swift - 条件绑定(bind)的初始值设定项必须具有可选类型

Question

我想创建一个 BLE 扫描器，并按照 developer.apple.com 上的开始开发 iOS 应用程序 (Swift) 进行操作，

当 session “Create a Table View”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]
}

错误发生，

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

如何解决？谢谢!

Answer 1

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

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]

}