没有花括号和没有参数标签的匿名函数? [英] Anonymous function with no curly braces and no argument labels?

问题描述

我在另一个问题上看到了一些代码，该代码似乎创建了具有某些异常语法的匿名函数(闭包表达式):

I saw some code on another question that seems to create an anonymous function (closure expression) with some unusual syntax:

let plus: (Int, Int) -> Int = (+)

我了解左侧-它声明了类型为(Int, Int) -> Int的常量(该函数接受两个Integer，并返回一个Integer).但是什么是(+)?如何声明没有花括号的函数?在没有任何类型的参数标签的情况下如何引用两个参数?

I understand the left side—that it's declaring a constant of type (Int, Int) -> Int (a function that takes two Integers and returns an Integer). But what is (+)? How can it declare a function without curly brackets, and how does it refer to the two arguments when there are no argument labels of any kind?

该函数接受两个参数，将它们加在一起，然后返回结果.如果将+运算符替换为其他运算符(例如*)，则操作会更改.那么{$0 + $1}的某种速记形式吗?如果是这样，此速记背后的逻辑是什么?

The function takes two arguments, adds them together, and returns the result. If I replace the + operator with a different one (say a *), the operation changes. So is it some kind of shorthand for {$0 + $1}? If so, what is the logic behind this shorthand?

推荐答案

实际上，这不是速记.

plus是类型(Int, Int) -> Int的变量.您可以为其分配任何该类型(或其任何子类型)的对象.文字lambda闭包肯定是这种类型的，但是实际上，命名函数或方法也可以.而这正是这里发生的事情.

plus is a variable of type (Int, Int) -> Int. You can assign it any object that is of this type (or any of its subtypes). A literal lambda closure is certainly of this type, but actually a named function or method would also do. And that is exactly what is happening here.

它正在将名为+的运算符方法对象分配给该变量.

It is assigning the operator method object named + to the variable.

关闭一章的语言指南:

This is mentioned sort-of implicitly in the Closures chapter of the language guide:

操作员方法

实际上，甚至还有一种更短的方式来编写上面的闭包表达式. Swift的String类型将其大于字符串运算符(>)的特定于字符串的实现定义为一种方法，该方法具有两个类型为String的参数，并返回类型为Bool的值.这与sorted(by:)方法所需的方法类型完全匹配.因此，您只需传递大于号运算符，Swift便会推断您要使用其特定于字符串的实现:

Operator Methods

There’s actually an even shorter way to write the closure expression above. Swift’s String type defines its string-specific implementation of the greater-than operator (>) as a method that has two parameters of type String, and returns a value of type Bool. This exactly matches the method type needed by the sorted(by:) method. Therefore, you can simply pass in the greater-than operator, and Swift will infer that you want to use its string-specific implementation:

reversedNames = names.sorted(by: >)

因此，代码正在执行的操作是将操作员方法+ 分配给变量plus. +只是分配给变量的函数的名称.没有魔术速记.

So, what the code is doing is assigning the Operator Method + to the variable plus. + is simply the name of the function assigned to the variable. No magic shorthand involved.

看到这个消息你会感到惊讶吗?

Would you be surprised to see this?

let plus: (Int, Int) -> Int = foo

这篇关于没有花括号和没有参数标签的匿名函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！