2016年9月更新
Swift 3.0:使用type(of:),例如type(of: someThing)(因为dynamicType关键字已被删除)
2015年10月更新:
我更新了下面的例子到新的Swift 2.0语法(例如println替换为print, toString()现在是String())。
Xcode 6.3发布说明:
@nschum在评论中指出,Xcode 6.3发布说明显示了另一种方式:
使用时,类型值现在打印为完整的需求类型名
Println或字符串插值。
import Foundation
class PureSwiftClass { }
var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"
print( "String(myvar0.dynamicType) -> \(myvar0.dynamicType)")
print( "String(myvar1.dynamicType) -> \(myvar1.dynamicType)")
print( "String(myvar2.dynamicType) -> \(myvar2.dynamicType)")
print( "String(myvar3.dynamicType) -> \(myvar3.dynamicType)")
print( "String(Int.self) -> \(Int.self)")
print( "String((Int?).self -> \((Int?).self)")
print( "String(NSString.self) -> \(NSString.self)")
print( "String(Array<String>.self) -> \(Array<String>.self)")
输出:
String(myvar0.dynamicType) -> __NSCFConstantString
String(myvar1.dynamicType) -> PureSwiftClass
String(myvar2.dynamicType) -> Int
String(myvar3.dynamicType) -> String
String(Int.self) -> Int
String((Int?).self -> Optional<Int>
String(NSString.self) -> NSString
String(Array<String>.self) -> Array<String>
Xcode 6.3更新:
你可以使用_stdlib_getDemangledTypeName():
print( "TypeName0 = \(_stdlib_getDemangledTypeName(myvar0))")
print( "TypeName1 = \(_stdlib_getDemangledTypeName(myvar1))")
print( "TypeName2 = \(_stdlib_getDemangledTypeName(myvar2))")
print( "TypeName3 = \(_stdlib_getDemangledTypeName(myvar3))")
并将其作为输出:
TypeName0 = NSString
TypeName1 = __lldb_expr_26.PureSwiftClass
TypeName2 = Swift.Int
TypeName3 = Swift.String
最初的回答:
在Xcode 6.3之前,_stdlib_getTypeName获取变量的类型名。伊万·斯维克(Ewan Swick)的博客有助于解读这些字符串:
例如:_TtSi代表Swift的内部Int类型。
Mike Ash有一篇很棒的博客文章涉及了同样的主题。