我已经搜索了Swift的书籍,但找不到@synchronized的Swift版本。我如何做互斥在Swift?
当前回答
细节
Xcode 8.3.1, Swift 3.1
Task
从不同的线程读写值(异步)。
Code
class AsyncObject<T>:CustomStringConvertible {
private var _value: T
public private(set) var dispatchQueueName: String
let dispatchQueue: DispatchQueue
init (value: T, dispatchQueueName: String) {
_value = value
self.dispatchQueueName = dispatchQueueName
dispatchQueue = DispatchQueue(label: dispatchQueueName)
}
func setValue(with closure: @escaping (_ currentValue: T)->(T) ) {
dispatchQueue.sync { [weak self] in
if let _self = self {
_self._value = closure(_self._value)
}
}
}
func getValue(with closure: @escaping (_ currentValue: T)->() ) {
dispatchQueue.sync { [weak self] in
if let _self = self {
closure(_self._value)
}
}
}
var value: T {
get {
return dispatchQueue.sync { _value }
}
set (newValue) {
dispatchQueue.sync { _value = newValue }
}
}
var description: String {
return "\(_value)"
}
}
使用
print("Single read/write action")
// Use it when when you need to make single action
let obj = AsyncObject<Int>(value: 0, dispatchQueueName: "Dispatch0")
obj.value = 100
let x = obj.value
print(x)
print("Write action in block")
// Use it when when you need to make many action
obj.setValue{ (current) -> (Int) in
let newValue = current*2
print("previous: \(current), new: \(newValue)")
return newValue
}
完整的样品
扩展DispatchGroup
extension DispatchGroup {
class func loop(repeatNumber: Int, action: @escaping (_ index: Int)->(), completion: @escaping ()->()) {
let group = DispatchGroup()
for index in 0...repeatNumber {
group.enter()
DispatchQueue.global(qos: .utility).async {
action(index)
group.leave()
}
}
group.notify(queue: DispatchQueue.global(qos: .userInitiated)) {
completion()
}
}
}
类ViewController
import UIKit
class ViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
//sample1()
sample2()
}
func sample1() {
print("=================================================\nsample with variable")
let obj = AsyncObject<Int>(value: 0, dispatchQueueName: "Dispatch1")
DispatchGroup.loop(repeatNumber: 5, action: { index in
obj.value = index
}) {
print("\(obj.value)")
}
}
func sample2() {
print("\n=================================================\nsample with array")
let arr = AsyncObject<[Int]>(value: [], dispatchQueueName: "Dispatch2")
DispatchGroup.loop(repeatNumber: 15, action: { index in
arr.setValue{ (current) -> ([Int]) in
var array = current
array.append(index*index)
print("index: \(index), value \(array[array.count-1])")
return array
}
}) {
print("\(arr.value)")
}
}
}
其他回答
使用Bryan McLemore的答案,我扩展了它,以支持使用Swift 2.0延迟能力扔进安全庄园的对象。
func synchronized( lock:AnyObject, block:() throws -> Void ) rethrows
{
objc_sync_enter(lock)
defer {
objc_sync_exit(lock)
}
try block()
}
是什么
final class SpinLock {
private let lock = NSRecursiveLock()
func sync<T>(action: () -> T) -> T {
lock.lock()
defer { lock.unlock() }
return action()
}
}
我喜欢并使用了这里的许多答案,所以我会选择最适合你的。也就是说,当我需要objective-c的@synchronized时,我更喜欢使用swift 2中引入的defer语句。
{
objc_sync_enter(lock)
defer { objc_sync_exit(lock) }
//
// code of critical section goes here
//
} // <-- lock released when this block is exited
这个方法的好处是,你的临界区可以以任何想要的方式退出包含块(例如,return, break, continue, throw),并且“不管程序控制如何转移,defer语句中的语句都将被执行”。1
斯威夫特4
在Swift 4中,你可以使用gcd调度队列来锁定资源。
class MyObject {
private var internalState: Int = 0
private let internalQueue: DispatchQueue = DispatchQueue(label:"LockingQueue") // Serial by default
var state: Int {
get {
return internalQueue.sync { internalState }
}
set (newState) {
internalQueue.sync { internalState = newState }
}
}
}
图我将发布我的Swift 5实现,建立在前面的答案。谢谢你们了!我发现有一个返回值的方法也很有用,所以我有两个方法。
这是一个简单的类首先创建:
import Foundation
class Sync {
public class func synced(_ lock: Any, closure: () -> ()) {
objc_sync_enter(lock)
defer { objc_sync_exit(lock) }
closure()
}
public class func syncedReturn(_ lock: Any, closure: () -> (Any?)) -> Any? {
objc_sync_enter(lock)
defer { objc_sync_exit(lock) }
return closure()
}
}
如果需要返回值,就像这样使用它:
return Sync.syncedReturn(self, closure: {
// some code here
return "hello world"
})
Or:
Sync.synced(self, closure: {
// do some work synchronously
})
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