how to check if a window visible or not using iswindowvisible() function.

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Hello,

I am writing a program in c++ to check weather a window (open by a user ex: notepad) is visible or not using "iswindowvisible(HWND hWnd)" function.
But it always returns a zero.

Need your help to resolve this problem.

Thank you
Suryakant tyagi

Posted 2-Nov-15 9:35am

Suryakant Turing

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jeron1 2-Nov-15 16:17pm

Maybe post some code that's relevant to your problem.

Suryakant Turing 3-Nov-15 2:17am

I got your point but I still getting zero output while using FindWindow() and then IsWindowvisible().

I also have process Id and execution application name if it can help in another way.

Thank you

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User 59241 · Answer 1 · 2015-11-02T12:53:00

Check these:
https://msdn.microsoft.com/en-us/library/windows/desktop/ms633530(v=vs.85).aspx[^]

http://stackoverflow.com/questions/1061246/how-can-i-check-if-a-window-has-ws-visible-to-set-or-if-is-visible[^]

This quote from first link is crucial:
"If the specified window, its parent window, its parent's parent window, and so forth, have the WS_VISIBLE style, the return value is nonzero. Otherwise, the return value is zero."

From the second link a solution:
"One nuance to be aware of. IsWindowVisible will return the true visibility state of the window, but that includes the visibility of all parent windows as well.

If you need to check the WS_VISIBLE flag for a specific window you can do GetWindowLong(hWnd, GWL_STYLE) and test for WS_VISIBLE."