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iOS 数组元素的深拷贝

2016-02-17  本文已影响2905人  我是C
实现 NSCopying,NSMutableCopying 协议
如果用NSCopying 实现:
 - (id)copyWithZone:(nullable NSZone *)zone
如果用NSMutableCopying 实现:
  - (id)mutableCopyWithZone:(nullable NSZone *)zone
Model
@interface Model : NSObject<NSCopying,NSMutableCopying>
@property (nonatomic, copy) NSString *str1;
@property (nonatomic, copy) NSString *str2;
@end 
@implementation Model
  - (id)copyWithZone:(nullable NSZone *)zone{
 Model *person = [[self class] allocWithZone:zone];
 person.str1 = [_str1 copy];
 person.str2 = [_str2 copy];
 return person;
}
  - (id)mutableCopyWithZone:(nullable NSZone *)zone{
 Model *person = [[self class] allocWithZone:zone];
 person.str1 = [_str1 copy];
 person.str2 = [_str2 copy];
 return person;
}
@end
//构造一个Model 数组
  NSMutableArray *arr = [NSMutableArray array];
  for (int i = 0; i< 3; i++) {
  Model *model = [[Model alloc] init];
  model.str1 = @"1";
  model.str2 = @"2";
  NSLog(@"数组%d:---%p",i,model);
  [arr addObject:model];
  }
那么我想拷贝数组对象,直接
NSMutableArray *arr1 = [arr mutableCopy];
是不是这样就可以了
arr内存地址:0x7f8d8b7972a0
arr1内存地址:0x7f8d8b796fe0
显然是不一样的,那么你会说这样就可以了,其实你想多了,断点后看两个数组的元素的地址改变了没有:
结果:po arr
<__NSArrayM 0x7f8d8b7972a0>(
<Model: 0x7f8d8b78f110>,
<Model: 0x7f8d8b52b470>,
<Model: 0x7f8d8b408830>
)
po arr1
<__NSArrayM 0x7f8d8b796fe0>(
<Model: 0x7f8d8b78f110>,
<Model: 0x7f8d8b52b470>,
<Model: 0x7f8d8b408830>
)
  NSMutableArray *arr = [NSMutableArray array];
  for (int i = 0; i< 3; i++) {
  Model *model = [[Model alloc] init];
  model.str1 = @"1";
  model.str2 = @"2";
  NSLog(@"数组%d:---%p",i,model);
  [arr addObject:model];
  }
  NSMutableArray *arr1 = [NSMutableArray array];
  for (Model *model in arr) {
  [arr1 addObject:[model copy]];//这里的copy,就是将arr数组里的每一个对象拷贝一份.
//所以Model对象实现- (id)copyWithZone:(nullable NSZone *)zone方法
  }
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