递增一个隐式展开的可选 [英] Incrementing an implicitly unwrapped optional

问题描述

我声明一个隐式展开的可选内容为:

I declare an implicitly unwrapped optional as:

var numberOfRows: Int!

并在init中初始化它:

and initialize it in init:

numberOfRows = 25

稍后我需要将其减一，所以我写:

Later I need to decrement it by one so I write:

numberOfRows--

但是不能编译.错误消息指出，递减运算符不能应用于隐式展开的可选.经过一些试验，我发现以下代码可以正确编译:

but this doesn't compile. The error message says the decrement operator can't be applied to an implicitly unwrapped optional. With a little experimentation I find that the following compiles without error:

numberOfRows!--

我想了解这一点.对于看起来像多余的！"的解释是什么?

I would like to understand this. What is the explanation for what seems like the extra '!'?

推荐答案

隐式展开的可选是一种单独的类型，与包装的类型不同.某些 optionals 和隐式解开的optionals 上的运算符是根据语言预先为您预定义的，但是对于其余操作符，您必须自己定义它们.

Implicitly unwrapped optional is a type on its own, and is different from the type it that wraps. Some operators on optionals and implicitly unwrapped optionals are pre-defined for you out of the box by the language, but for the rest you have to define them yourself.

在这种特殊情况下，只是未定义运算符postfix func --(inout value: Int!) -> Int!.如果要在Int!上使用后缀--运算符，就像在Int上使用后缀--运算符一样，则必须定义一个.

In this particular case an operator postfix func --(inout value: Int!) -> Int! is just not defined. If you want to use postfix -- operator on Int! just the same way you use it on Int then you will have to define one.

例如像这样:

postfix func --<T: SignedIntegerType>(inout value: T!) -> T! {
    guard let _value = value else { return nil }

    value = _value - 1
    return _value
}

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