Ios 错误:';没有可见的@接口;NSObject';声明选择器';copyWithZone:&x27;
我希望允许类对象的深度复制,并尝试实现copyWithZone,但调用Ios 错误:';没有可见的@接口;NSObject';声明选择器';copyWithZone:&x27;,ios,copywithzone,Ios,Copywithzone,我希望允许类对象的深度复制,并尝试实现copyWithZone,但调用[super copyWithZone:zone]会产生错误: error: no visible @interface for 'NSObject' declares the selector 'copyWithZone:' @interface MyCustomClass : NSObject @end @implementation MyCustomClass - (id)copyWithZone:(NSZone
[super copyWithZone:zone]
会产生错误:
error: no visible @interface for 'NSObject' declares the selector 'copyWithZone:'
@interface MyCustomClass : NSObject
@end
@implementation MyCustomClass
- (id)copyWithZone:(NSZone *)zone
{
// The following produces an error
MyCustomClass *result = [super copyWithZone:zone];
// copying data
return result;
}
@end
如何创建此类的深度副本?您应该将
NSCopying
协议添加到类的接口中
@interface MyCustomClass : NSObject <NSCopying>
NSObject
不符合NSCopying
协议。这就是为什么您不能调用超级copyWithZone:
编辑:根据罗杰的评论,我更新了
copyWithZone:
方法中的第一行代码。但根据其他评论,可以安全地忽略该区域。这是正确的。我写了一个类似的答案,但他打败了我:)你的答案完全忽略了区域。请参阅@RogerBinns查看已接受的答案
- (id)copyWithZone:(NSZone *)zone {
MyCustomClass *result = [[[self class] allocWithZone:zone] init];
// If your class has any properties then do
result.someProperty = self.someProperty;
return result;
}