如何在 Swift 中更改为类型显示的文本表示? [英] How can I change the textual representation displayed for a type in Swift?

问题描述

如何修改在字符串插值中显示的文本输出?

How can I modify the textual output that gets displayed in string interpolation?

Printable 协议看起来最明显，但它在字符串插值和打印实例时都被忽略，例如:

The Printable protocol looks the most obvious but it's ignored in both String Interpolation and when printing the instance, e.g:

struct Point : Printable
{
    var x = 0
    var y = 0

    var description : String {
        return "(\(x), \(y))"
    }

    func toString() -> String {
        return description
    }
}

同样，toString() 约定也没有效果:

Likewise the toString() convention has no effect either:

var p = Point(x: 10, y: 20)

println(p)                   // V11lldb_expr_05Point (has 2 children)
println("\(p)")              // V11lldb_expr_05Point (has 2 children)
println(p.description)       // (10, 20)
println("\(p.description)")  // (10, 20)

在 PlayGround 中的行为再次不同，它使用自己的字符串表示结构，即:

The behavior is different again in PlayGround which uses its own String representation for structs, i.e:

p // {x 10, y 20}

有没有办法改变实例的显示方式?

Is there a way I can change how an instance is displayed?

推荐答案

Swift 2 - 4

总结

符合CustomStringConvertible协议并添加description:

var description: String {
    return "description here"
}

<小时>

示例

您可以创建一些结构:

---

Example

You can create some structs:

struct Animal : CustomStringConvertible {
    let type : String

    var description: String {
        return type
    }
}

struct Farm : CustomStringConvertible {
    let name : String
    let animals : [Animal]

    var description: String {
        return "\(name) is a \(self.dynamicType) with \(animals.count) animal(s)."
    }
}

如果你初始化它们:

let oldMajor = Animal(type: "Pig")
let boxer = Animal(type: "Horse")
let muriel = Animal(type: "Goat")

let orwellsFarm = Farm(name: "Animal Farm", animals: [oldMajor, boxer, muriel])

自定义描述将出现在您的 Playground 中:

The custom descriptions will appear in your playground:

另见 CustomDebugStringConvertible，您可以在调试期间使用它来获得更详细的输出.

See also CustomDebugStringConvertible, which you can use for more verbose output during debugging.

您可以从任何类型初始化一个String，而无需实现此协议.例如:

You can initialize a String from any type without implementing this protocol. For example:

出于这个原因，文档说:

For this reason, the docs say:

因此不鼓励使用 CustomStringConvertible 作为通用约束，或直接访问符合类型的 description.

Using CustomStringConvertible as a generic constraint, or accessing a conforming type's description directly, is therefore discouraged.

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