条件绑定的初始值设定项必须具有可选类型
条件绑定的初始值设定项必须具有可选类型
提问于 2017-02-10 11:39:07
我想创建一个BLE扫描器，并在developer.apple.com上遵循开始开发iOS应用程序(Swift)，
当会话“创建一个表视图”时，我在这里得到一个错误，
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]
}和错误发生，
initializer for conditional binding must have optional type, not "BLEMember"如何修复它？谢谢!
回答 1
Stack Overflow用户
回答已采纳
发布于 2017-02-10 11:49:08
let member1 = BLEMember(rssi: 65, uuid: "testing ble 1")总是成功的。你不需要在这里使用卫士！
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]
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/42151095
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