Swift 4中如何将Type作为变量进行传递？

Great question! The key here is to use generics in Swift, which lets you pass a view controller type as a parameter while maintaining type safety—something your Any approach can't handle reliably. Here's how to refactor your code cleanly:

Basic Generic Function #

First, create a generic function that accepts a view controller type, handles the storyboard instantiation, and returns a strongly typed instance:

func instantiateViewController<T: UIViewController>(
    ofType type: T.Type,
    withIdentifier identifier: String,
    fromStoryboard storyboardName: String = "Main"
) -> T {
    let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
    // Use guard to safely cast and provide a meaningful error if something goes wrong
    guard let viewController = storyboard.instantiateViewController(withIdentifier: identifier) as? T else {
        fatalError("Failed to instantiate view controller with identifier \(identifier) as \(type) from \(storyboardName) storyboard")
    }
    return viewController
}

How to Use It #

Call the function by passing your view controller's type (using .self) and its storyboard identifier:

let mapVC = instantiateViewController(ofType: MapViewController.self, withIdentifier: "MapViewController")

Simplified Version (If Identifier Matches Class Name) #

If your storyboard identifiers match your view controller class names (a common best practice), you can eliminate the identifier parameter entirely by deriving it from the type:

func instantiateViewController<T: UIViewController>(
    ofType type: T.Type,
    fromStoryboard storyboardName: String = "Main"
) -> T {
    let identifier = String(describing: type)
    let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
    
    guard let viewController = storyboard.instantiateViewController(withIdentifier: identifier) as? T else {
        fatalError("Failed to instantiate \(identifier) from \(storyboardName) storyboard")
    }
    return viewController
}

Even Cleaner Call #

Now you don't need to repeat the identifier:

let mapVC = instantiateViewController(ofType: MapViewController.self)

Swifty Extension Approach #

For even better integration with Swift's style, extend UIStoryboard to add this functionality directly:

extension UIStoryboard {
    static func instantiateViewController<T: UIViewController>(
        ofType type: T.Type,
        fromStoryboard storyboardName: String = "Main"
    ) -> T {
        let identifier = String(describing: type)
        let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
        
        guard let viewController = storyboard.instantiateViewController(withIdentifier: identifier) as? T else {
            fatalError("Failed to instantiate \(identifier) from \(storyboardName) storyboard")
        }
        return viewController
    }
}

Usage with Extension #

let mapVC = UIStoryboard.instantiateViewController(ofType: MapViewController.self)

Why This Works #

• Generics: The T: UIViewController constraint ensures we only accept valid view controller types, and T.Type lets us pass the class itself as a parameter.
• Type Safety: Unlike using Any, this approach gives you compile-time checks and avoids unsafe as! casts without meaningful error handling.
• Reusability: You can now use this function/extension for any view controller in your app, eliminating repetitive storyboard instantiation code.

内容的提问来源于stack exchange，提问作者Scott Robinson