PostMessage unexpected behaviour

Question

I'm having a funny ol' time with PostMessage, wondering if anybody could enlighten me.

The method below processes the first left click but not the second, I did a bit of messing around and found that using a MessageBox and hitting OK even instantly provides the delay/release or whatever it is that's needed to make the second left click process.

public void MouseClick(IntPtr handle)
{
    PostMessage(handle, (uint)WMessages.WM_LBUTTONDOWN, 0, MAKELPARAM(521, 147));
    PostMessage(handle, (uint)WMessages.WM_LBUTTONUP, 0, MAKELPARAM(521, 147));
    //MessageBox.Show("WAITING...");
    PostMessage(handle, (uint)WMessages.WM_LBUTTONDOWN, 0, MAKELPARAM(675, 527));
    PostMessage(handle, (uint)WMessages.WM_LBUTTONUP, 0, MAKELPARAM(675, 527));
}

Thanks in advance.

http://blogs.msdn.com/b/oldnewthing/archive/2005/05/30/423202.aspx — ta.speot.is, Jan 26 '13 at 21:23
Your declaration for PostMessage is wrong, no way the 3rd argument can be 0. Other than that, surely it is quirk of the program you are trying to commandeer. — Hans Passant, Jan 26 '13 at 21:28
@HansPassant The third parameter for `PostMessage` is `WPARAM wParam`, why can't that be 0? — ta.speot.is, Jan 27 '13 at 01:08
Because the equivalent C# type is IntPtr and you can't pass 0 for an argument of type IntPtr. — Hans Passant, Jan 27 '13 at 01:16

score -2 · Answer 1 · answered Jan 26 '13 at 21:27

-2

That is because the main thread message pump is suspended while you are in the MouseClick function. The messagebox call re-enables the message pump. What you can do is call Application.DoEvents or do it in another thread.

answered Jan 26 '13 at 21:27

Mark PM

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PostMessage unexpected behaviour

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