Swift增量Int！不工作 [英] Swift increment Int! not working

问题描述

我了解可选控件的工作原理，但这使我陷入了循环.我有一个名为num的变量，我想增加它，所以我做了以下事情:

I understand how optionals work, but this is throwing me for a loop. I have a variable called num and I want to increment it, so I did the following:

var num:Int! = 0
num++             //ERROR - Unary operator ++ cannot be applied to an operand of type Int!

但是出于某种原因，Swift不允许我增加展开的Int的力，即使它被认为像常规的Int一样具有在后台无弹力的能力.因此，我尝试了以下方法，并且有效:

But for some reason Swift won't let me increment a force unwrapped Int, even though it is supposed to be treated like a regular Int with the capability for nil behind the scenes. So I tried the following, and it worked:

var num:Int! = 0
num = num + 1     //NO ERROR

但是，基于它给我的错误消息，我尝试了以下操作以使增量运算符仍然有效:

However, based on the error message it gave me, I tried the following to make the increment operator still work:

var num:Int! = 0
num!++            //NO ERROR

我的问题是，为什么第二和第三位代码没有中断，但是第一行代码却中断了?另外，由于num是Int!，我是否应该像普通的Int那样对待它?最后，由于Int!应该被当作常规的Int对待，因此在第三个示例中我该如何解包呢?谢谢.

My question is why does the first bit of code break, when the second and third bits of code don't? Also, since num is an Int!, shouldn't I be able to treat it like a regular Int? Lastly, since an Int! is supposed to be treated like a regular Int, how am I able to unwrap it in the third example? Thanks.

推荐答案

使用所有inout参数都会发生此错误，应将其视为错误.

This error occurs with all inout parameters and should be considered a bug.

inout参数通常的工作方式是调用它们的getter方法一次，并使用setter至少调用一次.在这种情况下，getter返回Int!:

The usual way how inout parameters work is that their getter gets called once and their setter at least once. In this case the getter returns an Int!:

let num: Int! = 0
let num2 = num // is inferred to be of type Int!

因此，getter/setter的签名与功能/运算符所期望的不同.但是，如果编译器将该值分配给Int或这样传递，则应隐式解开该值:

so the signature of the getter/setter is not the same as the function/operator expects. But the compiler should implicitly unwrap the value if it gets assigned to an Int or passed like so:

var num3 = 0 // is of type Int
num3 = num // gets automatically unwrapped

// also with functions
func someFunc(i: Int) {}
someFunc(num) // gets automatically unwrapped

旁注:

如果要将Int传递给参数类型为Int!的inout的函数，则会发生几乎相同的错误.但是在这里很明显为什么这在逻辑上不起作用:Int的设置者从不使用nil.

Almost the same error occurs if you want to pass an Int to a function with an inout parameter of type Int!. But here it is obvious why this doesn't work (logically): The setter of an Int never takes a nil.

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