问题
I\'m trying to declare an argument in Swift that takes an optional closure. The function I have declared looks like this:
class Promise {
func then(onFulfilled: ()->(), onReject: ()->()?){
if let callableRjector = onReject {
// do stuff!
}
}
}
But Swift complains that \"Bound value in a conditional must be an Optional type\" where the \"if let\" is declared.
回答1:
You should enclose the optional closure in parentheses. This will properly scope the ? operator.
func then(onFulfilled: ()->(), onReject: (()->())?){
if let callableRjector = onReject {
// do stuff!
}
}
回答2:
To make the code even shorter we can use nil as default value for onReject parameter and optional chaining ?() when calling it:
func then(onFulfilled: ()->(), onReject: (()->())? = nil) {
onReject?()
}
This way we can omit onReject parameter when we call then function.
then({ /* on fulfilled */ })
We can also use trailing closure syntax to pass onReject parameter into then function:
then({ /* on fulfilled */ }) {
// ... on reject
}
Here is a blog post about it.
回答3:
Since I assume, that this "optional" closure should simply do nothing, you could use a parameter with an empty closure as default value:
func then(onFulfilled: ()->(), onReject: ()->() = {}){
// now you can call your closures
onFulfilled()
onReject()
}
this function can now be called with or without the onReject callback
then({ ... })
then({ ... }, onReject: { ... })
No need for Swift's awesome Optionals? here!
回答4:
Maybe it's a cleaner way. Specially when the closure has complicated parameters.
typealias SimpleCallBack = () -> ()
class Promise {
func then(onFulfilled: SimpleCallBack, onReject: SimpleCallBack?){
if let callableRjector = onReject {
// do stuff!
}
}
}
来源：https://stackoverflow.com/questions/24395853/how-does-one-make-an-optional-closure-in-swift