IOS:向@selector添加一个参数

Question

当我有这行代码

UILongPressGestureRecognizer *downwardGesture = [[UILongPressGestureRecognizer alloc] initWithTarget:self action:@selector(dragGestureChanged:)];

还有这个

- (void)dragGestureChanged:(UILongPressGestureRecognizer*)gesture{
...
}

我想在"@selector(dragGestureChanged :)"中添加一个参数即"(UIScrollView*)scrollView",我该怎么办？

Answer 1

你不能直接UIGestureRecognizer知道如何调用只接受一个参数的选择器.为了完全一般,你可能希望能够传递一个块.Apple没有内置它,但它很容易添加,至​​少如果你愿意将手势识别器子类化,你想要解决添加新属性和正确清理的问题,而不深入研究运行时.

所以,例如(我去的时候,未经检查)

typedef void (^ recogniserBlock)(UIGestureRecognizer *recogniser);

@interface UILongPressGestureRecognizerWithBlock : UILongPressGestureRecognizer

@property (nonatomic, copy) recogniserBlock block;
- (id)initWithBlock:(recogniserBlock)block;

@end

@implementation UILongPressGestureRecognizerWithBlock
@synthesize block;

- (id)initWithBlock:(recogniserBlock)aBlock
{
    self = [super initWithTarget:self action:@selector(dispatchBlock:)];

    if(self)
    {
         self.block = aBlock;
    }

    return self;
}

- (void)dispatchBlock:(UIGestureRecognizer *)recogniser
{
    block(recogniser);
}

- (void)dealloc
{
    self.block = nil;
    [super dealloc];
}

@end

然后你可以这样做:

UILongPressGestureRecognizer = [[UILongPressGestureRecognizerWithBlock alloc] 
        initWithBlock:^(UIGestureRecognizer *recogniser)
        {
            [someObject relevantSelectorWithRecogniser:recogniser 
                      scrollView:relevantScrollView];
        }];