How to have an optional trailing closure?

Question

I'm trying to write a function that I can call with trailing closure syntax like so:

func hello( message: String, closure: (( msg: String ) -> Void)?) {

   println( "called hello with: \(message)" );

   closure?( msg: message );

}

I would expect to be able to call this function with a closure:

hello( "abc" ) { msg in

    let a = "def";

    println("Called callback with: \(msg) and I've got: \(a)");

};

And also without the closure, since it's optional:

hello( "abc" )

The latter doesn't work. It says I can't call hello with an argument list of (String).

I'm using XCode 6.3.2 and I tested this code within a playground.

ABakerSmith · Accepted Answer · 2015-06-19 00:32:07Z

4

I'm not sure you've got your definition of optional entirely correct. It doesn't mean you don't need to supply a value for the closure argument; it means closure can either have a value (of a closure in your case) or be nil. Therefore, if you wanted to call hello without providing a closure you would write:

hello("abc", nil)

However, you can achieve what you're after using default parameter values (I'd recommend you have a look at The Swift Programming Guide: Functions). Your function would be:

// Note the `= nil`:
func hello(message: String, closure: ((msg: String ) -> Void)? = nil) {
    println("called hello with: \(message)")

    closure?(msg: message)
}


// Example usage:
hello("abc")
hello("abc") { println($0) }

edited Jun 19, 2015 at 0:32

answered Jun 19, 2015 at 0:25

ABakerSmith

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1 Comment

0x6A75616E Over a year ago

That makes perfect sense, thank you. I was wrongly assuming the trailing closure would automatically be assumed to be nil if not provided.

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