我可以在Swift的guard语句中使用范围运算符吗？

发布时间：2020-12-14 05:38:48 所属栏目：百科来源：网络整理

导读：我正试图找出另一种方法来做这样的事情,使用范围运算符. guard let statusCode = (response as? HTTPURLResponse)?.statusCode,statusCode = 200 statusCode = 299 else {return} 也许是这样的： guard let statusCode = (response as? HTTPURLResponse)?.st

我正试图找出另一种方法来做这样的事情,使用范围运算符.

guard let statusCode = (response as? HTTPURLResponse)?.statusCode,statusCode >= 200 && statusCode <= 299 else {return}

也许是这样的：

guard let statusCode = (response as? HTTPURLResponse)?.statusCode where (200...299).contains(statusCode) else {return}

要么

guard let statusCode = (response as? HTTPURLResponse)?.statusCode,statusCode case 200...299 else {return}

这在Swift中可能吗？

随你心意：

guard
    let statusCode = (response as? HTTPURLResponse)?.statusCode,(200...299).contains(statusCode) else {return}

要么：

guard
    let statusCode = (response as? HTTPURLResponse)?.statusCode,case 200...299 = statusCode else {return}

要么：

guard
    let statusCode = (response as? HTTPURLResponse)?.statusCode,200...299 ~= statusCode else {return}

（编辑：李大同）

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