title: Prevent Xcode from preparing Apple Watch when running for iOS
body: <p>I have an iOS app that I want to test on my personal iOS device. I also wear an Apple Watch that's connected normally to this iPhone.</p>
  <p>I have not knowingly created any kind of Apple Watch target or code in my Xcode project. I do not want to run or test the app on my Apple Watch. I have not set up the Apple Watch as some kind of development target in Xcode... it's just connected to my iPhone.</p>
  <p>Despite this, Xcode continually tries to ready my Apple Watch for development alongside the iPhone. It delays running the iOS app on the iPhone so that it can &quot;prepare&quot; and &quot;fetch&quot; various things from the Watch. These spinners never end and as a result I am unable to deploy the iOS app to the iPhone.</p>
  <p><a href="https://i.stack.imgur.com/pr3R8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pr3R8.png" alt="iPhone is busy: Preparing Apple Watch for development via iPhone" /></a></p>
  <blockquote>
  <p>iPhone is busy: Preparing Apple Watch for development via iPhone<br />
  Xcode will continue when iPhone is finished.</p>
  </blockquote>
  <blockquote>
  <p>iPhone is busy: Fetching debug symbols for Apple Watch paired to iPhone</p>
  </blockquote>
  <blockquote>
  <p>Fetching debug symbols for the watch<br />
  Installing Symbols for watchOS 7.4</p>
  </blockquote>
  <blockquote>
  <p>Apple Watch via iPhone: Failed to prepare device for development.<br />
  This operation can fail if the version of the OS on the device is incompatible with the installed version of Xcode. You may also need to restart your mac and device in order to correctly detect compatibility.</p>
  </blockquote>
  <p>How can I fully hide this Apple Watch from being seen by Xcode? Disable any kind of development on the Watch? Deploy the iOS app to the iPhone without involving the connected Apple Watch at all?</p>
  <p>I'm not interested in &quot;fixing&quot; the errors and warnings... I want the Apple Watch to not be used <em>at all</em> by Xcode.</p>