Osc*_*and 12 objective-c nsobject ios
我想要完成的是类似的东西
Person *person1 = [[Person alloc]initWithDict:dict];
Run Code Online (Sandbox Code Playgroud)
然后在NSObject"人"中,有类似的东西:
-(void)initWithDict:(NSDictionary*)dict{
self.name = [dict objectForKey:@"Name"];
self.age = [dict objectForKey:@"Age"];
return (Person with name and age);
}
Run Code Online (Sandbox Code Playgroud)
然后,我允许我继续使用人物对象与那些参数.这是可能的,还是我必须正常做
Person *person1 = [[Person alloc]init];
person1.name = @"Bob";
person1.age = @"123";
Run Code Online (Sandbox Code Playgroud)
?
Hin*_*ndu 27
你的返回类型是空的instancetype.
你可以使用你想要的两种类型的代码....
更新:
@interface testobj : NSObject
@property (nonatomic,strong) NSDictionary *data;
-(instancetype)initWithDict:(NSDictionary *)dict;
@end
Run Code Online (Sandbox Code Playgroud)
.M
@implementation testobj
@synthesize data;
-(instancetype)initWithDict:(NSDictionary *)dict{
self = [super init];
if(self)
{
self.data = dict;
}
return self;
}
@end
Run Code Online (Sandbox Code Playgroud)
使用方法如下:
testobj *tt = [[testobj alloc] initWithDict:@{ @"key": @"value" }];
NSLog(@"%@",tt.ss);
Run Code Online (Sandbox Code Playgroud)
zt9*_*788 10
像这样改变你的代码
-(id)initWithDict:(NSDictionary*)dict
{
self = [super init];
if(self)
{
self.name = [dict objectForKey:@"Name"];
self.age = [dict objectForKey:@"Age"];
}
return self;
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
21353 次 |
| 最近记录: |