UITapGestureRecognizer gestureRecognizer :shouldReceiveTouch:? 中的 iOS6 错误

Question

我的 View Controller 中的以下代码在 touch.view 调用中导致 EXC_BAD_ACCESS:

- (BOOL)handleSingleTap:(UITapGestureRecognizer *)recognizer shouldReceiveTouch:(UITouch *)touch {

    if ([touch.view isKindOfClass:[UIControl class]]) {   // <<<< EXC_BAD_ACCESS HERE
        // we touched a button, slider, or other UIControl
        return NO; // ignore the touch
    }
    [self.view endEditing:YES];  // dismiss the keyboard
    return YES; // handle the touch    
}

touch 似乎是一个僵尸。具体来说，touch 被设置为一个地址，调试器认为这是一个 UITouch 指针，但它没有任何属性:

这在 iOS 4.x 中没有发生。那么 iOS 6 错误还是我的错？

手势识别器设置如下(在 ViewController 中):

UITapGestureRecognizer *tapRecognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(handleSingleTap:shouldReceiveTouch:)];
[tapRecognizer setDelegate:self];
[tapRecognizer setNumberOfTapsRequired:1];
[tapRecognizer setNumberOfTouchesRequired:1];
tapRecognizer.cancelsTouchesInView = NO;
[self.view addGestureRecognizer:tapRecognizer];
[tapRecognizer release];

更新/解决方案:

感谢 Rob 指出我的用户错误。不仅如此，我实际上并不需要 TapRecognizer，因为我只想知道发生了触摸。

下面是在 UIControl 以外的任何地方通过触摸关闭屏幕键盘的正确代码:

在 ViewController 的 viewDidLoad 中:

UIGestureRecognizer *myRecognizer = [[UIGestureRecognizer alloc] init];
[myRecognizer setDelegate:self];
myRecognizer.cancelsTouchesInView = NO;
[self.view addGestureRecognizer:myRecognizer];
[myRecognizer release];

和:

- (BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch  {

if ([touch.view isKindOfClass:[UIControl class]]) {
    // we touched a button, slider, or other UIControl
    return NO; // ignore the touch
}

[self.view endEditing:YES];  // dismiss the keyboard
return YES; // handle the touch    

Answer 1

您正在使用方法 handleSingleTap:shouldReceiveTouch:，其签名不符合允许的签名。查看Overview UIGestureRecognizer 类引用部分。 它应该只有一个参数，即手势识别器。

我认为您可能将处理程序与 UIGestureRecognizerDelegate 协议(protocol)方法混淆了 gestureRecognizer:shouldReceiveTouch: .