swift – 如何让x在x.hasSuffix(“pepper”)中工作

发布时间：2020-12-14 05:19:46 所属栏目：百科来源：网络整理

导读：在下面的代码块中,我无法理解let x其中x.hasSuffix(“pepper”). let vegetable = "red pepper"switch vegetable { case "celery": let vegetableComment = "Add some raisins and make ants on a log." case "cucumber","watercress": let vegetableComment

在下面的代码块中,我无法理解let x其中x.hasSuffix(“pepper”).

let vegetable = "red pepper"

switch vegetable {
    case "celery":
        let vegetableComment = "Add some raisins and make ants on a log."
    case "cucumber","watercress":
        let vegetableComment = "That would make a good tea sandwhich"
    case let x where x.hasSuffix("pepper"):
        let vegetableComment = "Is it a spicy (x)"
    default:
        let vegetableComment = "Everything tastes good in soup."
}

Console output

vegetableComment: Is it a spicy red pepper

似乎发生了以下逻辑.

x = vegetable
if (x's suffix == 'pepper') 
    run case

有人能帮我解释一下吗？

vegetable是一个隐式String.它和你写的一样：

var vegetable: String = "red pepper"

hasSuffix声明为func hasSuffix(后缀：String) – > Bool因此返回Bool. where关键字指定其他要求,并且只能在switch语句中使用.

因为所有这些都是泛滥的,所以将vegetable变量赋给x(let x).

您可以阅读有关where的更多信息并切换here.

（编辑：李大同）

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