1.如何破解Alfred3:程序员
https://www.jianshu.com/p/5b3f98b1f7b6学习
2.学习了一篇精品文章:我真的要作一生的程序员吗?spa
https://mp.weixin.qq.com/s/zeSgti_uxaSNzYkaAfct-gget
3.元组:程序
对元组中的元素进行修改,注意只能修改,不能添加或者删除,any类型能够改成任何类型d3
4.字典的合并: qq
//即便类型一致也不能相加进行合并dict
//let dict1 = ["name":"summer", "age" : 18] as [String : Any]di
//let dict2 = ["sex" : "男", "phoneNumber" : "+86 0393"] as [String : Any]co
//
//let resultDict = dict1 + dict2 //写法是错误的
//若是必须合并 那么只能
var dict1 = ["name":"summer", "age" : 18] as [String : Any]
let dict2 = ["sex" : "男", "phoneNumber" : "+86 0393"] as [String : Any]
for (key, value) in dict2{
dict1[key] = value as AnyObject?
}