我尝试使用协议,但是它需要为每个具有相同内容的2个类实现“ makeMeDraggable”方法。我不希望在每个2类中都实现,而不要从Shared Class继承实现。我该如何实现?
码:
//--(.h)--//
@interface SharedClass : UIView;
-(void) makeMeDraggable
@end
@interface DraggableUITextView: UITextView
@end
@interface DraggableUIImageView : UIImageView
@end
@implementaion SharedClass
-(void) makeMeDraggable {
//some code
}
@end
@implementation DraggableUITextView
@end
@implementation DraggableUIImageView
@end
//--(.m)--//
TextView *textView = [DraggableUITextView initWithFrame:CGRectMake(0,0,50,50)];
[textView makeMeDraggable];
ImageView *imageView = [DraggableUIImageView imageNamed:@"foo.png"];
[imageView makeMeDraggable];
我想避免以下情况:
@interface DraggableTextView: UITextView;
-(void) makeMeDraggable
@end
@interface DraggableUIImageView : UIImageView
-(void) makeMeDraggable
@end
@implementation DraggableUITextView
-(void) makeMeDraggable {
//same code...
}
@end
@implementation DraggableUIImageView
-(void) makeMeDraggable {
//same code...
}
@end
最佳答案
创建一个UIView
类别,然后在其中而不是SharedView
编写函数。
@interface UIView (UIViewCategory)
-(void) makeMeDraggable;
@end
@implementaion UIView (UIViewCategory)
-(void) makeMeDraggable {
//some code
}
@end
现在,您可以对从
makeMeDraggable
类继承的每个对象使用UIView
方法。