Swift 3 better way to unwrap an optional

Question

Right now I have some code that looks like the following:

let msg: String? = myStr.removingPercentEncoding ?? nil

print("msg \(msg!)")

I really don't like the use of ! in msg! as that can possibly throw an exception, I believe.

What's the best way to call myStr.removingPercentEncoding if myStr is not nil and then unwrap msg without throwing an exception?

By default, `removingPercentEncoding` returns an optional String. It should be: `let msg = myStr.stringByRemovingPercentEncoding` — Ahmad F, Jan 04 '17 at 17:04
@AhmadF `stringByRemovingPercentEncoding` is the old Swift syntax — shim, Jan 04 '17 at 17:06
@AhmadF yes, well read the question. It's written in Swift 3. — shim, Jan 04 '17 at 17:09
@shim Oh, I see... thanks for the note, but it's still the same :) it returns `String?`. You can check it from [the method documentation](https://developer.apple.com/reference/swift/string/1642934-removingpercentencoding). — Ahmad F, Jan 04 '17 at 17:13
Yes but it is confusing to comment with syntax that will not compile. — shim, Jan 04 '17 at 17:14

rmaddy · Accepted Answer · 2017-01-04T17:03:17.587

The proper way would be:

if let msg = myStr.removingPercentEncoding {
    print("msg \(msg)")
}

Here, msg is only valid inside the if statement and only if myStr.removingPercentEncoding isn't nil.

If you wish to print something if myStr.removingPercentEncoding is nil, then you could and an else:

if let msg = myStr.removingPercentEncoding {
    print("msg \(msg)")
} else {
    print("msg has no value")
}

score 2 · Answer 2 · answered Jan 04 '17 at 17:01

2

This line is completely useless, when myStr.removingPercentEncoding is nil, then nil (right hand side of ??) is assigned to msg:

let msg: String? = myStr.removingPercentEncoding ?? nil

Why don't you make it as:

let msg: String = myStr.removingPercentEncoding ?? ""

print("msg \(msg)")

answered Jan 04 '17 at 17:01

OOPer

Depending on the use case, it may be more desirable to do `let msg = myStr.removingPercentEncoding ?? myStr`, as seeing the full invalid percent-encoded string may be more useful than seeing nothing. – Hamish Jan 04 '17 at 17:09
@Hamish, a possible option, but depends on use case. In some cases, the desired output may be something like `msg --invalid input--`. I think most readers can choose their own alternative value for `""`. – OOPer Jan 04 '17 at 17:16

2 Answers2