条件绑定的初始化程序必须具有可选类型答案

【问题标题】：initializer for conditional binding must have optional type条件绑定的初始化程序必须具有可选类型
【发布时间】：2017-06-28 07:29:52
【问题描述】：

我想创建一个 BLE 扫描仪，并在 developer.apple.com 上关注 Start Developing iOS Apps (Swift)，

当会话“创建表格视图”时，我在这里遇到错误，

BLEMember.swift

import UIKit

class BLEMember {
    var rssi: Int
    var uuid: String

    init (rssi:Int, uuid:String){
      self.rssi = rssi
      self.uuid = uuid
    }
}

BLEScanTableViewController.swift

...
    private func loadBLEMembers() {

    var members = [BLEMember]()
    guard let member1 = BLEMember(rssi: 65, uuid: "testing ble 1") else {
        fatalError("Unable to instantiate meal1")
    }

    guard let member2 = BLEMember(rssi: 35, uuid: "testing ble 2") else {
        fatalError("Unable to instantiate meal2")
    }

    guard let member3 = BLEMember(rssi: 45, uuid: "testing ble 3") else {
        fatalError("Unable to instantiate meal2")
    }

    members += [member1, member2, member3]
}

然后发生错误，

 initializer for conditional binding must have optional type, not "BLEMember"

如何解决？谢谢！

【问题讨论】：

标签： swift

【解决方案1】：

let member1 = BLEMember(rssi: 65, uuid: "testing ble 1") 总是成功。这里不需要使用守卫！

private func loadBLEMembers() {

    var members = [BLEMember]()
    let member1 = BLEMember(rssi: 65, uuid: "testing ble 1")

    let member2 = BLEMember(rssi: 35, uuid: "testing ble 2")

    let member3 = BLEMember(rssi: 45, uuid: "testing ble 3")

    members += [member1, member2, member3]

}

【讨论】：

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