Swift: When you have an Optional String you need to remove

Swift: When you have an Optional String you need to remove

Screen Shot 2016-07-21 at 10.09.57 AM

Recently I have been developing an iOS app that comes with a companion Apple Watch extension. I wanted to play with the communication between iOS and WatchOS using

via

It was working well and I thought I had it nailed down. Nope.

At one point my watch extension makes a request to the iOS app for a preset index. That’s so I can display the preset number in an alert on the watch at another point. I was setting the title of the alert to display something like “Preset 2”. Instead I was getting “Preset Optional(2)”. Not good.

I thought the problem was in unwrapping an Optional on the watch extension side. I tried all sorts of things that did nothing. How can you unwrap a String since they are not Optional?

Finally I printed the full watch extension message and found the Optional was already in the String when it arrives. So I had to look at the iOS ViewController that created the message to send to the watch extension.

Here is a breakdown (simplified) of what it looks like for the iOS application.

I think the braces balance (I typed this by hand). They should anyway. Do you see anything obvious in there? I didn’t for a while until I posted to StackOverflow and the comments came pouring in. There was an optional being injected into the constructed string prior to being sent (as I discovered later).

The solution was an explicit unwrapping of something I didn’t think was an Optional to begin with.

Look at that. index!

This solved the problem on the other end where I was splitting the index off the String (into an array and used the number for the alert title). I was going mad, but the answer should have been more obvious.

Thanks to those on StackOverflow that helped me go through it. I don’t have code reviews for prototypes I usually build at work and that one sailed right by me for a while.

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