[英]Extend Optional and Non-Optional Type
我正在与 C API 进行交互。 For memory safety I chose to not keep the C types internally and just produce them when another C api needs them, so I can work with plain Swift structs instead. 在这个例子中我只是使用Foo
类型作为替代,这显然不是真正的类型。
现在,当我需要调用 C API 时,我在类型上有这两种方法,以生成 C 指针,我可以使用:
struct Foo {
let bar: Int
}
extension Optional where Wrapped == Foo {
func withUnsafePointer<T>(_ block: (Foo?) throws -> T) rethrows -> T {
guard let self = self else {
return try block(nil)
}
return try block(self)
}
}
extension Foo {
func withUnsafePointer<T>(_ block: (Foo) throws -> T) rethrows -> T {
return try block(self)
}
}
如您所见,我还需要扩展可选项以促进这一点。 否则对foo?.withUnsafePointer(...)
的调用将不会被执行。
我不觉得这种模式特别漂亮。 有没有更好的选择而不是两次实施该方法?
重点是,C object 应该只有有限的生命周期(这里在块内)。
您可以通过声明协议来避免重复使用withUnsafePointer
方法,同时使Foo
和Foo?
符合它,然后扩展协议。 这不一定像extension Foo, Optional<Foo>
那样漂亮,但是在撰写本文时,AFAIK 目前还没有办法在 Swift 中做到这一点。
struct Foo {
let myCPointer = UnsafeMutablePointer<UInt8>.allocate(capacity: 1)
}
protocol HasMyCPointer {
associatedtype MyCPointerType
var myCPointer: MyCPointerType { get }
}
extension Foo: HasMyCPointer {}
extension Optional: HasMyCPointer where Wrapped == Foo {
var myCPointer: UnsafeMutablePointer<UInt8>? { self?.myCPointer }
}
extension HasMyCPointer {
func withUnsafePointer<T>(_ block: (Self.MyCPointerType) throws -> T) rethrows -> T {
return try block(self.myCPointer)
}
}
let foo = Foo()
let bar: Foo? = nil
let baz: Foo? = Foo()
foo.withUnsafePointer {
print("Foo: \($0)")
}
bar.withUnsafePointer {
print("Bar: \($0 as Any)")
}
baz.withUnsafePointer {
print("Baz: \($0 as Any)")
}
印刷:
Foo: 0x00007fade3c05750
Bar: nil
Baz: Optional(0x00007fade3c00220)
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