在 Swift 5.5 中编写同步和异步函数

发布于07月08日

考虑以下(愚蠢的)示例:

func u () async -> UInt32 {
    let r =  UInt32(Int.random(in: 1...10))
    sleep(r)
    return r
}

func v(_ x: UInt32) -> UInt32 {
    x
}

Task.init{
    print(v(await u()))
    print(await v(u()))
    await print(v(u()))
}

任务中的所有三行似乎都有效.它们是等价的吗？或者我应该注意哪些trap ？谢谢

推荐答案

与try类似，Swift允许您在包含async方法的表达式的任何部分放置await关键字:

func ƒ<T>(_ v: T) -> T { v }


// All equivalent:
func somethingThatThrows() throws {}
print(ƒ(try somethingThatThrows()))
print(try ƒ(somethingThatThrows()))
try print(ƒ(somethingThatThrows()))


// All equivalent:
func someAsyncFunc() async {}
Task {
    print(ƒ(await someAsyncFunc()))
    print(await ƒ(someAsyncFunc()))
    await print(ƒ(someAsyncFunc()))
}

Swift grammar reference for the await keyword确实有一个例子，说明await的位置很重要(await只对需要awaited的整个表达式的子表达式进行awaiting)，但在这里并不太相关:

// await applies to both function calls
sum = await someAsyncFunction() + anotherAsyncFunction()

// await applies to both function calls
sum = await (someAsyncFunction() + anotherAsyncFunction())

// Error: await applies only to the first function call
sum = (await someAsyncFunction()) + anotherAsyncFunction()