ios - swift 中的尾随关闭问题？

Question

你好，我创建了一个函数，它接受最后一个参数作为闭包。

func sum(from: Int, to: Int, f: (Int) -> (Int)) -> Int {
    var sum = 0
    for i in from...to {
        sum += f(i)
    }
    return sum
}

现在我调用这个函数。调用这个函数的一种方法如下所示。

sum(from: 1, to: 10) { (num) -> (Int) in
return 10
}

我在 swift 中看到了尾随闭包的概念之一。有了尾随闭包，我可以像这样调用函数。

sum(from: 1, to: 10) {
    $0
}

但我不知道为什么它可以在没有任何return 语句的情况下调用。请告诉我这是怎么发生的？

Answer 1

除了“因为语言允许”之外，这里真的没有答案。如果闭包中只有一个表达式，则可以省略 return。

涵盖此内容的部分是 "Implicit Returns from Single-Expression Closures"来自 Swift 编程语言。

Single-expression closures can implicitly return the result of their single expression by omitting the return keyword from their declaration, as in this version of the previous example:

reversedNames = names.sorted(by: { s1, s2 in s1 > s2 } )

Here, the function type of the sorted(by:) method’s argument makes it clear that a Bool value must be returned by the closure. Because the closure’s body contains a single expression (s1 > s2) that returns a Bool value, there is no ambiguity, and the return keyword can be omitted.

然而，这与尾随闭包语法无关。如果闭包是单表达式，则所有闭包都有隐式返回。