IOS：在@selector中添加一个参数

概述当我有这行代码 UILongPressGestureRecognizer *downwardGesture = [[UILongPressGestureRecognizer alloc] initWithTarget:self action:@selector(dragGestureChanged:)]; 还有这个 - (void)dragGestureChanged:(UILongPressGe 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？

解决方法 你不能直接–UIGestureRecognizers知道如何发出一个只接受一个参数的选择器的调用.要完全一般,你可能希望能够传递一个块.苹果公司还没有这样做,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(我去的时候写的,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

然后你就可以这样做：

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

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