我需要为我自己的结构实现一个类型删除包装器,非常类似于 SequenceOf
、GeneratorOf
等。所以我开始尝试重新实现我自己的标准 SequenceOf
。
我刚刚复制并粘贴了 SequenceOf
的声明,将其重命名为 MySequenceOf
,并填写了一些 stub 以获得:
/// A type-erased sequence.
///
/// Forwards operations to an arbitrary underlying sequence with the
/// same `Element` type, hiding the specifics of the underlying
/// sequence type.
///
/// See also: `GeneratorOf<T>`.
struct MySequenceOf<T> : SequenceType {
/// Construct an instance whose `generate()` method forwards to
/// `makeUnderlyingGenerator`
init<G : GeneratorType where T == T>(_ makeUnderlyingGenerator: () -> G) {
fatalError("implement me")
}
/// Construct an instance whose `generate()` method forwards to
/// that of `base`.
init<S : SequenceType where T == T>(_ base: S) {
fatalError("implement me")
}
/// Return a *generator* over the elements of this *sequence*.
///
/// Complexity: O(1)
func generate() -> GeneratorOf<T> {
fatalError("implement me")
}
}
我收到编译器错误:“同一类型中的类型均未引用泛型参数或关联类型”。所以我假设 Xcode 生成的 SequenceOf
的“where T == T
”约束声明实际上意味着“where G.Element == T
”,这给了我以下可编译的结构:
struct MySequenceOf<T> : SequenceType {
init<G : GeneratorType where G.Element == T>(_ makeUnderlyingGenerator: () -> G) {
fatalError("implement me")
}
func generate() -> GeneratorOf<T> {
fatalError("implement me")
}
}
现在,很简单,我只需要从初始化程序中挂起 makeUnderlyingGenerator
并从 generate()
中调用它:
struct MySequenceOf<T> : SequenceType {
let maker: ()->GeneratorOf<T>
init<G : GeneratorType where G.Element == T>(_ makeUnderlyingGenerator: () -> G) {
self.maker = { return makeUnderlyingGenerator() }
}
func generate() -> GeneratorOf<T> {
return self.maker()
}
}
但这给了我错误:“‘G’不能转换为‘GeneratorOf’”
如果我强制转换,它会编译:
struct MySequenceOf<T> : SequenceType {
let maker: ()->GeneratorOf<T>
init<G : GeneratorType where G.Element == T>(_ makeUnderlyingGenerator: () -> G) {
self.maker = { return makeUnderlyingGenerator() as GeneratorOf<T> }
}
func generate() -> GeneratorOf<T> {
return self.maker()
}
}
但随后它会在运行时因动态转换而崩溃。
那么如何实现这样的类型删除呢?这一定是可能的,因为 Swift 标准库做了很多(SequenceOf、GeneratorOf、SinkOf)。
最佳答案
尝试:
struct MySequenceOf<T> : SequenceType {
private let _generate:() -> MyGeneratorOf<T>
init<G : GeneratorType where G.Element == T>(_ makeUnderlyingGenerator: () -> G) {
_generate = { MyGeneratorOf(makeUnderlyingGenerator()) }
}
init<S : SequenceType where S.Generator.Element == T>(_ base: S) {
_generate = { MyGeneratorOf(base.generate()) }
}
func generate() -> MyGeneratorOf<T> {
return _generate()
}
}
struct MyGeneratorOf<T> : GeneratorType, SequenceType {
private let _next:() -> T?
init(_ nextElement: () -> T?) {
_next = nextElement
}
init<G : GeneratorType where G.Element == T>(var _ base: G) {
_next = { base.next() }
}
mutating func next() -> T? {
return _next()
}
func generate() -> MyGeneratorOf<T> {
return self
}
}
实现的基本策略ProtocolOf<T>
是这样的:
protocol ProtocolType {
typealias Value
func methodA() -> Value
func methodB(arg:Value) -> Bool
}
struct ProtocolOf<T>:ProtocolType {
private let _methodA: () -> T
private let _methodB: (T) -> Bool
init<B:ProtocolType where B.Value == T>(_ base:B) {
_methodA = { base.methodA() }
_methodB = { base.methodB($0) }
}
func methodA() -> T { return _methodA() }
func methodB(arg:T) -> Bool { return _methodB(arg) }
}
添加在评论中回答@MartinR。
Is there a special reason that _generate is a closure and not the generator itself?
首先,我认为,这是规范或语义的问题。
不用说,区别在于“何时创建生成器”。
考虑这段代码:
class Foo:SequenceType {
var vals:[Int] = [1,2,3]
func generate() -> Array<Int>.Generator {
return vals.generate()
}
}
let foo = Foo()
let seq = MySequenceOf(foo)
foo.vals = [4,5,6]
let result = Array(seq)
问题是:result
应该是 [1,2,3]
或 [4,5,6]
?我的MySequenceOf
和内置 SequenceOf
结果是后者。我只是将行为与内置行为相匹配。
关于swift - 类型删除的通用包装器是如何实现的?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26832498/