UIApplication.registerForRemoteNotifications() must be called from main thread only

Question

Xcode 9 (iOS 11) showing me an error/warning while registering for Push (remote) notification.

Here is error message

And here is code, I've tried:

let center  = UNUserNotificationCenter.current()
center.delegate = self
center.requestAuthorization(options: [.sound, .alert, .badge]) { (granted, error) in
        if error == nil{
              UIApplication.shared.registerForRemoteNotifications()
        }
 }

Error/Warning Line:

UIApplication.shared.registerForRemoteNotifications()

How to resolve this?

Answer 1

In swift4

You can solve this issue with

DispatchQueue.main.async {
  UIApplication.shared.registerForRemoteNotifications()
}

Hope this will help...

Answer 2

For Objective C, the below code works

    dispatch_async(dispatch_get_main_queue(), ^{
        [[UIApplication sharedApplication] registerForRemoteNotifications];
    });

Answer 3

TL;DR:
All UI manipulations should be done in the Main Thread to avoid problems. If failed to do so, Main Thread Checker (Newly introduced debugging feature in Xcode 9) will produce issues at Runtime. So wrap your code in Main Thread block like below to avoid glitches and runtime warnings.

DispatchQueue.main.async {
    UIApplication.shared.registerForRemoteNotifications()
}

In Xcode releases before ver. 9, the warnings related to main thread would get printed in the console area textually. Anyway, you can optionally disable (not a recommended approach) the Main Thread Checker in the Diagnostic settings in Edit Scheme.

Explanation:

Apple introduced a new debugging option in Xcode 9 for checking issues at Runtime for UIKit and other API's that manipulate UI elements. If there's any change to the UI elements from UIKit API at Runtime, without a Main thread block, it is highly likely to cause UI glitches and crashes. The Main Thread Checker is enabled by default to catch those issues at runtime. You can disable Main Thread Checker in the Edit Scheme window just like below, although it is not really recommended to do so:

If you have any older SDK's or Frameworks, when updating to Xcode 9, you may face this warning since some of the UIKit method calls wouldn't have been wrapped in Main Thread. Updating them to latest version would fix the issue (if the developer is aware of it and fixed it).

Quote from Xcode 9 beta release notes:

• New in Xcode 9 – Main Thread Checker.

• **Enable detection of UI API

misuse from background thread**

• Detects AppKit, UIKit, and WebKit method calls that are not made on the main thread.
• Automatically enabled during debugging, and can be disabled in the Diagnostic tab of the scheme editor.
• Main Thread Checker works with Swift and C languages.

Answer 4

The error message is pretty clear: dispatch registerForRemoteNotifications to the main thread.

I would use the granted parameter and handle the error accordingly

center.requestAuthorization(options: [.sound, .alert, .badge]) { (granted, error) in
        if granted {
              DispatchQueue.main.async {
                  UIApplication.shared.registerForRemoteNotifications()
              }
        } else {
           print(error!)
           // handle the error
        }
}

Answer 5

This is also correct way to do in Swift 4.0

UNUserNotificationCenter.current().delegate = self
        UNUserNotificationCenter.current().requestAuthorization(options: [.alert,.sound,.badge], completionHandler: {(granted,error) in
            if granted{
                DispatchQueue.main.async {
                    application.registerForRemoteNotifications()
                }
            }
        })

Answer 6

In swift 4 or 5

DispatchQueue.main.async {
  UIApplication.shared.registerForRemoteNotifications()
}

on Objectiv-C

dispatch_async(dispatch_get_main_queue(), ^{
        [[UIApplication sharedApplication] registerForRemoteNotifications];
});

Answer 7

Hope this will help

DispatchQueue.main.async(execute: {
  UIApplication.shared.registerForRemoteNotifications()
})

Answer 8

This is what worked for me. Courtesy of @Mason11987 in the accepted comment above.

DispatchQueue.main.async() { code }