小编Har*_*sha的帖子

每次,当我在xcode中打开一个新项目时,它会将默认的模拟器设备显示为iPhone 8plus.所以,我每次都在运行之前将设备更改为iPhoneX.有没有办法将它作为任何项目的默认值.

我正在尝试为我的视图控制器实现屏幕边缘平移手势。但问题是，如果尝试为两个边缘（UIRectEdge.left、UIRectEdge.right）添加边缘平移手势，

let screenEdgePanGesture = UIScreenEdgePanGestureRecognizer.init(target: self, action: #selector(self.didPanningScreen))
screenEdgePanGesture.edges = [.right, .left]
screenEdgePanGesture.delegate = self
self.view.addGestureRecognizer(screenEdgePanGesture)

选择器方法未调用。但边缘平移手势适用于一个边缘，即，

let screenEdgePanGesture = UIScreenEdgePanGestureRecognizer.init(target: self, action: #selector(self.didPanningScreen))
screenEdgePanGesture.edges = .right
screenEdgePanGesture.delegate = self
self.view.addGestureRecognizer(screenEdgePanGesture)

如何根据结构（名称）中的键合并两个（第一个和第二个）结构数组。合并时，如果该元素中的任何值发生更改，我需要用第二个数组元素替换第一个数组的现有元素。

struct Example: Codable {
    var name: String
    var dob: String
    var address: String
}

var first: [Example] = []
var second: [Example] = []

first.append(Example(name: "Arun", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
first.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))
    first.append(Example(name: "Satish", dob: "01-01-1994", address: "Tirupati"))


second.append(Example(name: "Arun", dob: "01-01-1994", address: "Kadapa"))
second.append(Example(name: "Balaji", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Prasanth", dob: "01-01-1994", address: "Tirupati"))
second.append(Example(name: "Harsha", dob: "01-01-1994", address: "Tirupati"))

/*    let merged: [Example] = merge(first, second, …