quitting xcode cocoa swift app

Question

I have written my first swift OS/X application in XCode 6. It all works except I cannot figure out how to exit the app. I have a button to exit and the ExitNow function defined as follows:

@IBAction func ExitNow(sender: AnyObject) {
    // ???
}

I cannot figure out what the code would be. By searching online I've found various options, but they were either in Objective C or too general for me to comprehend. I would appreciate an example which would behave the same way as cmd-Q.

Solution 1

You should be able to just call terminate on the global NSApp object.

---

Swift 4 & 5:

@IBAction func ExitNow(sender: AnyObject) {
    NSApplication.shared.terminate(self)
}

---

Swift 3:

@IBAction func ExitNow(sender: AnyObject) {
    NSApplication.shared().terminate(self)
}

---

Swift 2:

@IBAction func ExitNow(sender: AnyObject) {
    NSApplication.sharedApplication().terminate(self)
}

Solution 2

Or we could simply exit from the app like this:

@IBAction func ExitNow(sender: AnyObject) {    
    exit(0)
}

As a side note you can exit because of an error:

fatalError("reason for exiting")

Unconditionally prints a message and stops execution. iOS 8.1 and later.

Solution 3

In Xcode 9.0 you might use NSApp.terminate(nil).

Comments

[Lanny Rosicky]

This I tried and got the following error NSApp.terminate(self)'terminate' is unavailale. APIs deprecated as of OS X 10.9 and earlier are unavailable in Swift.

[Mike S]

Oops, apparently you can't use that with NSApp, sorry. I edited my answer so that it uses NSApplication.sharedApplication() instead. I also tested it real quick in a brand new OSX Swift project targeting 10.10 and it worked fine.

[Amr Angry]

error in syntax --> for swift 3 solution ---> use if unravel identifier

[SaganRitual]

As of Swift 4, it's NSApplication.shared.terminate(self)

[Zeeshan Suleman]

If you terminate using this, then the app would not open again if you want to open it programmatically