CODEWITHSUNDEEP
iosswiftclosuresswift3
Artem Stepanenko
Artem Stepanenko

Reputation: 3501

"Segmentation fault: 11" in Xcode 8.1

Is it really forbidden to use if...let inside closures in Swift 3? Xcode constantly gives me Swift Compiler Error / Command failed due to signal: Segmentation fault: 11 when I put it in a closure.

import Foundation

class ItemsManager: NSObject {

    fileprivate let networking: Networking

    init(networking: Networking) {
        self.networking = networking
    }

    func fetchItems() {
        networking.fetchItemsSuccess({ [weak self] (responseObject: Any) in

            // the line below causes "Segmentation fault: 11"
            if let itemDictionaries = responseObject as [Dictionary] {
                // ...
            }
            // ...                    
        }, failure: { [weak self] in
            // ...
        })
    }
}

UPDATE 1

While the current class is in swift, Networking is written in Objective-C, where responseObject is declared as id. Swift treats it as Any.

UPDATE 2

I checked a type of the responseObject in runtime, it appears to be __NSArrayI. Is it normal?

Upvotes: 1

Views: 1201

Answers (1)

Sethmr
Sethmr

Reputation: 3084

import Foundation

class ItemsManager: NSObject {

    fileprivate let networking: Networking

    init(networking: Networking) {
        self.networking = networking
    }

    func fetchItems() {
        networking.fetchItemsSuccess({ [weak self] responseObject in

            if let itemDictionaries = responseObject as [[String: AnyObject]] {
                // ...
            }

        }) { [weak self] in
            // ...
        }
    }
}

This is how I would write this function. (I like using trailing closures like what I did.)

The type of what is returned is inferred from the function definition so placing anything there might just confuse the compiler. Removing extraneous details became a big thing in Swift 3, so that could be your error.

The next thing is that Dictionary is a [String: AnyObject] by nature. There are ways of type defining a Dictionary using the word Dictionary, but why bother when we can define it directly.

Upvotes: 1

Related Questions