
“Swift automatically brIDges between the String type and the Nsstring class. This means that anywhere you use an Nsstring object,you can use a Swift String type instead and gain the benefits of both types—the String type’s interpolation and Swift-designed APIs and the Nsstring class’s broad functionality. For this reason,you should almost never need to use the Nsstring class directly in your own code. In fact,when Swift imports Objective-C APIs,it replaces all of the Nsstring types with String types. When your Objective-C code uses a Swift class,the importer replaces all of the String types with Nsstring in imported API.
要启用字符串桥接,只需导入基础“.
我做了这个…考虑:
import Foundationvar str = "Hello World"var range = str.rangeOfString("e")// returns error: String does not contain member named: rangeOfString() 然而:
var str = "Hello World" as Nsstringvar range = str.rangeOfString("e")// returns correct (2,1) 我错过了什么吗?
解决方法 你已经有你的问题的答案.你错过了演员.写Swift代码时,这样一个语句var str = "Hello World"
创建一个Swift字符串,而不是Nsstring.要使其作为Nsstring工作,您应该在使用之前使用as *** 作符将其转换为Nsstring.
这不同于调用在Objective-C中编写的方法,并提供一个String而不是一个Nsstring作为参数.
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