您如何解开Swift可选项? [英] How do you unwrap Swift optionals?

问题描述

您如何正确地包装普通和隐式可选件?

How do you properly unwrap both normal and implicit optionals?

这个主题似乎有些混乱，我只想对所有方法以及它们如何有用提供参考.

There seems to be confusion in this topic and I would just like to have a reference for all of the ways and how they are useful.

当前有两种创建可选内容的方法:

There are currently two ways to create optionals:

var optionalString: String?

var implicitOptionalString: String!

有两种方法都可以解开包装吗?另外，在展开过程中使用!和?有什么区别?

What are all the ways to unwrap both? Also, what is the difference between using ! and ? during the unwrapping?

推荐答案

有很多相似之处，但有一些区别.

There are many similarities and just a handful of differences.

• 声明:var opt: Type?

不安全地展开:let x = opt!.property // error if opt is nil

安全地测试存在性:if opt != nil { ... someFunc(opt!) ... } // no error

通过绑定安全地展开:if let x = opt { ... someFunc(x) ... } // no error

Safely unwrapping via binding: if let x = opt { ... someFunc(x) ... } // no error

安全链接:var x = opt?.property // x is also Optional, by extension

安全合并nil值:var x = opt ?? nonOpt

• 声明:var opt: Type!

不安全地展开(隐式):let x = opt.property // error if opt is nil

Unsafely unwrapping (implicit): let x = opt.property // error if opt is nil

• 通过分配不安全地展开:
  let nonOpt: Type = opt // error if opt is nil

• Unsafely unwrapping via assignment:
  let nonOpt: Type = opt // error if opt is nil

通过参数传递不安全地展开:
func someFunc(nonOpt: Type) ... someFunc(opt) // error if opt is nil

Unsafely unwrapping via parameter passing:
func someFunc(nonOpt: Type) ... someFunc(opt) // error if opt is nil

安全测试存在:if opt != nil { ... someFunc(opt) ... } // no error

安全链接:var x = opt?.property // x is also Optional, by extension

安全合并nil值:var x = opt ?? nonOpt

这篇关于您如何解开Swift可选项?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！