Replace Sleep until Window on desktop opened - c++

Question

When, I open some of the software applications I have to wait 2-3 seconds until window will show on desktop. I have to use Sleep(2000); and then call method set always on top. I'm trying to replace Sleep in my code. I would like to get signal from opened window and after this, call a method, which allows opened window be always on top.
Here's my code:
BOOL CALLBACK EnumWindowsProc(HWND windowHandle, LPARAM lParam)
{
DWORD searchedProcessId = (DWORD)lParam;
DWORD windowProcessId = 0;
GetWindowThreadProcessId(windowHandle, &windowProcessId);
cout << "Process id: " << windowProcessId << endl;
if(searchedProcessId == windowProcessId) {
HWND hwnd = windowHandle;
Sleep(2000);
SetWindowPos(hwnd, HWND_TOPMOST, 0, 0, 0, 0, SWP_NOMOVE|SWP_NOSIZE);
cout << "Process ID found!" << endl;
return TRUE;
}
return TRUE;
}
void AlwaysOnTop(int processId)
{
EnumWindows(&EnumWindowsProc, (LPARAM)processId);
}
void AlwaysOnTop(char *name)
{
cout << "String: " << name << endl;
Sleep(2000);
HWND h = FindWindow(NULL, (LPCSTR) name);
SetActiveWindow(h);
SetForegroundWindow(h);
SetWindowPos(h, HWND_TOPMOST, 0,0,0,0, SWP_NOSIZE | SWP_NOMOVE);
}
int main()
{
char s[] = {"Application"};
AlwaysOnTop(s);
//AlwaysOnTop(2307);
system("PAUSE");
return 0;
}

Answer 1

Probably the best you can do is to call WaitForInputIdle:
Waits until the specified process has finished processing its initial input and is waiting for user input with no input pending, or until the time-out interval has elapsed.
This is the closest you can get to a general way to wait until a process is showing its UI. It won't always do what you want, but it's the best there is.