Swift 3 NSError 期望调用类型之后? [英] Swift 3 NSError expecting call type after?

问题描述

我正在将 google 登录添加到我的 Xcode 项目中.我正在设置一个错误配置器，以便登录可以工作，但是当我尝试将错误定义为 NSError? 时，它一直给我错误.我做错了什么?

I am adding google sign in to my Xcode project. I am setting up an error configurator so the sign in will work, but it keeps giving me errors when I try to define the error as NSError?. What am I doing wrong?

class logInController: UIViewController, GIDSignInUIDelegate {

    func CGRectMake(_ x: CGFloat, _ y: CGFloat, _ width: CGFloat, _ height: CGFloat) -> CGRect {
        return CGRect(x: x, y: y, width: width, height: height)
    }

    override func viewDidLoad(){
        super.viewDidLoad()

        var error = NSError?
        GGLContext.sharedInstance().configureWithError(&error)

        if error != nil {
            print(error)
            return
        }

        GIDSignIn.sharedInstance().uiDelegate = self

        let signInButton = GIDSignInButton(frame: CGRectMake(0, 0, 100, 50))
        signInButton.center = view.center
        view.addSubview(signInButton)
    }    
}

Xcode 说，名称后的预期成员名称或构造函数调用类型"，但是当我添加域或 userInfo 之类的内容时，它告诉我，参数标签 [] 不匹配任何可用的重载"

Xcode says, "expected member name or constructor call type after name", but when I add something like a domain or userInfo it tells me, "argument labels [] do not match any available overloads"

推荐答案

要么使用构造函数声明类型

Either you declare a type with a constructor

var error = NSError(domain:...

或作为可选指针

var error : NSError?

要将变量用作 inout 参数，您必须使用后者.

To use the variable as an inout parameter you have to use the latter.

PS:习惯 Swift 3 的变化，比如 CGRect 传递所有参数标签...

PS: Get used to the Swift 3 changes like CGRect passing all parameter labels...

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