迅速.使用扩展函数替换变量的值
发布时间:2020-12-14 04:49:35 所属栏目:百科 来源:网络整理
导读:我想创建一个函数来更改变量的原始值.例如 class Something { var name:String = " John Diggle " name.trim() print(name) // prints out " John Diggle " // what I wanna do is to make it so that I don't do this name = name.trim() print(name) // pr
我想创建一个函数来更改变量的原始值.例如
class Something { var name:String = " John Diggle " name.trim() print(name) // prints out " John Diggle " // what I wanna do is to make it so that I don't do this name = name.trim() print(name) // prints out "John Diggle" } extension String { func trim() -> String{ return self.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()) } } 有没有办法在不执行name = name.trim()的情况下更改函数内部变量的值? 解决方法
也许是这样的?
extension String { mutating func trim() { self = self.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines) } } 然后你可以使用它作为name.trim() (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |