;call CreateDevice
.text:0043F4C7                 mov     eax, [ebp+lpParam]
.text:0043F4CA                 mov     ecx, [eax+730h];LPDIRECT3DDEVICE9 in return

Using experimental way I figured out LPDIRECT3DDEVICE9 variable presents in:

.text:0043F48D                 push    eax  ;LPDIRECT3DDEVICE9

Also, using debugger I figured out edx register:

.text:0043F4C1                 push    edx  ;??????????????

has value d3d9.dll:6521F191 db 0B8h what in expand view is:

d3d9.dll:6521F191 db 0B8h ; ¬
d3d9.dll:6521F192 db  48h ; H
d3d9.dll:6521F193 db  77h ; w
d3d9.dll:6521F194 db  23h ; #

My question, why before CreateDevice function calling is there "push edx" line of code? In my example I used OOP and ecx is this pointer:

.text:0043F4C2                 mov     ecx, [eax+40h]   ;this pointer

But what purpose is this line of code:

.text:0043F4C1                 push    edx  ;d3d9.dll:6521F191 db 0B8h ; ¬

Is it like this pointer for DirectX COM interface or what?

Thanks in advance!

1 Answer 1


DirectX interfaces like IDirect3D9/IDirect3DDevice9 are COM interfaces and not C++ classes.

COM (OLE2) predates the wide use of C++ and uses a C-compatible calling convention, __stdcall, instead of the C++specific __thiscall. The use of ecx is a red herring; here it's used to load the function pointer (IDirect3D9::CreateDevice) and jump to it, not as the this pointer. The first argument to the call (push edx) is the This pointer (the interface itself).

In C syntax, the call would look like this:

IDirect3D9 *d3d = Direct3DCreate9(D3D_SDK_VERSION);
IDirect3DDevice9 *pDevice;
d3d->lpVtbl->CreateDevice(d3d, adapter, D3DADAPTER_DEFAULT, D3DDEVTYPE_HAL, hWnd, BehaviorFlags, &presentationParameters, &pDevice);

All arguments to CreateDevice are pushed on the stack according to the __stdcall convention.


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