Swift 3闭包重载解析 [英] Swift 3 closure overload resolution

问题描述

我对函数重载解析和Swift 3中的闭包感到困惑。

I'm confused by function overload resolution with closures in Swift 3.

例如，在代码中：

func f<T>(_ a: T) {
    print("Wide")
}

func f(_ a: (Int)->(Int)) {
    print("Narrow")
}

f({(a: Int) -> Int in return a + 1})

我希望 Narrow ，而不是宽打印到控制台。谁能解释为什么为非闭包参数选择了更具体的重载而不是闭包，或者这是编译器错误吗？

I expect Narrow, not Wide, to be printed to the console. Can anyone explain why the more specific overload gets chosen for non-closure arguments but not for closures or is this a compiler bug?

Swift 2表现出了预期的行为。

Swift 2 exhibited the expected behavior.

推荐答案

这可能是由于关闭参数的默认转义行为发生了变化。

This is probably due to the change in the default "escaping" behaviour for closure parameters.

如果将特定功能更改为：

If you change the specific function to :

func f(_ a:@escaping (Int)->Int) 
{
    print("Narrow")
}

它将按预期方式打印 Narrow（窄，这可能是您可能在其他几个更明显的地方所做的相同更改）

it will print "Narrow" as expected (this this the same change that you probably had to make in several other places that were more obvious)

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