C++有判断窗口是否显示隐藏的api吗？

[wangsky2]

如题。。。。。。。。。。。。。。

[烟袅许伊]

应该有吧？Windows编程。。。怎么会没有窗口的说法。。。
[Quote=引用 1 楼 的回复:]

C++一般没窗口的说法吧
如果你是说MFC，就是 IsWindowVisible
[/Quote]

[Piao_Polar]

恩，windows也有，但问题是人家说C++，好吧，我执着了

[W170532934]

C++是语言，API归API啦。

[wangsky2]

大家多原谅啊，菜鸟。我的意思是，自己创建的Button，可以用ShowWindow(m_hWnd, SW_HIDE)来隐藏，用ShowWindow(m_hWnd, SW_SHOW)来显示，不知道有没有判断函数来判断它是否隐藏？

[zhanshen2891]

[Quote=引用 5 楼 的回复:]
大家多原谅啊，菜鸟。我的意思是，自己创建的Button，可以用ShowWindow(m_hWnd, SW_HIDE)来隐藏，用ShowWindow(m_hWnd, SW_SHOW)来显示，不知道有没有判断函数来判断它是否隐藏？
[/Quote]

有啊，IsWindowVisible（m_hWnd）

[赵4老师]

查MSDN是Windows程序员必须掌握的技能之一。

IsWindowVisible
The IsWindowVisible function retrieves the visibility state of the specified window.

BOOL IsWindowVisible(
HWND hWnd // handle to window
);

Parameters
hWnd
Handle to the window to test.
Return Values
If the specified window and its parent window have the WS_VISIBLE style, the return value is nonzero.

If the specified window and its parent window do not have the WS_VISIBLE style, the return value is zero. Because the return value specifies whether the window has the WS_VISIBLE style, it may be nonzero even if the window is totally obscured by other windows.

Remarks
The visibility state of a window is indicated by the WS_VISIBLE style bit. When WS_VISIBLE is set, the window is displayed and subsequent drawing into it is displayed as long as the window has the WS_VISIBLE style.

Any drawing to a window with the WS_VISIBLE style will not be displayed if the window is obscured by other windows or is clipped by its parent window.

QuickInfo
Windows NT: Requires version 3.1 or later.
Windows: Requires Windows 95 or later.
Windows CE: Requires version 1.0 or later.
Header: Declared in winuser.h.
Import Library: Use user32.lib.

See Also
Windows Overview, Window Functions, ShowWindow

[Piao_Polar]

C++一般没窗口的说法吧
如果你是说MFC，就是 IsWindowVisible