通常需要实现对模型的拷贝都需要先实现NSCopying、 NSMutableCopying协议，在这里我一直有个误区，以为实现了copying协议，数组使用拷贝操作都会对数组内实现copy协议的对象进行拷贝。
创建两个model并实现copying协议
// Dog.h
@interface Dog : NSObject <NSCopying, NSMutableCopying>
@property (nonatomic, copy) NSString *name;
@end
// Dog.m
@implementation Dog
- (nonnull id)copyWithZone:(nullable NSZone *)zone {
Dog *copy = [[Dog allocWithZone:zone]init];
copy.name = self.name;
return copy;
}
- (id)mutableCopyWithZone:(NSZone *)zone {
Dog *copy = [[Dog allocWithZone:zone]init];
copy.name = self.name;
return copy;
}
@end
// Person.h
#import "Dog.h"
@interface Person : NSObject <NSCopying, NSMutableCopying>
@property (nonatomic, copy) NSString *name;
@property (nonatomic, assign) NSInteger age;
@property (nonatomic, copy) Dog *dog;
@end
// Person.m
@implementation Person
- (nonnull id)copyWithZone:(nullable NSZone *)zone {
Person *copy = [[Person allocWithZone:zone]init];
copy.name = self.name;
copy.age = self.age;
copy.dog = [self.dog copy];
return copy;
}
- (id)mutableCopyWithZone:(NSZone *)zone {
Person *copy = [[Person allocWithZone:zone]init];
copy.name = self.name;
copy.age = self.age;
copy.dog = [self.dog copy];
return copy;
}
@end
对模型数组使用mutablCopy操作并不是深拷贝
NSMutableArray <Person *>*dataSourceAry = [NSMutableArray new];
for (int i = 0; i < 2; i++) {
Dog *dog = [[Dog alloc]init];
dog.name = @"拉不拉多不多就拉倒";
Person *item = [[Person alloc]init];
item.name = @"杰克";
item.age = 18;
item.dog = [dog copy];
[dataSourceAry addObject:item];
}
NSArray <Person *>*array = [dataSourceAry mutableCopy];
NSLog(@"<dataSourceAry: %@>", dataSourceAry);
NSLog(@"<array: %@>", array);
[array enumerateObjectsUsingBlock:^(Person *obj, NSUInteger idx, BOOL * _Nonnull stop) {
obj.name = @"肉丝";
}];
NSLog(@"dataSourceAry[0].name = %@", dataSourceAry[0].name);
NSLog(@"array[0].name = %@", array[0].name);
/** 打印结果
<dataSourceAry: ("<Person: 0x60000043ec00>", "<Person: 0x60000043e9c0>")>
<array: ("<Person: 0x60000043ec00>","<Person: 0x60000043e9c0>")>
dataSourceAry[0].name = 肉丝
array[0].name = 肉丝
*/
可以从打印的结果看出，数组内元素的内存地址是相同的，所以出现了修改拷贝后数组的第一个元素，导致原数组的第一个元素也发生了同样的改变。虽然两个数组的内存地址不一样，但是内部元素内存地址还是同一地址，不是我们想要的结果。
实现模型数组深拷贝的方法
1、最笨的方法就是通过遍历逐个拷贝元素
NSMutableArray *array = [NSMutableArray array];
for (Person *person in dataSourceAry) {
[array addObject:[person copy]];
}
2、也有人使用归档解档实现数组内部元素拷贝
3、这么好用的一个方法现在才发现（推荐）
- (instancetype)initWithArray:(NSArray<ObjectType> *)array copyItems:(BOOL)flag
NSArray <Person *>*deepCopyAry = [[NSArray alloc]initWithArray:dataSourceAry copyItems:YES];
NSLog(@"<dataSourceAry: %@>", dataSourceAry);
NSLog(@"<deepCopyAry: %@>", deepCopyAry);
[deepCopyAry enumerateObjectsUsingBlock:^(Person *obj, NSUInteger idx, BOOL * _Nonnull stop) {
obj.name = @"弗兰克";
obj.dog.name = @"弗兰克的dog";
}];
NSLog(@"dataSourceAry[0].name = %@", dataSourceAry[0].name);
NSLog(@"deepCopyAry[0].name = %@", deepCopyAry[0].name);
NSLog(@"dataSourceAry[0].dog.name = %@", dataSourceAry[0].dog.name);
NSLog(@"deepCopyAry[0].dog.name = %@", deepCopyAry[0].dog.name);
/** 打印结果
<dataSourceAry: ("<Person: 0x604000427680>", "<Person: 0x604000425220>")>
<deepCopyAry: ("<Person: 0x60000042cb80>", "<Person: 0x60000042cae0>")>
dataSourceAry[0].name = 肉丝
deepCopyAry[0].name = 弗兰克
dataSourceAry[0].dog.name = 拉不拉多不多就拉倒
deepCopyAry[0].dog.name = 弗兰克的dog
*/
总结
1、模型数组内元素中模型必须要实现copying协议，模型内如果有嵌套模型，也需要实现copying协议，否则执行对对象拷贝操作会出现崩溃；
2、使用- (instancetype)initWithArray:(NSArray<ObjectType> *)array copyItems:(BOOL)flag实现模型数组拷贝；
