swift - 是否有 swiftc 编译器优化来删除不必要的大变量？

Question

当我用高阶函数编写代码时，我发现像这样一次编写一个转换的表达式会更清晰:

let children = objects.map { $0.children }
let validChildren = children.filter { $0.isValid }
let sortedChildren = validChildren.sorted { $0.count < $1.count }

但是，我知道这些函数中的每一个都会返回一个新的 Array 对象，我将其存储到一个变量中，因此理论上我每次都会创建并保留一个新的 Array 并浪费大量内存。最好将调用写成一行，以便在使用后销毁不需要的数组。

let sortedChildren = objects.map { $0.children } .filter { $0.isValid } .sorted { $0.count < $1.count }

但稍后阅读起来会更烦人，因为一行代码中发生了太多事情。所以我的问题是:Swift 的编译器是否有优化来原谅我的挑剔并在编译时删除未使用的变量？

Answer 1

@MartinR 在问题评论中提到了这一点，但值得官方回答和来源。

As per the Swift repository docs:

All standard library containers in Swift are value types that use COW (copy-on-write) [4] to perform copies instead of explicit copies. In many cases this allows the compiler to elide unnecessary copies by retaining the container instead of performing a deep copy. This is done by only copying the underlying container if the reference count of the container is greater than 1 and the container is mutated. For instance in the following, no copying will occur when d is assigned to c, but when d undergoes structural mutation by appending 2, d will be copied and then 2 will be appended to d:

var c: [Int] = [ ... ]
var d = c        // No copy will occur here.
d.append(2)      // A copy *does* occur here.

由于高阶函数不会修改调用它们的对象，因此可以很有把握地说结果针对大小进行了优化。