如何在扩展中使用Swift返回Int中的变异函数？

Why doesn't this code work?

为什么这段代码不起作用?

protocol ExampleProtocol {
    var simpleDescription: String { get }
    mutating func adjust()
}
extension Int: ExampleProtocol {
    var simpleDescription: String {
    return "The number \(self)"
    }
    mutating func adjust() {
        self += 42
    }
}
var x:Int = 7
let y:Int = x.adjust()

here is what I get on XCODE

这是我在XCODE上得到的

is there a way to make adjust() return Int without changing its definition in the protocol?

有没有办法让adjust()返回Int而不改变协议中的定义?

3 个解决方案

#1

Yes, you can give adjust a return value. Define it to return an Int in the protocol and class, then have it return itself in the mutating method:

是的,您可以给出调整返回值。定义它以在协议和类中返回Int,然后让它在变异方法中返回:

protocol ExampleProtocol {
    var simpleDescription: String { get }
    mutating func adjust() -> Int
}
extension Int: ExampleProtocol {
    var simpleDescription: String {
    return "The number \(self)"
    }
    mutating func adjust() -> Int {
        self += 42
        return self
    }
}
var x:Int = 7
let y:Int = x.adjust() //49

#2

Because the adjust() function does not return value (it just change the value of its instance), you can achieve this by this orders:

因为adjust()函数不返回值(它只是更改其实例的值),您可以通过以下命令实现此目的:

var x:Int = 7
x.adjust() //adjust x self value
let y:Int = x //assigne x value to y

Hope this helps.

希望这可以帮助。

#3

Ques : is there a way to make adjust() return Int without changing its definition in the protocol?

问:有没有办法让adjust()返回Int而不改变协议中的定义?

Ans : No

答:没有

The definition/signature of a method tells us what it's going to take as input and what data type it's going to return.

方法的定义/签名告诉我们它将作为输入采用什么以及它将返回什么数据类型。

You need to return an Int from a method then it's signature should end with -> Int. That's the syntax ! Can't change it.

您需要从方法返回一个Int,然后它的签名应以 - > Int结尾。这就是语法!无法改变它。

connor has provided the right piece of code you need.

connor提供了您需要的正确代码。