Ошибка WinApi C lang

Язык C
Visual Studio 19
SDK 10.0
Код:

#include <stdio.h>
#include <stdlib.h>

//#include "main.h"

#include <Windows.h>
wchar_t classname = L'window';

LRESULT WndProc(HWND hwnd, UINT message, WPARAM wparam, LPARAM lparam) {
    if (message == WM_DESTROY) {
        PostQuitMessage(0);
        return 0;
    }
    else if (message == WM_KEYDOWN)
        printf("code = %d \n", wparam);
    
    return DefWindowProcA(hwnd, message, wparam, lparam);
}
int main() {

    HDC hDCScreen = GetDC(NULL);
    int Horres = GetDeviceCaps(hDCScreen, HORZRES);
    int Vertres = GetDeviceCaps(hDCScreen, VERTRES);

    int cordX = (Horres / 2) - 120;
    int cordY = (Vertres / 2) - 160;

    WNDCLASSA wcl;
    memset(&wcl, 0, sizeof(WNDCLASSA));
    wcl.lpszClassName = classname;
    wcl.lpfnWndProc = WndProc;

    RegisterClassA(&wcl);

    HWND hwnd;
    hwnd = CreateWindow(classname, L"simple window", WS_OVERLAPPEDWINDOW, cordX, cordY, 240, 320, NULL, NULL, NULL, NULL);

    ShowWindow(hwnd, SW_SHOWNORMAL);
    MSG msg;
    while (GetMessage(&msg, NULL, 0, 0)) {
        DispatchMessage(&msg);
    }

    return 0;
}

Ошибка:

Ошибка  C2440   =: невозможно преобразовать "LRESULT (__cdecl *)(HWND,UINT,WPARAM,LPARAM)" в "WNDPROC"  Interface   C:\Users\pepe\source\repos\Interface\Interface\main.c   31  ```