#Objective-C 枚举ENUMcode
##简单枚举 ###定义it
typedef NS_ENUM(NSUInteger, Type) { //用户未持有券的状态 TypeA = 0, TypeB = 1, TypeC = 2, }
###使用io
Type type = TypeA; //if 语句 if (status9==CouponStatus_UserGet) { //your code } //switch 语句 switch (type) { case TypeA: //your code break; case TypeB: //your code break; default: break; }
##高级枚举原理
###定义request
typedef NS_ENUM(NSUInteger, Type) { TypeA = 1 << 0, TypeB = 1 << 1, TypeC = 1 << 2, TypeD = TypeA|TypeB, }
使用<<属于枚举的高级用法 a<<N: 右移符号,按二级制向右右移N位,多出来的位置,由0补充 3 << 2 = 0b11<<3 即0b1100
###使用error
下面一种常见的用法推送
[[UNUserNotificationCenter currentNotificationCenter] requestAuthorizationWithOptions:(UNAuthorizationOptionBadge|UNAuthorizationOptionSound|UNAuthorizationOptionAlert) completionHandler:^(BOOL granted, NSError * _Nullable error) { NSLog(@"iOS10注册消息推送:%@",granted?@"成功":@"失败"); }];
参数中出现了这样一个参数(UNAuthorizationOptionBadge|UNAuthorizationOptionSound|UNAuthorizationOptionAlert)消息推送
很显然是想同时实现3种状况。那么它的内部实现是怎么的?ant
- (void)options:(Type)type { if(type&TypeA){ //your code } if(type&TypeB){ //your code } if(type&TypeC){ //your code } } //原理解析 TypeA = 1 << 0 , 0b1<<0 = 0b1 = 1 TypeB = 1 << 1 , 0b1<<1 = 0b10 = 2 TypeC = 1 << 2 , 0b1<<2 = 0b100 = 4 TypeD = TypeA|TypeB = 1|2 = 0b1| 0b10= 0b11 = 3 //下面计算 TypeD&TypeA = 3 & 1 = 0b11 & 0b1 = 0b01 = 1 TypeD&TypeC = 3 & 4 = 0b11 & 0b100 = 0b000 = 0