Iphone 如何找到未记录版本的HandleTesture选择器（适用于UIWebView）需要哪些类型的参数？

我已将自定义弹出菜单添加到UIWebView实例：

- (void)viewDidLoad
{
    UILongPressGestureRecognizer* gesture = [[[UILongPressGestureRecognizer alloc] 
    initWithTarget:self action:@selector(handleGesture::)] autorelease];
}

- (void)handleGesture
{

}

- (void)handleGesture:(UIGestureRecognizer*)gestureRecognizer
{
    if (gestureRecognizer.state == UIGestureRecognizerStateEnded) {
        [gestureRecognizer.view becomeFirstResponder];
        UIMenuController* mc = [UIMenuController sharedMenuController];
        [mc setTargetRect: gestureRecognizer.view.frame inView: gestureRecognizer.view.superview];
        [mc setMenuVisible: YES animated: YES];
}

而且它有效！直到我将textarea（CodeMirror编辑器）聚焦到网页上。在这种情况下，我有以下例外：

-[FirstViewController handleGesture::]: unrecognized selector sent to instance 0x20369c00
*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[FirstViewController handleGesture::]: unrecognized selector sent to instance 0x20369c00'
*** First throw call stack:
(0x3608c2a3 0x3439c97f 0x3608fe07 0x3608e531 0x35fe5f68 0x3747ad31 0x374423dd 0x3762f479 0x37366837 0x3736529b 0x360616cd 0x3605f9c1 0x3605fd17 0x35fd2ebd 0x35fd2d49 0x3650f2eb 0x373b1301 0x140bb 0x14060)
libc++abi.dylib: terminate called throwing an exception
(lldb) 

我认为需要实现带有外来参数的HandleTesture选择器的无文档版本。是这样吗？如何找到所需的参数类型？

正确的代码：

initWithTarget:self action:@selector(handleGesture:)] autorelease];

---

我希望您没有打算将此提交到应用商店。使用未记录的API是立即拒绝的理由：“2.5使用非公共API的应用将被拒绝”，但这是你的应用。如果你想抓住机会，希望他们不会注意到，那是你的权利。我只是想确保当你被拒绝时你不会感到惊讶。