ios - 为什么CALayer的类别可以声明一个属性而不需要实现并直接使用它？

Question

据我所知，你不应该在类别中定义实例变量，如果你声明一个属性，那只意味着你声明了“setter & getter”方法。

@interface CALayer (XQ)
    @property NSString *demoVar;
    - (void)demoFunc;
@end
@implementation CALayer (XQ)
    - (void)demoFunc {
        self.demoVar = @"cuteeeee";
        NSLog(@"%@", self.demoVar);// when i call this method, it should crash, but the output is normal, why?
    }
@end

抱歉我的英语和短语不好，谢谢。

Answer 1

CALayer.h 中的注释指出:

/** Property methods. **/

/* CALayer implements the standard NSKeyValueCoding protocol for all
 * Objective C properties defined by the class and its subclasses. It
 * dynamically implements missing accessor methods for properties
 * declared by subclasses.

显然，CALayer 对待类别中声明的属性与对待子类中声明的属性相同，因此它会为您动态实现访问器方法。