八月第四周:

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?

}

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