如何确定Swift enum中的案例数?
(我希望避免手动枚举所有值,或者如果可能的话使用旧的“enum_count技巧”。)
如何确定Swift enum中的案例数?
(我希望避免手动枚举所有值,或者如果可能的话使用旧的“enum_count技巧”。)
当前回答
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().enumCount
}
}
enum E {
case A
case B
case C
}
E.enumCases() // [A, B, C]
E.enumCount // 3
但是在非enum类型上使用时要小心。一些变通办法可以是:
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
guard sizeof(T) == 1 else {
return nil
}
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().count
}
}
enum E {
case A
case B
case C
}
Bool.enumCases() // [false, true]
Bool.enumCount // 2
String.enumCases() // []
String.enumCount // 0
Int.enumCases() // []
Int.enumCount // 0
E.enumCases() // [A, B, C]
E.enumCount // 4
其他回答
或者你可以在枚举之外定义_count,并静态地附加它:
let _count: Int = {
var max: Int = 0
while let _ = EnumName(rawValue: max) { max += 1 }
return max
}()
enum EnumName: Int {
case val0 = 0
case val1
static let count = _count
}
这样,不管你创建了多少个枚举,它只会被创建一次。
(如果是静态的,就删除这个答案)
这种函数能够返回枚举的计数。
斯威夫特2:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(&i) { UnsafePointer<T>($0).memory }).hashValue != 0 {
i += 1
}
return i
}
斯威夫特3:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: T.self, capacity: 1, { return $0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}
enum WeekDays : String , CaseIterable
{
case monday = "Mon"
case tuesday = "Tue"
case wednesday = "Wed"
case thursday = "Thu"
case friday = "Fri"
case saturday = "Sat"
case sunday = "Sun"
}
var weekdays = WeekDays.AllCases()
print("\(weekdays.count)")
此函数依赖于2个未记录的当前(Swift 1.1) enum行为:
枚举的内存布局只是一个大小写索引。如果case count为2 ~ 256,则为UInt8。 如果枚举从无效的大小写索引进行位转换,则其hashValue为0
所以请自担风险使用:)
func enumCaseCount<T:Hashable>(t:T.Type) -> Int {
switch sizeof(t) {
case 0:
return 1
case 1:
for i in 2..<256 {
if unsafeBitCast(UInt8(i), t).hashValue == 0 {
return i
}
}
return 256
case 2:
for i in 257..<65536 {
if unsafeBitCast(UInt16(i), t).hashValue == 0 {
return i
}
}
return 65536
default:
fatalError("too many")
}
}
用法:
enum Foo:String {
case C000 = "foo"
case C001 = "bar"
case C002 = "baz"
}
enumCaseCount(Foo) // -> 3
扩展Matthieu Riegler的回答,这是一个Swift 3的解决方案,不需要使用泛型,可以很容易地使用枚举类型EnumType.elementsCount调用:
extension RawRepresentable where Self: Hashable {
// Returns the number of elements in a RawRepresentable data structure
static var elementsCount: Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: self, capacity: 1, { return
$0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}