C++ 运行一个异步函数而不阻塞 ui

Question

我有一个检查 dotnet 框架的小应用程序，如果没有安装它会安装它

现在，当应用程序启动时，我想弹出一个 gif 图像，其中包含加载之类的内容，并在后台检查框架并安装。

这里的问题是它不能有任何先决条件ui站点到运行应用程序

这是我目前所拥有的

int APIENTRY wWinMain(_In_ HINSTANCE hInstance,
                  _In_opt_ HINSTANCE hPrevInstance,
                  _In_ LPWSTR lpCmdLine,
                  _In_ int nCmdShow)
{
       int exitCode = -1;
       showPic(hInstance);
       MessageBox(0L, L"Dotnet will installed", L"Alert", 0);
       auto fut = std::async(std::launch::async, DoWork, hInstance, lpCmdLine);
       fut.wait();
       exitCode = fut.get();
       return exitCode;
}

showPic()

void showPic(HINSTANCE hInstance)
{
    loadImage(hInstance);
    // create window
    wnd = createWindow(hInstance);
    SetWindowLong(wnd, GWL_STYLE, 0);
    ShowWindow(wnd, SW_SHOW);
}

loadImage(HINSTANCE hInstance)

void loadImage(HINSTANCE hInstance)
{
    imageDC = CreateCompatibleDC(NULL);
    imageBmp = LoadBitmap(hInstance, MAKEINTRESOURCE(IDB_BITMAP1));
    imageBmpOld = (HBITMAP)SelectObject(imageDC, imageBmp);
}

现在这里发生的事情是，如果我不显示消息框，图片不会加载到 window，而且 window 仍然进入无响应模式，我也无法得到它适用于 gif，仅适用于 bmp 图像任何帮助都适用

现在因为我等待 fut 很明显它会阻塞 ui 直到它有值，解决方法是什么

Answer 1

这应该很简单。创建 window，显示它，调用线程，转到主消息循环。线程完成后，它将销毁 window.

struct T_data {
    HWND hWnd;
    HINSTANCE hInstance;
    LPTSTR cmdline;
    int exitCode;
};

DWORD WINAPI taking_too_long(LPVOID p) {
    Sleep(2000); //wait at least 2 seconds!
    T_data* data = reinterpret_cast<T_data*>(p);
    auto fut = std::async(std::launch::async, DoWork, data->hInstance, data->lpCmdLine);
    fut.wait();
    data->exitCode = fut.get();
    //make sure the window handles IDM_EXIT to close itself
    PostMessage(data->hWnd, WM_COMMAND, IDM_EXIT, 0);
}

int APIENTRY wWinMain(HINSTANCE hInstance, HINSTANCE, LPTSTR lpCmdLine, int) {
    T_data data;
    data.exitCode = -1;
    data.hWnd = hWnd;
    data.hInstance = hInstance;
    data.cmdline = lpCmdLine;

    data.hWnd = showPic(hInstance);

    CreateThread(NULL, 0, taking_too_long, &data, 0, NULL);

    MSG msg;
    while (GetMessage(&msg, NULL, 0, 0)) {
        TranslateMessage(&msg);
        DispatchMessage(&msg);
    }
    return data.exitCode;
}

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C++ 运行一个异步函数而不阻塞 ui