类型擦除的通用包装器是如何实现的? [英] How is a type-erased generic wrapper implemented?
问题描述
我需要为我自己的结构实现一个类型擦除包装,非常类似于 SequenceOf
, GeneratorOf
,等等。所以我开始试图只重新实现标准 SequenceOf
我自己。
我刚刚复制&粘贴 SequenceOf
的声明,将其重命名为 MySequenceOf
,并填写一些存根以获取:
///一个类型擦除序列。
///
///使用
///相同的`Element`类型将操作转向任意的基础序列,隐藏底层
///序列的细节类型。
///
///另请参阅:`GeneratorOf< T>`。
struct MySequenceOf< T> :SequenceType {
///构造一个实例,其`generate()`方法转发给
///`makeUnderlyingGenerator`
init< G:GeneratorType其中T == T> ;(_ makeUnderlyingGenerator:() - > G){
fatalError(implement我)
}
///构造一个其generate()方法的实例转发给
///base。
init< S:SequenceType where T == T>(_ base:S){
fatalError(implement我)
}
///返回一个*发生器*在这个*序列*的元素上。
///
///复杂度:O(1)
func generate() - > GeneratorOf< T> {
fatalError(实施我)
}
}
我得到的编译器错误:两种类型都不指向泛型参数或关联类型。所以我假设Xcode生成的 SequenceOf
的其中T == T
约束的声明真的意味着 其中G.Element == T
,它给了我下面的可编译结构:
struct MySequenceOf< T> :SequenceType {
init< G:GeneratorType where G.Element == T>(_ makeUnderlyingGenerator:() - > G){
fatalError(implement我)
}
func generate() - > GeneratorOf< T> {
fatalError(实施我)
}
}
所以,现在很简单,我只需要从初始化程序挂到 makeUnderlyingGenerator
,并从 generate()
:
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()
}
}
但是这给了我错误:'G'不能转换为'GeneratorOf'
编译如果我强制转换:
struct MySequenceOf< T> :SequenceType {
let maker:() - > GeneratorOf< T>
init< G:GeneratorType其中G.Element == T>(_ makeUnderlyingGenerator:() - > G){
self.maker = {返回makeUnderlyingGenerator()as GeneratorOf< T> ; }
}
func generate() - > GeneratorOf< T> {
return self.maker()
}
}
但是,它会在运行时从动态转换崩溃。
那么类型擦除如何实现呢?它必须是可能的,因为Swift标准库做了很多工作(SequenceOf,GeneratorOf,SinkOf)。解析方案
试试:
struct MySequenceOf< T> :SequenceType {
private let _generate :() - > MyGeneratorOf< T>
init< G:GeneratorType where G.Element == T>(_ makeUnderlyingGenerator:() - > G){
_generate = {MyGeneratorOf(makeUnderlyingGenerator())}
}
init< S:SequenceType其中S.Generator.Element == T>(_ base:S){
_generate = {MyGeneratorOf(base.generate())}
}
func generate() - > MyGeneratorOf< T> {
return _generate()
}
}
struct MyGeneratorOf< T> :GeneratorType,SequenceType {
private让_next :() - > T'
init(_ nextElement:() - > T?){
_next = nextElement
}
init< G:GeneratorType其中G.元素== T>(var _ base:G){
_next = {base.next()}
}
mutating func next() - > T' {
return _next()
}
func generate() - > MyGeneratorOf< T> {
return self
}
}
实现 ProtocolOf< T>
的基本策略是这样的:
protocol ProtocolType {
typealias Value
func methodA() - >价值
func methodB(arg:Value) - > Bool
}
struct ProtocolOf< T>:ProtocolType {
private let _methodA:() - > T
private let _methodB:(T) - > Bool
init< B:ProtocolType其中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)}
}
是否有一个特殊原因,_generate是一个闭包而不是生成器本身?
首先,我认为,这是一个规范或语义问题。
$ b
不用说,不同之处在于何时创建生成器。
请考虑以下代码:
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
和内置的 SequenceOf
产生后者。我只是将行为与内置行为进行了匹配。
I need to implement a type-erasing wrapper for my own structure, very similar to SequenceOf
, GeneratorOf
, etc. So I started by trying to just re-implement the standard SequenceOf
myself.
I just copied & pasteed the the declaration for SequenceOf
, renamed it to MySequenceOf
, and filled in some stubs to get:
/// 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")
}
}
I get the compiler error: "Neither type in same-type refers to a generic parameter or associated type". So I assume that the Xcode-generated declaration of SequenceOf
's "where T == T
" constraint really means "where G.Element == T
", which gives me the following compilable struct:
struct MySequenceOf<T> : SequenceType {
init<G : GeneratorType where G.Element == T>(_ makeUnderlyingGenerator: () -> G) {
fatalError("implement me")
}
func generate() -> GeneratorOf<T> {
fatalError("implement me")
}
}
So now, easy enough, I just need to hang on to makeUnderlyingGenerator
from the initializer and invoke it from generate()
:
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()
}
}
but that gives me the error: "'G' is not convertible to 'GeneratorOf'"
It does compiles if I force a cast:
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()
}
}
But then it crashes at runtime from the dynamic cast.
So how can type-erasure like this be implemented? It must be possible, because the Swift standard library does it a bunch (SequenceOf, GeneratorOf, SinkOf).
Try:
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
}
}
The basic strategy of implementing ProtocolOf<T>
is, like this:
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) }
}
Added to answering @MartinR in comment.
Is there a special reason that _generate is a closure and not the generator itself?
First of all, I think, It's a matter of specification or semantics.
Needless to say, the difference is "when to create the generator".
Consider this code:
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)
The problem is: result
should be [1,2,3]
or [4,5,6]
? My MySequenceOf
and built-in SequenceOf
results the latter. I just matched the behaviors with built-in one.
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