Swift 相当于@encode [英] Swift equivalent of @encode

问题描述

是否有与 Objective-C 的 @encode 等效的 Swift?

例如

@encode(void *)//->@"^v"

搜索一无所获.

解决方案

不，没有 - 因为在幕后 Swift 类不使用 Objective-C 自省来完成他们的工作.没有必要计算这个(就像在 Objective-C 中那样)来传递/调用数据.

但是，如果您需要在运行时动态使用它(例如，为了与现有的 Objective-C 方法进行互操作)，那么您可以创建一个 Objective-C 调用并传递对象，或者(对于简单类型)编写查找表.

类型编码列于 https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/ObjCRuntimeGuide/Articles/ocrtTypeEncodings.html 有地图，并且可以编写一个开关类型语句来进行查找.>

但从根本上说，如果你有一个类型想要传入并且发现它是客观的c编码类型，你可以使用NSObject的objCType方法:

var i = 1 作为 NSNumberString.fromCString(i.objCType)！==q"

如果您无论如何都需要将其作为未受干扰的 C 字符串传递，您甚至可能不需要将其转换回 Swift 字符串类型.

Is there a Swift equivalent to Objective-C's @encode?

For instance

@encode(void *) // -> @"^v"

Searching yielded nothing.

解决方案

No, there isn't - because under the hood Swift classes don't use Objective-C introspection to do their work. There's no need to calculate this (like there is in Objective-C) in order to pass/call data.

However, if you need to use it dynamically at runtime (say, for interoperation with existing Objective-C methods) then you can either create an Objective-C call and pass the object through or (for simple types) write a lookup table.

The type encodings are listed at https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/ObjCRuntimeGuide/Articles/ocrtTypeEncodings.html which have the map, and it's possible to write a switch type statement that does the lookup.

But fundamentally if you have a type that you want to pass in and find it's objective c encoding type, you can use the NSObject's objCType method:

var i = 1 as NSNumber
String.fromCString(i.objCType)! == "q" 

If you need to pass it through as an unmolested C string anyway, you may not even need to convert it back to a Swift string type.

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